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Gauss's lemma (polynomials)

In algebra, Gauss's lemma, named after Carl Friedrich Gauss, is a theorem about polynomials over the integers, or, more generally, over a unique factorization domain (that is, a ring that has a unique factorization property similar to the fundamental theorem of arithmetic). Gauss's lemma underlies all the theory of factorization and greatest common divisors of such polynomials.

Gauss's lemma asserts that the product of two primitive polynomials is primitive. (A polynomial with integer coefficients is primitive if it has 1 as a greatest common divisor of its coefficients. )

A corollary of Gauss's lemma, sometimes also called Gauss's lemma, is that a primitive polynomial is irreducible over the integers if and only if it is irreducible over the rational numbers. More generally, a primitive polynomial has the same complete factorization over the integers and over the rational numbers. In the case of coefficients in a unique factorization domain $R$, "rational numbers" must be replaced by "field of fractions of $R$". This implies that, if $R$ is either a field, the ring of integers, or a unique factorization domain, then every polynomial ring (in one or several indeterminates) over $R$ is a unique factorization domain. Another consequence is that factorization and greatest common divisor computation of polynomials with integers or rational coefficients may be reduced to similar computations on integers and primitive polynomials. This is systematically used (explicitly or implicitly) in all implemented algorithms (see Polynomial greatest common divisor and Factorization of polynomials).

Gauss's lemma, and all its consequences that do not involve the existence of a complete factorization remain true over any GCD domain (an integral domain over which greatest common divisors exist). In particular, a polynomial ring over a GCD domain is also a GCD domain. If one calls primitive a polynomial such that the coefficients generate the unit ideal, Gauss's lemma is true over every commutative ring. However, some care must be taken when using this definition of primitive, as, over a unique factorization domain that is not a principal ideal domain, there are polynomials that are primitive in the above sense and not primitive in this new sense.

The lemma over the integers
If $$F(X) = a_0 + a_1 X + \dots + a_n X^n$$ is a polynomial with integer coefficients, then $$F$$ is called primitive if the greatest common divisor of all the coefficients $$a_0, a_1, \dots, a_n$$ is 1; in other words, no prime number divides all the coefficients.

Proof: Clearly the product f(x)g(x) of two primitive polynomials has integer coefficients. Therefore, if it is not primitive, there must be a prime p which is a common divisor of all its coefficients. But p cannot divide all the coefficients of either f(x) or g(x) (otherwise they would not be primitive). Let arxr be the first term of f(x) not divisible by p and let bsxs be the first term of g(x) not divisible by p. Now consider the term xr+s in the product, whose coefficient is

 * $$\cdots + a_{r+2} b_{s-2} + a_{r+1} b_{s-1} + a_r b_s + a_{r-1} b_{s+1} + a_{r-2} b_{s+2} + \cdots .$$

The term arbs is not divisible by p (because p is prime), yet all the remaining ones are, so the entire sum cannot be divisible by p. By assumption all coefficients in the product are divisible by p, leading to a contradiction. Therefore, the coefficients of the product can have no common divisor and are thus primitive. $$\square$$

The proof is given below for the more general case. Note that an irreducible element of Z (a prime number) is still irreducible when viewed as constant polynomial in Z[X]; this explains the need for "non-constant" in the statement.

Statements for unique factorization domains
Gauss's lemma holds more generally over arbitrary unique factorization domains. There the content $c(P)$ of a polynomial $P$ can be defined as the greatest common divisor of the coefficients of $P$ (like the gcd, the content is actually a set of associate elements). A polynomial $P$ with coefficients in a UFD $R$ is then said to be primitive if the only elements of $R$ that divide all coefficients of $P$ at once are the invertible elements of $R$; i.e., the gcd of the coefficients is one.

Primitivity statement: If $R$ is a UFD, then the set of primitive polynomials in $R[X]$ is closed under multiplication. More generally, the content of a product $$fg$$ of polynomials is the product $$c(f)c(g)$$ of their individual contents.

Irreducibility statement: Let $R$ be a unique factorization domain and $F$ its field of fractions. A non-constant polynomial $$f$$ in $$R[x]$$ is irreducible in $$R[x]$$ if and only if it is both irreducible in $$F[x]$$ and primitive in $$R[x]$$.

(For the proofs, see below.)

Let $$R$$ be a unique factorization domain with field of fractions $$F$$. If $$f \in F[x]$$ is a polynomial over $$F$$ then for some $$d$$ in $$R$$, $$df$$ has coefficients in $$R$$, and so – factoring out the gcd $$q$$ of the coefficients – we can write $$df = qf'$$ for some primitive polynomial $$f' \in R[x]$$. As one can check, this polynomial $$f'$$ is unique up to the multiplication by a unit and is called the primitive part (or primitive representative) of $$f$$ and is denoted by $$\operatorname{pp}(f)$$. The procedure is compatible with product: $$\operatorname{pp}(fg) = \operatorname{pp}(f)\operatorname{pp}(g)$$.

The construct can be used to show the statement: Indeed, by induction, it is enough to show $$R[x]$$ is a UFD when $$R$$ is a UFD. Let $$f \in R[x]$$ be a non-zero polynomial. Now, $$F[x]$$ is a unique factorization domain (since it is a principal ideal domain) and so, as a polynomial in $$F[x]$$, $$f$$ can be factorized as:
 * A polynomial ring over a UFD is a UFD.
 * $$f = g_1 g_2 \dots g_r$$

where $$g_i$$ are irreducible polynomials of $$F[x]$$. Now, we write $$f = c f'$$ for the gcd $$c$$ of the coefficients of $$f$$ (and $$f'$$ is the primitive part) and then:
 * $$f = c f' = c \operatorname{pp}(g_1) \operatorname{pp}(g_2) \cdots \operatorname{pp}(g_r).$$

Now, $$c$$ is a product of prime elements of $$R$$ (since $$R$$ is a UFD) and a prime element of $$R$$ is a prime element of $$R[x]$$, as $$R[x]/(p) \cong R/(p)[x]$$ is an integral domain. Hence, $$c$$ admits a prime factorization (or a unique factorization into irreducibles). Next, observe that $$f' = \operatorname{pp}(g_1) \cdots \operatorname{pp}(g_r)$$ is a unique factorization into irreducible elements of $$R[x]$$, as (1) each $$\operatorname{pp}(g_i)$$ is irreducible by the irreducibility statement and (2) it is unique since the factorization of $$f'$$ can also be viewed as a factorization in $$F[x]$$ and factorization there is unique. Since $$c$$ and $$f'$$ are uniquely determined by $$f$$ up to unit elements, the above factorization of $$f$$ is a unique factorization into irreducible elements. $$\square$$

The condition that "R is a unique factorization domain" is not superfluous because it implies that every irreducible element of this ring is also a prime element, which in turn implies that every non-zero element of R has at most one factorization into a product of irreducible elements and a unit up to order and associate relationship. In a ring where factorization is not unique, say $pa = qb$ with p and q irreducible elements that do not divide any of the factors on the other side, the product $(p + qX)(a + qX) = pa + (p+a)qX + q^{2}X^{2} = q(b + (p+a)X + qX^{2})$ shows the failure of the primitivity statement. For a concrete example one can take $R = Z[i√5]$, $p = 1 + i√5$, $a = 1 − i√5$, $q = 2$, $b = 3$. In this example the polynomial $3 + 2X + 2X^{2}$ (obtained by dividing the right hand side by $q = 2$) provides an example of the failure of the irreducibility statement (it is irreducible over R, but reducible over its field of fractions $Q[i√5]$). Another well-known example is the polynomial $X^{2} − X − 1$, whose roots are the golden ratio $φ = (1 + √5)/2$ and its conjugate $(1 − √5)/2$ showing that it is reducible over the field $Q[√5]$, although it is irreducible over the non-UFD $Z[√5]$ which has $Q[√5]$ as field of fractions. In the latter example the ring can be made into an UFD by taking its integral closure $Z[φ]$ in $Q[√5]$ (the ring of Dirichlet integers), over which $X^{2} − X − 1$ becomes reducible, but in the former example R is already integrally closed.

General version
Let $$R$$ be a commutative ring. If $$f$$ is a polynomial in $$R[x_1, \dots, x_n]$$, then we write $$\operatorname{cont}(f)$$ for the ideal of $$R$$ generated by all the coefficients of $$f$$; it is called the content of $$f$$. Note that $$\operatorname{cont}(af) = a\operatorname{cont}(f)$$ for each $$a$$ in $$R$$. The next proposition states a more substantial property.

A polynomial $$f$$ is said to be primitive if $$\operatorname{cont}(f)$$ is the unit ideal $$(1)$$. When $$R = \mathbb{Z}$$ (or more generally when $$R$$ is a Bézout domain), this agrees with the usual definition of a primitive polynomial. (But if $$R$$ is only a UFD, this definition is inconsistent with the definition of primitivity in .)

Proof: This is easy using the fact that $$\sqrt{I} = (1)$$ implies $$I = (1).$$ $$\square$$

Proof: ($$\Rightarrow$$) First note that the gcd of the coefficients of $$f$$ is 1 since, otherwise, we can factor out some element $$c \in R$$ from the coefficients of $$f$$ to write $$f = cf'$$, contradicting the irreducibility of $$f$$. Next, suppose $$f = gh$$ for some non-constant polynomials $$g, h$$ in $$F[x]$$. Then, for some $$d \in R$$, the polynomial $$dg$$ has coefficients in $$R$$ and so, by factoring out the gcd $$q$$ of the coefficients, we write $$dg = qg'$$. Do the same for $$h$$ and we can write $$f = c g'h'$$ for some $$c \in F$$. Now, let $$c = a/b$$ for some $$a, b \in R$$. Then $$bf = a g'h'$$. From this, using the proposition, we get:
 * $$(b) \supset \operatorname{gcd}(\operatorname{cont}(bf)) = (a)$$.

That is, $$b$$ divides $$a$$. Thus, $$c \in R$$ and then the factorization $$f = c g'h'$$ constitutes a contradiction to the irreducibility of $$f$$.

($$\Leftarrow$$) If $$f$$ is irreducible over $$F$$, then either it is irreducible over $$R$$ or it contains a constant polynomial as a factor, the second possibility is ruled out by the assumption. $$\square$$

Proof of the proposition: Clearly, $$\operatorname{cont}(fg) \subset \operatorname{cont}(f) \operatorname{cont}(g)$$. If $$\mathfrak{p}$$ is a prime ideal containing $$\operatorname{cont}(fg)$$, then $$fg \equiv 0$$ modulo $$\mathfrak{p}$$. Since $$R/\mathfrak{p}[x_1, \dots, x_n]$$ is a polynomial ring over an integral domain and thus is an integral domain, this implies either $$f \equiv 0$$ or $$g \equiv 0$$ modulo $$\mathfrak{p}$$. Hence, either $$\operatorname{cont}(f)$$ or $$\operatorname{cont}(g)$$ is contained in $$\mathfrak{p}$$. Since $$\sqrt{\operatorname{cont}(fg)}$$ is the intersection of all prime ideals that contain $$\operatorname{cont}(fg)$$ and the choice of $$\mathfrak p$$ was arbitrary, $$\operatorname{cont}(f) \operatorname{cont}(g) \subset \sqrt{\operatorname{cont}(fg)}$$.

We now prove the "moreover" part. Factoring out the gcd's from the coefficients, we can write $$f = a f'$$ and $$g = b g'$$ where the gcds of the coefficients of $$f', g'$$ are both 1. Clearly, it is enough to prove the assertion when $$f, g$$ are replaced by $$f', g'$$; thus, we assume the gcd's of the coefficients of $$f, g$$ are both 1. The rest of the proof is easy and transparent if $$R$$ is a unique factorization domain; thus we give the proof in that case here (and see Proof for the GCD case: The proof here is adopted from We need the following simple lemma about gcd: (The proof of the lemma is not trivial but is by elementary algebra.)
 * If $$\gcd(a, b) = \gcd(a, c) = 1$$, then $$\gcd(a, bc) = 1$$.

We argue by induction on the sum of the numbers of the terms in $$f, g$$; that is, we assume the proposition has been established for any pair of polynomials with one less total number of the terms. Let $$(c) = \gcd(\operatorname{cont}(fg))$$; i.e., $$c$$ is the gcd of the coefficients of $$fg$$. Assume $$(c) \ne (1)$$; otherwise, we are done. Let $$f_0, g_0$$ denote the highest-degree terms of $$f, g$$ in terms of lexicographical monomial ordering. Then $$f_0g_0$$ is precisely the leading term of $$fg$$ and so $$c$$ divides the (unique) coefficient of $$f_0g_0$$ (since it divides all the coefficients of $$fg$$). Now, if $$c$$ does not have a common factor with the (unique) coefficient of $$f_0$$ and does not have a common factor with that of $$g_0$$, then, by the above lemma, $$\gcd(c, \operatorname{cont}(f_0g_0)) = (1)$$. But $$c$$ divides the coefficient of $$f_0g_0$$; so this is a contradiction. Thus, either $$c$$ has a common factor with the coefficient of $$f_0$$ or does with that of $$g_0$$; say, the former is the case. Let $$(d) = \operatorname{gcd}(c, \operatorname{cont}(f_0))$$. Since $$d$$ divides the coefficients of $$fg - f_0g = (f - f_0)g$$, by inductive hypothesis,
 * $$(d) \supset \operatorname{gcd}(\operatorname{cont}((f - f_0)g)) = \operatorname{gcd}(\operatorname{cont}(f - f_0)) \operatorname{gcd}(\operatorname{cont}(g)) = \operatorname{gcd}(\operatorname{cont}(f - f_0))$$.

Since $$(d)$$ contains $$\operatorname{cont}(f_0)$$, it contains $$\operatorname{cont}(f)$$; i.e., $$(d) = (1)$$, a contradiction. $$\square$$ for the proof for the GCD case). If $$\gcd(\operatorname{cont}(fg)) = (1)$$, then there is nothing to prove. So, assume otherwise; then there is a non-unit element dividing the coefficients of $$fg$$. Factorizing that element into a product of prime elements, we can take that element to be a prime element $$\pi$$. Now, we have:
 * $$(\pi) = \sqrt{(\pi)} \supset \sqrt{\operatorname{cont}(fg)} \supset \operatorname{cont}(f) \operatorname{cont}(g)$$.

Thus, either $$(\pi)$$ contains $$\operatorname{cont}(f)$$ or $$\operatorname{cont}(g)$$; contradicting the gcd's of the coefficients of $$f, g$$ are both 1. $$\square$$

 * Remark: Over a GCD domain (e.g., a unique factorization domain), the gcd of all the coefficients of a polynomial $$f$$, unique up to unit elements, is also called the content of $$f$$.

Applications
It follows from Gauss's lemma that for each unique factorization domain $$R$$, the polynomial ring $$R[X_1, X_2,. .., X_n]$$ is also a unique factorization domain (see ). Gauss's lemma can also be used to show Eisenstein's irreducibility criterion. Finally, it can be used to show that cyclotomic polynomials (unitary units with integer coefficients) are irreducible.

Gauss's lemma implies the following statement: If $$R = \mathbb{Z}$$, then it says a rational root of a monic polynomial over integers is an integer (cf. the rational root theorem). To see the statement, let $$a/b$$ be a root of $$f$$ in $$F$$ and assume $$a, b$$ are relatively prime. In $$F[x]$$ we can write $$f = (x - a/b)g$$ with $$cg \in R[x]$$ for some $$c \in R$$. Then
 * If $$f(x)$$ is a monic polynomial in one variable with coefficients in a unique factorization domain $$R$$ (or more generally a GCD domain), then a root of $$f$$ that is in the field of fractions $$F$$ of $$R$$ is in $$R$$.
 * $$cb f = (bx - a)cg$$

is a factorization in $$R[x]$$. But $$bx - a$$ is primitive (in the UFD sense) and thus $$cb$$ divides the coefficients of $$cg$$ by Gauss's lemma, and so
 * $$f = (bx - a)h$$

with $$h$$ in $$R[x]$$. Since $$f$$ is monic, this is possible only when $$b$$ is a unit.

A similar argument shows:
 * Let $$R$$ be a GCD domain with the field of fractions $$F$$ and $$f \in R[x]$$. If $$f = gh$$ for some polynomial $$g \in R[x]$$ that is primitive in the UFD sense and $$h \in F[x]$$, then $$h \in R[x]$$.

The irreducibility statement also implies that the minimal polynomial over the rational numbers of an algebraic integer has integer coefficients.

title: Gauss's lemma (polynomials)


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Gauss's lemma (number theory)

Gauss's lemma in number theory gives a condition for an integer to be a quadratic residue. Although it is not useful computationally, it has theoretical significance, being involved in some proofs of quadratic reciprocity.

It made its first appearance in Carl Friedrich Gauss's third proof (1808) of quadratic reciprocity and he proved it again in his fifth proof (1818).

Statement of the lemma
For any odd prime $p$ let $a$ be an integer that is coprime to $p$.

Consider the integers
 * $$a, 2a, 3a, \dots, \frac{p-1}{2}a$$

and their least positive residues modulo $p$. These residues are all distinct, so there are ($p − 1)/2$ of them.

Let $n$ be the number of these residues that are greater than $p/2$. Then
 * $$\left(\frac{a}{p}\right) = (-1)^n,$$

where $$\left(\frac{a}{p}\right)$$ is the Legendre symbol.

Example
Taking $p$ = 11 and $a$ = 7, the relevant sequence of integers is
 * 7, 14, 21, 28, 35.

After reduction modulo 11, this sequence becomes
 * 7, 3, 10, 6, 2.

Three of these integers are larger than 11/2 (namely 6, 7 and 10), so $n$ = 3. Correspondingly Gauss's lemma predicts that
 * $$\left(\frac

{7}{11}\right) = (-1)^3 = -1.$$ This is indeed correct, because 7 is not a quadratic residue modulo 11.

The above sequence of residues
 * 7, 3, 10, 6, 2

may also be written
 * −4, 3, −1, −5, 2.

In this form, the integers larger than 11/2 appear as negative numbers. It is also apparent that the absolute values of the residues are a permutation of the residues
 * 1, 2, 3, 4, 5.

Proof
A fairly simple proof, reminiscent of one of the simplest proofs of Fermat's little theorem, can be obtained by evaluating the product
 * $$Z = a \cdot 2a \cdot 3a \cdot \cdots \cdot \frac{p-1}2 a$$

modulo p in two different ways. On one hand it is equal to
 * $$Z = a^{(p-1)/2} \left(1 \cdot 2 \cdot 3 \cdot \cdots \cdot \frac{p-1}2 \right)$$

The second evaluation takes more work. If $x$ is a nonzero residue modulo $p$, let us define the "absolute value" of $x$ to be
 * $$|x| = \begin{cases} x & \mbox{if } 1 \leq x \leq \frac{p-1}2, \\ p-x & \mbox{if } \frac{p+1}2 \leq x \leq p-1. \end{cases}$$

Since $n$ counts those multiples $ka$ which are in the latter range, and since for those multiples, $−ka$ is in the first range, we have
 * $$Z = (-1)^n \left(|a| \cdot |2a| \cdot |3a| \cdot \cdots \cdots \left|\frac{p-1}2 a\right|\right).$$

Now observe that the values $|ra|$ are distinct for $r = 1, 2, …, (p − 1)/2$. Indeed, we have

\begin{align} ra &\equiv \pm sa &\pmod{p} \\ r &\equiv \pm s &\pmod{p} \end{align} $$ because $a$ is coprime to $p$.
 * ra| &\equiv |sa| &\pmod{p} \\

This gives $r$ = $s$, since $r$ and $s$ are positive least residues. But there are exactly $(p − 1)/2$ of them, so their values are a rearrangement of the integers $1, 2, …, (p − 1)/2$. Therefore,
 * $$Z = (-1)^n \left(1 \cdot 2 \cdot 3 \cdot \cdots \cdot \frac{p-1}2\right).$$

Comparing with our first evaluation, we may cancel out the nonzero factor
 * $$1 \cdot 2 \cdot 3 \cdot \cdots \cdot \frac{p-1}2$$

and we are left with
 * $$a^{(p-1)/2} = (-1)^n.$$

This is the desired result, because by Euler's criterion the left hand side is just an alternative expression for the Legendre symbol $$\left (\frac{a}{p}\right )$$.

Generalization
For any odd prime $p$ let $a$ be an integer that is coprime to $p$.

Let $$I \subset(\mathbb{Z} / p \mathbb{Z})^{\times}$$ be a set such that $$(\mathbb{Z} / p \mathbb{Z})^{\times}$$ is the disjoint union of $$I$$ and $$-I=\{-i: i \in I\}$$.

Then $$\left(\frac{a}{p}\right)=(-1)^t$$, where $$t=\#\{j \in I: a j \in-I\}$$.

In the original statement, $$I=\{1,2,\dots,\frac{p-1}2\}$$.

The proof is almost the same.

Applications
Gauss's lemma is used in many,  but by no means all, of the known proofs of quadratic reciprocity.

For example, Gotthold Eisenstein used Gauss's lemma to prove that if $p$ is an odd prime then
 * $$\left(\frac{a}{p}\right)=\prod_{n=1}^{(p-1)/2}\frac{\sin{(2\pi an/p)}}{\sin{(2\pi n/p)}},$$

and used this formula to prove quadratic reciprocity. By using elliptic rather than circular functions, he proved the cubic and quartic reciprocity laws.

Leopold Kronecker used the lemma to show that
 * $$\left(\frac{p}{q}\right)=\sgn\prod_{i=1}^{\frac{q-1}{2}}\prod_{k=1}^{\frac{p-1}{2}}\left(\frac{k}{p}-\frac{i}{q}\right).$$

Switching $p$ and $q$ immediately gives quadratic reciprocity. It is also used in what are probably the simplest proofs of the "second supplementary law"
 * $$\left(\frac{2}{p}\right) = (-1)^{(p^2-1)/8} = \begin{cases} +1\text{ if }p\equiv \pm 1\pmod {8}\\-1\text{ if }p\equiv \pm 3\pmod {8}\end{cases}$$

Higher powers
Generalizations of Gauss's lemma can be used to compute higher power residue symbols. In his second monograph on biquadratic reciprocity, Gauss used a fourth-power lemma to derive the formula for the biquadratic character of $1 + i$ in $Z[i]$, the ring of Gaussian integers. Subsequently, Eisenstein used third- and fourth-power versions to prove cubic and quartic reciprocity.

nth power residue symbol
Let k be an algebraic number field with ring of integers $$\mathcal{O}_k,$$ and let $$\mathfrak{p} \subset \mathcal{O}_k $$ be a prime ideal. The ideal norm $$ \mathrm{N} \mathfrak{p} $$ of $$\mathfrak{p}$$ is defined as the cardinality of the residue class ring. Since $$\mathfrak{p} $$ is prime this is a finite field $$\mathcal{O}_k / \mathfrak{p}$$, so the ideal norm is $$ \mathrm{N} \mathfrak{p} = |\mathcal{O}_k / \mathfrak{p}|$$.

Assume that a primitive $n$th root of unity $$\zeta_n\in\mathcal{O}_k,$$ and that $n$ and $$\mathfrak{p} $$ are coprime (i.e. $$n\not\in \mathfrak{p}$$). Then no two distinct $n$th roots of unity can be congruent modulo $$\mathfrak{p}$$.

This can be proved by contradiction, beginning by assuming that $$\zeta_n^r\equiv\zeta_n^s$$ mod $$\mathfrak{p}$$, $0 < r < s ≤ n$. Let $t = s − r$ such that $$\zeta_n^t\equiv 1$$ mod $$\mathfrak{p}$$, and $0 < t < n$. From the definition of roots of unity,
 * $$x^n-1=(x-1)(x-\zeta_n)(x-\zeta_n^2)\dots(x-\zeta_n^{n-1}),$$

and dividing by $x − 1$ gives
 * $$x^{n-1}+x^{n-2}+\dots +x + 1 =(x-\zeta_n)(x-\zeta_n^2)\dots(x-\zeta_n^{n-1}).$$

Letting $x = 1$ and taking residues mod $$\mathfrak{p}$$,
 * $$n\equiv(1-\zeta_n)(1-\zeta_n^2)\dots(1-\zeta_n^{n-1})\pmod{\mathfrak{p}}.$$

Since $n$ and $$ \mathfrak{p}$$ are coprime, $$ n\not\equiv 0$$ mod $$\mathfrak{p},$$ but under the assumption, one of the factors on the right must be zero. Therefore, the assumption that two distinct roots are congruent is false.

Thus the residue classes of $$ \mathcal{O}_k / \mathfrak{p}$$ containing the powers of $ζ_{n}$ are a subgroup of order $n$ of its (multiplicative) group of units, $$(\mathcal{O}_k/\mathfrak{p}) ^\times = \mathcal{O}_k /\mathfrak{p}- \{0\}.$$ Therefore, the order of $$(\mathcal{O}_k/\mathfrak{p})^ \times$$ is a multiple of $n$, and

\begin{align} \mathrm{N} \mathfrak{p} &= |\mathcal{O}_k / \mathfrak{p}| \\ &= \left |(\mathcal{O}_k / \mathfrak{p} )^\times \right| + 1 \\ &\equiv 1 \pmod{n}. \end{align}$$

There is an analogue of Fermat's theorem in $$\mathcal{O}_k$$. If $$\alpha \in \mathcal{O}_k$$ for $$\alpha\not\in \mathfrak{p}$$, then
 * $$\alpha^{\mathrm{N} \mathfrak{p} -1}\equiv 1 \pmod{\mathfrak{p} },$$

and since $$\mathrm{N} \mathfrak{p} \equiv 1 $$ mod $n$,
 * $$\alpha^{\frac{\mathrm{N} \mathfrak{p} -1}{n}}\equiv \zeta_n^s\pmod{\mathfrak{p} }$$

is well-defined and congruent to a unique $n$th root of unity ζns.

This root of unity is called the $n$th-power residue symbol for $$\mathcal{O}_k,$$ and is denoted by

\begin{align} \left(\frac{\alpha}{\mathfrak{p} }\right)_n &= \zeta_n^s \\ &\equiv \alpha^{\frac{\mathrm{N} \mathfrak{p} -1}{n}}\pmod{\mathfrak{p}}. \end{align} $$

It can be proven that
 * $$\left(\frac{\alpha}{\mathfrak{p} }\right)_n= 1$$

if and only if there is an $$\eta \in\mathcal{O}_k$$ such that $α ≡ η^{n}$ mod $$\mathfrak{p}$$.

1/n systems
Let $$\mu_n = \{1,\zeta_n,\zeta_n^2,\dots,\zeta_n^{n-1}\} $$ be the multiplicative group of the $n$th roots of unity, and let $$A=\{a_1, a_2,\dots,a_m\}$$ be representatives of the cosets of $$(\mathcal{O}_k / \mathfrak{p})^\times/\mu_n.$$ Then $A$ is called a $1/n$ system mod $$\mathfrak{p}.$$

In other words, there are $$mn=\mathrm{N} \mathfrak{p} -1 $$ numbers in the set $$A\mu=\{ a_i \zeta_n^j\;:\; 1 \le i \le m, \;\;\;0 \le j \le n-1\},$$ and this set constitutes a representative set for $$(\mathcal{O}_k / \mathfrak{p})^\times.$$

The numbers $1, 2, … (p − 1)/2$, used in the original version of the lemma, are a 1/2 system (mod $p$).

Constructing a $1/n$ system is straightforward: let $M$ be a representative set for $$(\mathcal{O}_k / \mathfrak{p})^\times.$$ Pick any $$a_1\in M $$ and remove the numbers congruent to $$a_1, a_1\zeta_n, a_1\zeta_n^2, \dots, a_1\zeta_n^{n-1}$$ from $M$. Pick $a_{2}$ from $M$ and remove the numbers congruent to $$a_2, a_2\zeta_n, a_2\zeta_n^2, \dots, a_2\zeta_n^{n-1}$$ Repeat until $M$ is exhausted. Then $\{a_{1}, a_{2}, … a_{m}\}$ is a $1/n$ system mod $$\mathfrak{p}.$$

The lemma for nth powers
Gauss's lemma may be extended to the $n$th power residue symbol as follows. Let $$\zeta_n\in \mathcal{O}_k $$ be a primitive $n$th root of unity, $$\mathfrak{p} \subset \mathcal{O}_k $$ a prime ideal, $$\gamma \in \mathcal{O}_k, \;\;n\gamma\not\in\mathfrak{p},$$ (i.e. $$\mathfrak{p}$$ is coprime to both $γ$ and $n$) and let $A = \{a_{1}, a_{2}, …, a_{m}\}$ be a $1/n$ system mod $$\mathfrak{p}.$$

Then for each $i$, $1 ≤ i ≤ m$, there are integers $π(i)$, unique (mod $m$), and $b(i)$, unique (mod $n$), such that
 * $$\gamma a_i \equiv \zeta_n^{b(i)}a_{\pi(i)} \pmod{\mathfrak{p}},$$

and the $n$th-power residue symbol is given by the formula
 * $$\left(\frac{\gamma}{\mathfrak{p} }\right)_n = \zeta_n^{b(1)+b(2)+\dots+b(m)}.$$

The classical lemma for the quadratic Legendre symbol is the special case $n = 2$, $ζ_{2} = −1$, $A = \{1, 2, …, (p − 1)/2\}$, $b(k) = 1$ if $ak > p/2$, $b(k) = 0$ if $ak < p/2$.

Proof
The proof of the $n$th-power lemma uses the same ideas that were used in the proof of the quadratic lemma.

The existence of the integers $π(i)$ and $b(i)$, and their uniqueness (mod $m$) and (mod $n$), respectively, come from the fact that $Aμ$ is a representative set.

Assume that $π(i)$ = $π(j)$ = $p$, i.e.
 * $$\gamma a_i \equiv \zeta_n^r a_p \pmod{\mathfrak{p}}$$

and
 * $$\gamma a_j \equiv \zeta_n^s a_p \pmod{\mathfrak{p}}.$$

Then
 * $$\zeta_n^{s-r}\gamma a_i \equiv \zeta_n^s a_p \equiv \gamma a_j\pmod{\mathfrak{p}}$$

Because $γ$ and $$\mathfrak{p}$$ are coprime both sides can be divided by $γ$, giving
 * $$\zeta_n^{s-r} a_i \equiv a_j\pmod{\mathfrak{p}},$$

which, since $A$ is a $1/n$ system, implies $s = r$ and $i = j$, showing that $π$ is a permutation of the set $\{1, 2, …, m\}$.

Then on the one hand, by the definition of the power residue symbol,

\begin{align} (\gamma a_1)(\gamma a_2)\dots(\gamma a_m) &= \gamma^{\frac{\mathrm{N} \mathfrak{p} -1}{n}} a_1 a_2\dots a_m \\ &\equiv \left(\frac{\gamma}{\mathfrak{p} }\right)_n a_1 a_2\dots a_m \pmod{\mathfrak{p}}, \end{align} $$ and on the other hand, since $π$ is a permutation,

\begin{align} (\gamma a_1)(\gamma a_2)\dots(\gamma a_m) &\equiv {\zeta_n^{b(1)}a_{\pi(1)}} {\zeta_n^{b(2)}a_{\pi(2)}}\dots{\zeta_n^{b(m)}a_{\pi(m)}} &\pmod{\mathfrak{p}}\\ &\equiv \zeta_n^{b(1)+b(2)+\dots+b(m)}a_{\pi(1)} a_{\pi(2)}\dots a_{\pi(m)} &\pmod{\mathfrak{p}}\\ &\equiv \zeta_n^{b(1)+b(2)+\dots+b(m)} a_1 a_2\dots a_m &\pmod{\mathfrak{p}}, \end{align} $$ so

\left(\frac{\gamma}{\mathfrak{p} }\right)_n a_1 a_2\dots a_m \equiv \zeta_n^{b(1)+b(2)+\dots+b(m)} a_1 a_2\dots a_m \pmod{\mathfrak{p}}, $$ and since for all $1 ≤ i ≤ m$, $a_{i}$ and $$\mathfrak{p}$$ are coprime, $a_{1}a_{2}…a_{m}$ can be cancelled from both sides of the congruence,
 * $$\left(\frac{\gamma}{\mathfrak{p} }\right)_n \equiv \zeta_n^{b(1)+b(2)+\dots+b(m)}

\pmod{\mathfrak{p}}, $$ and the theorem follows from the fact that no two distinct $n$th roots of unity can be congruent (mod $$\mathfrak{p}$$).

Relation to the transfer in group theory
Let $G$ be the multiplicative group of nonzero residue classes in $Z/pZ$, and let $H$ be the subgroup {+1, −1}. Consider the following coset representatives of $H$ in $G$,
 * $$1, 2, 3, \dots, \frac{p-1}{2}.$$

Applying the machinery of the transfer to this collection of coset representatives, we obtain the transfer homomorphism
 * $$\phi : G \to H,$$

which turns out to be the map that sends $a$ to $(−1)^{n}$, where $a$ and $n$ are as in the statement of the lemma. Gauss's lemma may then be viewed as a computation that explicitly identifies this homomorphism as being the quadratic residue character.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: Gauss's lemma (number theory)


text:
File:Translucentmudhoney.jpg

Fair use in Since We've Become Translucent
Though this image is subject to copyright, its use is covered by the U.S. fair use laws because:
 * 1) It's a low resolution copy of a CD/album cover.
 * 2) It doesn't limit the copyright owner's rights to sell the CD/album in any way, in fact, it may encourage sales.
 * 3) Because of the low resolution, copies could not be used to make illegal copies of the album artwork on another CD.
 * 4) The image is itself a subject of discussion in the article or used in the infobox thereof.
 * 5) The image on the cover is significant because it was made by a famous artist.

Source

 * 1) Derived from a digital capture (scan/photo) of the album/CD cover (creator of this digital version is irrelevant as the copyright in all equivalent images is still held by the same party). Copyright held by the record company or the artist. Claimed as fair use regardless.

title: File:Translucentmudhoney.jpg


text: Template:Urban rail transit in China

title: Template:Urban rail transit in China


text:
Zeeman

Zeeman may refer to:
 * "Zeeman (Je Verlangen Is De Zee)", Dutch rendering of the 1960-61 German-language hit single ""Seemann (Deine Heimat ist das Meer)"
 * Zeeman (crater), a lunar impact crater located on the far side of the Moon near its south pole
 * Zeeman effect, the splitting of a spectral line into several components in the presence of a static magnetic field
 * Zeeman (store), a European clothing chain based in the Netherlands
 * Zeeman slower, a scientific apparatus that is commonly used in quantum optics to cool a beam of atoms from room temperature or above to a few kelvins
 * Zeeman (surname)
 * SS Zeeman (1943) a Dutch Liberty ship in service 1946-50

title: Zeeman


text:
The Super 6

The Super 6 is an animated cartoon series which was produced by DePatie–Freleng Enterprises and Mirisch-Rich Television Productions in 1966, and shown on NBC from 1966 to 1969. This was DePatie–Freleng's first vehicle for Saturday morning. Only one season of the show was produced, but NBC ran the series for three years in the same timeslot.

The show was a superhero spoof which featured six diverse characters working for Super Services Inc., offering "heroes for hire" under the supervision of the cranky Super Chief. Each episode consisted of three five- to six-minute segments, with the introductory segment featuring Super Bwoing and the last featuring one of the other five heroes. The middle segment featured the unrelated The Brothers Matzoriley.

The Super 6 theme song was performed by Gary Lewis & the Playboys.

The Super 6 membership
The Super 6 consisted of the following:

 * Super Bwoing: Noted for his rather unremarkable physique, extreme clumsiness, and his winged red helmet, echoing the helmet worn by the Greco-Roman deity Mercury. His preferred mode of transportation is to fly through the air on his guitar (the "Super Bwoinger"), using the strings like reins on a horse. His primary superpower is the ability to emit a super "laser beam" from his eyes, although he has superhuman strength as well. Due to his being an apprentice hero, in addition to being the clumsiest of the Super Service heroes, he seems to only get jobs if the others were busy or on holidays; in essence, he is the last choice and, in any case, better than nothing. Despite his bumbling, however, he almost always manages to catch his villains. His voice is a caricature of Jimmy Stewart's.
 * Granite Man: Composed entirely of gray rock, and possessing the expected powers of near-invulnerability, incredible strength and a powerful uppercut. When not working for the Super Service, Granite Man is a statue in a local park. His assistant is Percival, a messenger pigeon who wakes him from his statue mode with the words "Granite Man, oh rock of power, awake and face this dangerous hour!".
 * Magneto Man: As the name implies, his powers are based on magnetism. His assistant is a young sidekick named "Cal", the brains of the pair. Both heroes are from London, and have exaggerated English accents; Magneto Man sounds like actor Cary Grant.
 * Elevator Man: Ironically the shortest of the Six, he wears a pale gray safari suit. By pressing one of two elevator buttons on his belt buckle, he can shrink to the size of an ant or grow into a giant.
 * Super Scuba: An ocean-based hero who speaks like Dean Martin, Super Scuba wears green scuba gear and lives in an underwater cave with his secretary Bubbles, a mermaid.
 * Captain Whammo/Zammo: This muscular, long-haired blond hero is usually dispatched to fight villains of a more military nature. His powers include flight and incredible strength, and he uses the catchphrase of "Thither, Yonder and Away!" as he leaps to the skies. The Captain and his partner Private Hammo also have the power to travel back in time to various historical events as part of their assignments.

Captain Whammo vs. Captain Zammo
This character appeared in the smallest number of segments of any of the other members of the Super 6. After his first appearance, however, his name was changed from "Whammo" to "Zammo". According to Friz Freleng in a [1982] interview, the name change occurred when Wham-O, creators of such toys as the Super Ball, filed a legal grievance against DePatie-Freleng over trademark infringement. Reportedly, the first commercial to air after the first televised "Captain Whammo" segment was, ironically, for the Super Ball.

The Brothers Matzoriley
This third segment was unrelated to the Super 6, and starred a bizarre, three-headed character known collectively as "The Brothers Matzoriley" a parody in name, of The Brothers Karamazov. They were "Siamese triplets" who were Three Stooges-style bumblers, always looking for a new job. Their names (Weft, Wight and Wong) were a play on "left", "right" and "wrong", respectively. Although occupying one body, each head had its own distinct personality, one head (Weft) being that of a tough guy with a Brooklyn accent, another (Wight) being a wimpy coward with a very nervous delivery, and the middle head (Wong) being a smart alecky Chinese, who dispensed absurd parodies of Confucius proverbs beginning with "Confusion say...". All three personalities were of a broad, stereotypical nature.

A prototype version of the Brothers Matzoriley first appeared in the credits of A Shot in the Dark. Another version appeared in The Great De Gaulle Stone Operation (the first short in The Inspector series). In the cartoon, the brothers steal a precious diamond in the oddest ways from the Inspector. In a comic plot twist ending, they wind up winning the day after the Inspector accidentally drinks a glass of water with the diamond in it. The brothers (disguised as a nurse) steal it after a surgeon removes it during an emergency operation. In The Great De Gaulle Stone Operation, the first head was a suave British-type (Pat Harrington Jr. was the original voice for the left head) and the third head had a Soviet-Russian accent (Paul Frees was the voice of the right head, as well as the middle Chinese head). In contrast to their depiction in The Super Six, they had different voices, but similar personalities - although the middle Chinese head bore some resemblance to the version that would later appear in The Super 6.

Data
The series was broadcast by NBC from September 10, 1966 to January 21, 1967. NBC continued to air reruns until August 31, 1969.

20 episodes (each containing three cartoons) were made.

TGG Direct (under license from MGM Home Entertainment) has released the entire series on DVD.

Voices

 * Daws Butler - Brother Matzoriley #2, Magneto Man
 * Pat Carroll - Dispatcher "Super Chief", Additional Voices
 * Paul Frees - Brother Matzoriley #1 and #3, Captain Whammo/Zammo
 * June Foray - Bubbles
 * Joan Gerber - Additional Voices
 * Arte Johnson - Super Scuba
 * Lyn Johnson - Granite Man, Percival
 * Diana Maddox - Cal
 * Charles Smith - Super Bwoing
 * Paul Stewart - Elevator Man

Staff

 * Supervising Director: Gerry Chiniquy
 * Directors: Steven Clark, Hawley Pratt, Norm McCabe, George Singer, Robert McKimson, John Walker
 * Story Supervision: John Dunn
 * Writers: Tony Benedict, Alan Dinehart, Don Jurwich, Walter Black, John Freeman, Lee Mishkin, Homer Brightman, Dale Hale, Jack Miller, Bill Danch, Bill Hamilton, Michael O'Conner, Art Diamond, Cal Howard, Jim Ryan
 * Storyboard & Layout: Corny Cole, Burt Freund, Bob Givens, Norm Gottfredson, Jan Green, Dave Hanan, Lin Larsen, Marty Murphy, Tony Rivera, Jacques Rupp, Al Wilson
 * Animation: Frank Andrina, Bob Goe, Bob Matz, Dale Case, Manny Gould, Murray McClellan, Herman Cohen, Lee Halpern, Morey Redan, Jim Davis, Bill Hutten, Ed Soloman, Xenia Demattia, Art Leonardi, Ken Southworth, John Garling, Ed Love, Don Williams
 * Backgrounds: Shelia Brown, Gloria Wood, Roger Collins, Tom Yakutis
 * Supervising Film Editor: Lee Gunther
 * Film Editors: Lou Gordon, Chuck McCann, Joseph J. Reitano
 * Voices: Daws Butler, Pat Carroll, Paul Frees, June Foray, Joan Gerber, Artie Johnson, Lyn Johnson, Diana Maddox, Charles Smith, Paul Stewart
 * Music: Bill Lava
 * Production Managers: Basil Cox, Bill Orcutt
 * Asst. Production Managers: Armand Shaw, Harry Love
 * Title Song by: Gary Lewis
 * Produced by: David H. DePatie & Friz Freleng

title: The Super 6


text:
User talk:82.45.5.160

Hi. Thanks for creating Paranormal television and Category:Paranormal Television. I thought I would just point out that it isn't usually Wikipedia practise to have material in the Category namespace that would be in the article namespace. As such I've changed Paranormal television back to a normal article and created a shorter one sentence intro for Category:Paranormal Television. Evil Monkey∴Hello 08:03, 26 July 2005 (UTC)

User:Cj.23
Hi, thanks for all of you contributions. I am assuming that you are User:Cj.23, and just wanted to ask you to think about logging in. It is not required, just makes it easier for us to tell if a user is vandalising your user page or not, and all of your edits will be assigned to your username. Thanks. ∞ Who ? ¿ ? 10:32, 26 July 2005 (UTC)

Speedy deletion of Jason Karl
A tag has been placed on Jason Karl requesting that it be speedily deleted from Wikipedia. This has been done under section G12 of the criteria for speedy deletion, because the article appears to be a blatant copyright infringement. For legal reasons, we cannot accept copyrighted text or images borrowed from other web sites or printed material, and as a consequence, your addition will most likely be deleted. You may use external websites as a source of information, but not as a source of sentences. This part is crucial: say it in your own words.

If the external website belongs to you, and you want to allow Wikipedia to use the text — which means allowing other people to modify it — then you must verify that externally by one of the processes explained at Donating copyrighted materials. If you are not the owner of the external website but have permission from that owner, see Requesting copyright permission. You might want to look at Wikipedia's policies and guidelines for more details, or ask a question here.

If you think that this notice was placed here in error, you may contest the deletion by adding  to the top of the page that has been nominated for deletion (just below the existing speedy deletion or "db" tag), coupled with adding a note on the talk page explaining your position, but be aware that once tagged for speedy deletion, if the page meets the criterion it may be deleted without delay. Please do not remove the speedy deletion tag yourself, but don't hesitate to add information to the page that would render it more in conformance with Wikipedia's policies and guidelines.  MBisanz  talk 19:58, 1 May 2009 (UTC)

Proposed deletion of Paranormal?

The article Paranormal? has been proposed for deletion because of the following concern: "Tagged for notability since 2009. Only lasted 3 episodes so it was unlikely to have attracted media attention"

While all constructive contributions to Wikipedia are appreciated, pages may be deleted for any of several reasons.

You may prevent the proposed deletion by removing the notice, but please explain why in your edit summary or on the article's talk page.

Please consider improving the page to address the issues raised. Removing will stop the proposed deletion process, but other deletion processes exist. In particular, the speedy deletion process can result in deletion without discussion, and articles for deletion allows discussion to reach consensus for deletion. Ten Pound Hammer • (What did I screw up now?) 18:59, 15 May 2022 (UTC)

title: User talk:82.45.5.160


text:
The Masque of Anarchy

The Masque of Anarchy (or The Mask of Anarchy) is a British political poem written in 1819 (see 1819 in poetry) by Percy Bysshe Shelley following the Peterloo Massacre of that year. In his call for freedom, it is perhaps the first modern statement of the principle of nonviolent resistance.

The poem was not published during Shelley's lifetime and did not appear in print until 1832 (see 1832 in poetry), when published by Edward Moxon in London with a preface by Leigh Hunt. Shelley had sent the manuscript in 1819 for publication in The Examiner. Hunt withheld it from publication because he "thought that the public at large had not become sufficiently discerning to do justice to the sincerity and kind-heartedness of the spirit that walked in this flaming robe of verse". The epigraph on the cover of the first edition is from Shelley's The Revolt of Islam (1818): "Hope is strong; Justice and Truth their winged child have found."

The poem’s use of masque and mask has been discussed by Morton Paley; Shelley used mask in the manuscript but the first edition uses masque in the title. The poem has 372 lines, largely in four-line quatrains; two more quatrains appear in some manuscript versions.

Synopsis
Shelley begins his poem, written on the occasion of the Peterloo Massacre, Manchester 1819, with the powerful images of the unjust forms of authority of his time, "God, and King, and Law" – and then imagines the stirrings of a radically new form of social action: "Let a great assembly be, of the fearless, of the free". The crowd at this gathering is met by armed soldiers, but the protesters do not raise an arm against their assailants:

Shelley elaborates on the psychological consequences of violence met with pacifism. The guilty soldiers, he says, will return shamefully to society, where "blood thus shed will speak / In hot blushes on their cheek". Women will point out the murderers on the streets, their former friends will shun them, and honourable soldiers will turn away from those responsible for the massacre, "ashamed of such base company". A version was taken up by Henry David Thoreau in his essay Civil Disobedience, and later by Mahatma Gandhi in his doctrine of Satyagraha. Gandhi's passive resistance was influenced and inspired by Shelley's nonviolence in protest and political action. It is known that Gandhi would often quote Shelley's The Masque of Anarchy to vast audiences during the campaign for a free India.

The poem mentions several members of Lord Liverpool's government by name: the Foreign Secretary, Castlereagh, who appears as a mask worn by Murder, the Home Secretary, Lord Sidmouth, whose guise is taken by Hypocrisy, and the Lord Chancellor, Lord Eldon, whose ermine gown is worn by Fraud. Led by Anarchy, a skeleton with a crown, they try to take over England, but are slain by a mysterious armoured figure who arises from a mist. The maiden Hope, revived, then calls to the people of England:

Literary criticism
Political authors and campaigners such as Richard Holmes and Paul Foot, among others, describe it as "the greatest political poem ever written in English". In his book An Encyclopedia of Pacifism, Aldous Huxley noted the poem's exhortation to the English to resist assault without fighting back, stating "The Method of resistance inculcated by Shelley in The Mask of Anarchy is the method of non-violence".

Author, educator, and activist Howard Zinn refers to the poem in A People's History of the United States. In a subsequent interview, he underscored the power of the poem, suggesting: "What a remarkable affirmation of the power of people who seem to have no power. Ye are many, they are few. It has always seemed to me that poetry, music, literature, contribute very special power." In particular, Zinn uses "The Mask of Anarchy" as an example of literature that members of the American labour movement would read to other workers to inform and educate them.

Use in politics
The rallying language of the poem had led to elements of it being used by political movements. It was recited by students at the Tiananmen Square protests of 1989 and by protesters in Tahrir Square during the Egyptian revolution of 2011. The phrase "like lions after slumber, in unvanquishable number" from the poem was used as a motto/slogan by the International Socialist Organization in their organ. The line "Ye are many-they are few" inspired the campaign slogan "We are many, they are few" used by protesters during the Poll tax riots of 1989–90 in the United Kingdom, and also inspired the title of the 2014 documentary film We Are Many, which focused on the global 15 February 2003 anti-war protests.

Jeremy Corbyn, the former leader of the British Labour Party, quoted the final stanza from the poem at his rally in Islington, on the final day of campaigning for the 2017 general election. Corbyn subsequently quoted the final stanza again during his speech at the Pyramid stage at the 2017 Glastonbury Festival. Academic and writer John Sutherland has suggested that the title of the party's 2017 manifesto, "For the Many, Not the few", was derived from the poem. The phrase 'a community in which power, wealth and opportunity are in the hands of the many, not the few' also appears in the revised version of Clause IV of the Labour Party Constitution.

The same variation, "For The Many, Not The Few", was the sub-title to Robert Reich's 2016 book, Saving Capitalism.

The poem was also quoted on the back cover of The Jam's 1980 album Sound Affects.

title: The Masque of Anarchy


text:
Talk:Jatts

Unwanted page
This page should be merged with jat.

- Manjunatha 26-Jul-2005 13:35 (IST)

title: Talk:Jatts


text:
Since We've Become Translucent

Since We've Become Translucent is the sixth studio album by American rock band Mudhoney, released in 2002. The album was the first to be recorded after the departure of their original bassist Matt Lukin, three years earlier. It was also the first to be released through Sub Pop after the band returned to the label.

Since We've Become Translucent marked a prominent change in the band's sound. The album departed from their typical grunge sound and features a relatively accessible rock sound. However, on tracks such as "Baby, Can You Dig the Light?", psychedelica, synthpop, and jazz are explored.

The bulk of the album was recorded over three weekends with three different producers – Johnny Sangster, Martin Feveyear and Scott Colburn – between November 2001 and February 2022. The band focused on three songs at a time, which they rehearsed and recorded before moving on to the next songs. This approach would also be used on the band's next album Under a Billion Suns. One track, "Inside Job", was recorded with Jack Endino in a single day in April 2000 with former MC5 Wayne Kramer on bass, before new bassist Guy Maddison joined the band in 2001.

Track listing
All tracks are written by Mudhoney

 * 1) "Baby, Can You Dig the Light?" - 8:26
 * 2) "The Straight Life" - 3:33
 * 3) "Where the Flavor Is" - 3:34
 * 4) "In the Winner's Circle" - 4:27
 * 5) "Our Time Is Now" - 3:39
 * 6) "Dyin' for It" - 4:54
 * 7) "Inside Job" - 2:52
 * 8) "Take It Like a Man" - 2:35
 * 9) "Crooked and Wide" - 4:54
 * 10) "Sonic Infusion" - 7:40

Mudhoney

 * Mark Arm – vocals, guitar (except 8), organ (1, 6), piano (3, 8), backing vocals (4, 8)
 * Steve Turner – guitar, backing vocals (2, 4, 5, 8)
 * Guy Maddison – bass guitar (except 7)
 * Dan Peters – drums

Additional personnel

 * Craig Flory – tenor saxophone (1, 3, 8), baritone saxophone (3, 8), horn arrangements (3, 8)
 * Jeff McGrath – trumpet (3, 8)
 * Greg Powers – trombone (3, 8)
 * Wayne Kramer – bass guitar (7)
 * Jo Claxton – violin (10)
 * Miho Takekawa – vibraphone (10)
 * Johnny Sangster – backing vocals (2, 5), producer, engineer, mixing (2, 5, 6)
 * Martin Feveyear – backing vocals (8), producer, engineer, mixing (1, 3, 8)
 * Scott Colburn – infused sonics (10), producer, engineer, mixing (4, 9, 10)
 * Julian Martlew – backing vocals (4), assistant engineer, mixing assistant (4, 9, 10)
 * Jack Endino – producer, engineer, mixing (7)
 * Steve Hall – mastering
 * Jeff Kleinsmith – design
 * Jesse LeDoux – design
 * Lance Hammond – cover photos
 * Emily Rieman – inside photo

title: Since We've Become Translucent


text:
Category:Soubrettes

A soubrette is a lyric female voice that is light, mellow and sweet. The term normally applies to operatic sopranos who sing the comic roles, often of young girls.

title: Category:Soubrettes


text:
Thomas M'Crie

Thomas M'Crie was the name of two Scottish Seceder ministers and historians, father and son:

 * Thomas M'Crie the Elder (1772–1835)
 * Thomas M'Crie the Younger (1797–1875)

title: Thomas M'Crie


text:
Blackwater Estuary

The  Blackwater Estuary is the estuary of the River Blackwater between Maldon and West Mersea in Essex. It is a 5,538 hectare biological Site of Special Scientific Interest (SSSI). An area of 4,395 hectares is also designated a Ramsar Wetland of International Importance, and a Special Protection Area 1,099 hectares is a National Nature Reserve. Tollesbury Wick and part of Abbotts Hall Farm, both nature reserve managed by the Essex Wildlife Trust, are in the SSSI.

Oysters have been harvested from the estuary for more than a thousand years and there are remains of fish weirs from the Anglo-Saxon era. At the head of the estuary is the town of Maldon, which is a centre of salt production. The other major settlement is the town West Mersea, of Mersea Island, on the northern seaward side. Numerous other villages are on its banks.

Within the estuary is Northey Island which was the location for the first experiments in the UK in 'managed retreat', i.e. creating saltmarsh by setting sea walls back from what are perceived to be unsustainable positions. The area is notable as a breeding area for little tern (Sternula albifrons) and as a transit point for ringed plover (Charadrius hiaticula).

 * Over-wintering species:
 * Pied avocet (Recurvirostra avosetta)
 * Black-tailed godwit (Limosa limosa islandica)
 * Dark-bellied brent goose (Branta bernicla bernicla)
 * Dunlin (Calidris alpina alpina)
 * Eurasian golden plover (Pluvialis apricaria)
 * Grey plover (Pluvialis squatarola)
 * Hen harrier (Circus cyaneus)
 * Common redshank (Tringa totanus)
 * Ringed plover (Charadrius hiaticula)
 * Ruff (Philomachus pugnax)
 * Common shelduck (Tadorna tadorna)

The estuary is also the current mooring location for the radio ship Ross Revenge, the home of former pirate station Radio Caroline

title: Blackwater Estuary


text:
Talk:Graphic

Untitled
This is not a disambiguation page. If is to be one, then what should it link to? As far as I know it doesn't differ from Graphics, so I suggest a redirect to there. DirkvdM 08:06, 26 July 2005 (UTC)

title: Talk:Graphic


text:
Jaringan Islam Liberal

Jaringan Islam Liberal (JIL) or the Liberal Islam Network is a loose forum for discussing and disseminating the concept of Islamic liberalism in Indonesia. One reason for its establishment is to counter the growing influence and activism of militant and Islamic extremism in Indonesia. The "official" description of JIL is "a community which is studying and bringing forth a discourse on Islamic vision that is tolerant, open and supportive for the strengthening of Indonesian democratization."

It was started from several meetings and discussions among young Muslim intellectuals in ISAI (Institut Studi Arus Informasi; Institute for the Studies on Free Flow of Information), Jakarta, and then extended through discussion using a mailing list in early 2001. The founders held the first discussion on February 21, 2001, at Teater Utan Kayu, Jakarta, on Akar-Akar Liberalisme Islam: Pengalaman Timur Tengah (The Roots of Islamic Liberalism: The Middle East Experience), presented by a young progressive scholar, Luthfi Assyaukanie. That meeting was followed by other discussions, either in the form of face-to-face meetings or through the mailing list.

Since mid-2001, the "official" name of Jaringan Islam Liberal has been used on their website, which displays their activities, articles, discussions, and relevant sources for the dissemination of liberal Islam. Their place of meeting and secretariat is at Teater Utan Kayu, Jakarta, a complex owned by Goenawan Mohamad, a leading journalist and author, and used for arts performances and by non-governmental organizations. By now, the network is led by liberal young thinker, Ulil Abshar Abdalla, who is also the director of ICRP - Indonesian Conference on Religion and Peace.

Mission
JIL interpretation of Islam is based on the principles follows:

 * 1) Open to all forms of intellectual exploration on all dimensions of Islam
 * 2) Prioritizing religion ethics, not literal textual readings
 * 3) Believing that truth is relative, open for interpretations and plural
 * 4) Siding with oppressed minorities
 * 5) Believing in the freedom to practice religious beliefs
 * 6) Separation of world and heavenly authorities, religious and political authorities

For the proponents of JIL, using liberal Islam in the name of their forum is intended to make a clear point. Liberal Islam represents an Islam that emphasizes individual freedom in accordance with the Mu'tazila doctrine of human freedom and liberation of socio-political structure from oppressive and undemocratic domination. The adjective "liberal" for the activists in JIL has two meanings: "being liberal" and "liberating." Islam has always many different interpretations, so liberal Islam is another variant of Islam and an alternative to literal Islam, Islamic fundamentalism, Islamic extremism, and many others.

Activities
Since its inception, JIL has conducted many regular activities concerning public education. With the assistance from funding agencies such as The Asia Foundation, JIL is able to voice liberal, tolerant aspirations and interpretations of Islam in Indonesia. Activities and programs of JIL include:

 * Syndication of liberal Islam writers. This is the most important program undertaken by JIL. This program is intended to collect writings from authors who defend pluralism and inclusivism and disseminate them to local mass media which have difficulties in finding good writings and writers on those issues. This syndication provides selected articles, interviews, and sources every week for local newspapers.
 * Talk-shows in the news office of Radio 68H, Jakarta. The talk-show, which interviews those who promote pluralism and an inclusive understanding of religion, is broadcast through 40 radio stations on the Namlapanha radio network across Indonesia. This program is the most effective effort to disseminate liberal Islam.
 * Publication of books on liberal thoughts in religion.
 * Publication of booklets or leaflets which contain a short article, interview, or abstract from books on controversial issues in religion.
 * Publication of the JIL website, http://www.islamlib.com, which displays information and relevant writings on liberal Islam.
 * Public service advertisements on television which contain messages for religious toleration and peaceful co-existence among different religious followers and pluralism in Islam.
 * Discussion of Islam. With cooperation from other institutions, JIL arranges discussions on Islam with distinguished speakers from all over the world.
 * Road-show doctrinate the idea of liberal Islam which are held on campuses in Indonesia with the cooperation of student organizations.

Some figures who support or follow liberalism are Ulil Abshar Abdalla, Nong Darol Mahmada, Burhanuddin, Hamid Basyaib, Taufiq Adnan Amal, Saiful Mujani, and Luthfi Assyaukanie.

Conflicts
Ulil in his book "Becoming Liberal Islam" rejects the types of religious interpretation that are not pluralist, or against democracy which he said potentially undermines Islamic thinking. Ulil Abshar candidly criticizes the MUI has monopolized the interpretation of Islam in Indonesia. MUI fatwas stating that Pluralism, Liberalism, and Secularism are misguided ideologies and Ahmadiyah is out of Islam, have kindled Ulil's emotion.

Some Muslims respond to the emergence of liberal Islamic movement in various ways, either with no respond, respond and criticize it intellectually through medias, or with serious response because it has challenged the basic principles of Islam. There are even groups of radicals who justify killing of Ulil and other members of the Liberal Islam Network (JIL), such as Islamic Ulema Forum (Forum Ulama Ummat Islam, FUUI) in Bandung which issued a death fatwa for Ulil Abshar as the chairman of JIL.

Among other controversies deviating from mainstream Islam are JIL believes al-Qur'an is a text needs to be studied Hermeneutically; exegesis based on classical Tafsir books is no longer needed; polygamy must be banned; dowry in marriage may be paid by either the husband or wife; prescribed period ('Iddah) should be imposed on men too, either because of divorce or death; predetermined period Marriage (Nikah mut‘ah) is legal, inter-religious marriage shall be allowed to men and women equally; portion of inheritance to daughters should be equal (1:1) to portion given to sons; a bastard child is eligible for inheritance if his biological father is known.

In 2003, Ulil Abshar is sentenced to death by a fatwa issued by several ulama in FUUI declaring Liberal Islam has misguided and derailed from the teachings of Islam due to JIL's article which is considered an insult and distort the truth of religion. Habib Rizieq mentions that Liberal Islam is a "Plagiarist of thinking", because its ideas simply imitate the ideas of previous orientalists who began the reforms, especially in understanding and assessing Islam.

According to Habib Rizieq, The Indonesian's Liberal Islam movement rants excitedly to the application of Hermeneutics in the studies of the Qur'an. In fact, long before JIL resonating the matter, an Assyrian reverend named Alphonse Mingana (1881-1937), who was also a lecturer of Christian Theology at Birmingham University, in his book "Syriac Influence on The Style of The Koran" published in 1927 states: "It is time to criticize the text of the Qur'an, as we have done to the Aramaic Jewish Torah and Greek Christian bibles." Rizieq furthermore says that the JIL is just a group of inferior people with  "intellectual backwardness" disease, which according to him, has explained its various statements and actions that are often inconsequentially baseless, ignorant and even tend to get lost, like crazy lunatics.

On May, 26 to July 29, 2005 the Indonesian Ulema Council (MUI) at its 7th Mu'tamar Nasional (مؤتمرالوطنية; National Conference) issued eleven non-legally-binding fatwas, where one of them is to forbid pluralism, secularism and liberalism in Indonesia. JIL responded by reasserting its position not to dissolve and stating the fatwas had restricted the freedom of speech in the democratic country. The JIL movement began to fade out with the departure of Ulil Abshar Abdalla to the United States to continue his education, which also dampened the overreactions from its opposing parties. In mid 2005, FPI attacked JIL headquarters due to the fatwa.

Founders

 * Luthfi Assyaukanie
 * Ulil Abshar Abdalla
 * Nong Darol Mahmada

title: Jaringan Islam Liberal


text: User talk:146.182.9.9

title: User talk:146.182.9.9


text:
Intentional programming

In computer programming, Intentional Programming is a programming paradigm developed by Charles Simonyi that encodes in software source code the precise intention which programmers (or users) have in mind when conceiving their work. By using the appropriate level of abstraction at which the programmer is thinking, creating and maintaining computer programs become easier. By separating the concerns for intentions and how they are being operated upon, the software becomes more modular and allows for more reusable software code.

Intentional Programming was developed by former Microsoft chief architect Charles Simonyi, who led a team in Microsoft Research, which developed the paradigm and built an integrated development environment (IDE) called IP (for Intentional Programming) that demonstrated the paradigm. Microsoft decided not to productize the Intentional Programming paradigm, as in the early 2000s Microsoft was rolling out C# and .NET to counter Java adoption. Charles Simonyi decided, with approval of Microsoft, to take his idea out from Microsoft and commercialize it himself. He founded the company Intentional Software to pursue this. Microsoft licensed the Intentional Programming patents Simonyi had acquired while at Microsoft, but no source code, to Intentional Software.

An overview of Intentional Programming as it was developed at Microsoft Research is given in Chapter 11 of the book Generative Programming: Methods, Tools, and Applications.

Development cycle
As envisioned by Simonyi, developing a new application via the Intentional Programming paradigm proceeds as follows. A programmer builds a WYSIWYG-like environment supporting the schema and notation of business knowledge for a given problem domain (such as productivity applications or life insurance). Users then use this environment to capture their intentions, which are recorded at high level of abstraction. The environment can operate on these intentions and assist the user to create semantically richer documents that can be processed and executed, similar to a spreadsheet. The recorded knowledge is executed by an evaluator or is compiled to generate the final program. Successive changes are done at the WYSIWYG level only. As opposed to word processors, spreadsheets or presentation software, an Intentional environment has more support for structure and semantics of the intentions to be expressed, and can create interactive documents that capture more richly what the user is trying to accomplish. A special case is when the content is program code, and the environment becomes an intelligent IDE.

Separating source code storage and presentation
Key to the benefits of Intentional Programming is that domain code which capture the intentions are not stored in source code text files, but in a tree-based storage (could be binary or XML). Tight integration of the environment with the storage format brings some of the nicer features of database normalization to source code. Redundancy is eliminated by giving each definition a unique identity, and storing the name of variables and operators in exactly one place. This makes it easier to intrinsically distinguish declarations from references, and the environment can show them differently.

Whitespace in a program is also not stored as part of the source code, and each programmer working on a project can choose an indentation display of the source. More radical visualizations include showing statement lists as nested boxes, editing conditional expressions as logic gates, or re-rendering names in Chinese.

The system uses a normalized language for popular languages like C++ and Java, while letting users of the environment mix and match these with ideas from Eiffel and other languages. Often mentioned in the same context as language-oriented programming via domain-specific languages, and aspect-oriented programming, IP purports to provide some breakthroughs in generative programming. These techniques allow developers to extend the language environment to capture domain-specific constructs without investing in writing a full compiler and editor for any new languages.

Programming Example
A Java program that writes out the numbers from 1 to 10, using a curly bracket syntax, might look like this:

The code above contains a common construct of most programming languages, the bounded loop, in this case represented by the  construct. The code, when compiled, linked and run, will loop 10 times, incrementing the value of i each time after printing it out.

But this code does not capture the intentions of the programmer, namely to "print the numbers 1 to 10". In this simple case, a programmer asked to maintain the code could likely figure out what it is intended to do, but it is not always so easy. Loops that extend across many lines, or pages, can become very difficult to understand, notably if the original programmer uses unclear labels. Traditionally the only way to indicate the intention of the code was to add source code comments, but often comments are not added, or are unclear, or drift out of sync with the source code they originally described.

In intentional programming systems the above loop could be represented, at some level, as something as obvious as " ". The system would then use the intentions to generate source code, likely something very similar to the code above. The key difference is that the intentional programming systems maintain the semantic level, which the source code lacks, and which can dramatically ease readability in larger programs.

Although most languages contain mechanisms for capturing certain kinds of abstraction, IP, like the Lisp family of languages, allows for the addition of entirely new mechanisms. Thus, if a developer started with a language like C, they would be able to extend the language with features such as those in C++ without waiting for the compiler developers to add them. By analogy, many more powerful expression mechanisms could be used by programmers than mere classes and procedures.

Identity
IP focuses on the concept of identity. Since most programming languages represent the source code as plain text, objects are defined by names, and their uniqueness has to be inferred by the compiler. For example, the same symbolic name may be used to name different variables, procedures, or even types. In code that spans several pages – or, for globally visible names, multiple files – it can become very difficult to tell what symbol refers to what actual object. If a name is changed, the code where it is used must carefully be examined.

By contrast, in an IP system, all definitions not only assign symbolic names, but also unique private identifiers to objects. This means that in the IP development environment, every reference to a variable or procedure is not just a name – it is a link to the original entity.

The major advantage of this is that if an entity is renamed, all of the references to it in the program remain valid (known as referential integrity). This also means that if the same name is used for unique definitions in different namespaces (such as " "), references with the same name but different identity will not be renamed, as sometimes happens with search/replace in current editors. This feature also makes it easy to have multi-language versions of the program; it can have a set of English-language names for all the definitions as well as a set of Japanese-language names which can be swapped in at will.

Having a unique identity for every defined object in the program also makes it easy to perform automated refactoring tasks, as well as simplifying code check-ins in versioning systems. For example, in many current code collaboration systems (e.g. Git), when two programmers commit changes that conflict (i.e. if one programmer renames a function while another changes one of the lines in that function), the versioning system will think that one programmer created a new function while another modified an old function. In an IP versioning system, it will know that one programmer merely changed a name while another changed the code.

Levels of detail
IP systems also offer several levels of detail, allowing the programmer to "zoom in" or out. In the example above, the programmer could zoom out to get a level that would say something like:

< >

Thus IP systems are self-documenting to a large degree, allowing the programmer to keep a good high-level picture of the program as a whole.

Similar works
There are projects that exploit similar ideas to create code with higher level of abstraction. Among them are:
 * Concept programming
 * Language-oriented programming (LOP)
 * Domain-specific language (DSL)
 * Program transformation
 * Semantic-oriented programming (SOP)
 * Literate programming
 * Model-driven architecture (MDA)
 * Software factory
 * Metaprogramming
 * Lisp (programming language)

title: Intentional programming


text:
Talk:Chakri dynasty

The Third King
There're also the 3rd king too. It's originated by ancient army. The king lead grand army at the center. Upraja, vice king, lead the front line army and other royalty cover the rear. At the period of Ayutthaya, Upraja also called as 'Bantoon' and usually be the oldest son of king. Some time other son might elevated as 'Bantoon Noi' or Little Upraja and also be a second heir of the throne. In Bangkok period there was only one 'Bantoon Noi'. He was a nephew of King Rama I. His palace placed on the the ground that now is Siriraj Hospital, Thai people may call this area as 'Wang Lang' or the rear palace, originated by his duty to cover back in the army in the ancient time. --Pudtipong Nawasornyuttana 08:17, 26 July 2005 (UTC)

Naming
Rama III didn't name himself, Rama IV Named him. --Pudtipong Nawasornyuttana 08:28, 26 July 2005 (UTC)

Ananda Mahidol
His name should be Anandha Mahidol. --Pudtipong Nawasornyuttana 08:33, 26 July 2005 (UTC)

Re: Naming
Agreed there. The 4th king is the one who named all the other kings, not the 3rd. The 3rd was called "last". If the 4th wants to rule, there would be no place for him. That is why he renamed the whole thing. Actually, it is more like "First Land", "Middle Land", and "Last Land".

name of article
Shouldn't it be written in small letters, "Chakri dynasty", considering nouns in English are normally written small? Gryffindor 16:17, 19 February 2007 (UTC)

Succession
Are women allowed to inherit the kingship? What is the succession law? ♦Drachenfyre♦ · Talk 13:28, 17 March 2009 (UTC)

News
Mom Rajawongse Narisa Chakrabongse: The Royal Family of Thailand chronicles the descendants of King Rama V to the present day. This book is regarded as the bible of Thai royalty.Also Katya and the Prince of Siam tells the life of the Prince and Mom Catherine.--Felix Folio Secundus (talk) 20:39, 24 May 2009 (UTC)

The Great
I am wondering if the translation of "great" is truly correct here. By western standards, calling three kings in just over two hundred years by such an honour, including the current one in his lifetime, would seem as if the Thais are self-aggrandizing their own nation and institutions. Alterntatively it could be down to translation. Words often have technically correct translation which doesn't always translate well and this means we can interpret things in more than one way e.g. orangutan (Malay language) can be equally tranlated as "old man in the woods" or "man of the forest" yet the imagery inferred to an english speaker in each case is quite different. At times finding a true translation cannot be donw and we simply use the foreign word e.g. "kitsch". Of course, I may be entirely wrong and Great may have been appropriately used in each case, if so please convince me. Dainamo (talk) 10:27, 8 January 2010 (UTC)

Where was this dynasty?
A map would tell a thousand words. I came here after reading that the kingdom was once much larger and that it once included parts of Burma and other regions. If anybody has this info or a map (or cares to make one) please include it in the article. Much thanks.BrianAlex (talk) 00:18, 4 October 2010 (UTC)

Regents Ninth reign
Prince Dhanivat Bidyalabh Bridhyakon (1946) has an article on Thai Wikipedia title "พระวรวงศ์เธอ พระองค์เจ้าธานีนิวัต กรมหมื่นพิทยลาภพฤฒิยากร" also spelled H.H. Prince Dhani Nivat Kromamun Bidyalabh Bridhya]

--Pawyilee (talk) 16:45, 17 June 2013 (UTC)

Proposed merge with Chakri Dynasty's Family Tree
The family tree shouldn't be a stand-alone article. It would be better to have it as a collapsible box in the main article. Paul_012 (talk) 18:46, 8 December 2013 (UTC)
 * This sounds completely reasonable to me. I don't envisage that a large amount of debate would accompany this proposal. It would be good to act on this as soon as possible.--Soulparadox (talk) 05:59, 23 February 2014 (UTC)
 * I have implemented the merger. Sodacan (talk) 15:18, 27 December 2014 (UTC)
 * Excellent work! Regards,--Soulparadox (talk) 15:33, 28 December 2014 (UTC)

Requested move 2 November 2016
Chakri Dynasty → Chakri dynasty – This is the only dynasty article in Category:Asian dynasties, Category:European dynasties and Category:Middle Eastern dynasties using a capital D. Celia Homeford (talk) 09:15, 2 November 2016 (UTC) --Relisting. — Andy W. ( talk ) 01:17, 10 November 2016 (UTC)
 * Comment: The above claim is untrue. The articles under Category:Dynasties of ancient Egypt, for example, are all capitalised. There have been previous discussions regarding the capitalisation of dynasty, but their scope was limited to Chinese history. I don't think they apply to this article, because unlike the use of Chinese dynasty names, which mainly refer to historical periods, this article is about the ruling house known as the Chakri Dynasty. (Compare to e.g. House of Windsor, which is spelled with a capital H in the article text.) There appears to be much variation among book and scholarly sources. This proposal needs more policy-based arguments than claiming consistency alone. --Paul_012 (talk) 10:32, 2 November 2016 (UTC)
 * I've amended the opening statement. Using sentence case and consistency are part of Article titles, which is a policy. Celia Homeford (talk) 15:14, 2 November 2016 (UTC)
 * So is capitalising proper nouns. This discussion needs to establish whether or not dynasty is part of the subject's proper name. --Paul_012 (talk) 10:12, 3 November 2016 (UTC)
 * WP:IAR I've removed the requested move/dated transclusion above to remove this page from bot listing, as it appears there is not a consensus forming for this move. I have not closed this discussion, as seems to be leaving this open-ended encouraging further discussion. If a consensus is reached, consider opening another RM. — Andy W. ( talk ) 06:20, 12 November 2016 (UTC)

Requested move 1 December 2016

 * The following is a closed discussion of a requested move. Please do not modify it. Subsequent comments should be made in a new section on the talk page. Editors desiring to contest the closing decision should consider a move review. No further edits should be made to this section.

The result of the move request was: moved. There is a general consensus that WP:NCCAPS and MOS:CAPS apply here, that these suggest rendering 'dynasty' in the lowercase, and that WP:CONSISTENCY with other similar articles should be a priority. Therefore, the article will be moved to the proposed title. RGloucester — ☎ 19:39, 21 December 2016 (UTC)

Chakri Dynasty → Chakri dynasty – Consistency with other pages in Category:Asian dynasties and use in reliable sources, e.g. Celia Homeford (talk) 11:48, 1 December 2016 (UTC) --Relisting. Armbrust The Homunculus 05:10, 8 December 2016 (UTC)

 * Oppose until .... I cannot read the references in the article, but the external links all treat Chakri Dynasty as a proper noun with both words capitalised.  Perhaps better references in the article would be the first step before deciding on any change?    D b f i r s   12:15, 1 December 2016 (UTC)
 * The three better references above all treat it as lower case. Celia Homeford (talk) 12:27, 1 December 2016 (UTC)
 * Yes, I noticed that fact after I'd opposed the move, so I changed my comment. I suggest that you put the references into the article, then you would have some grounds for your case.  We are inconsistent in our capitalisation of Dynasty as shown in our list.    D b f i r s   12:31, 1 December 2016 (UTC)
 * I don't quite know what you mean by your final comment. I've just looked at the first 10 instances of "Dynasty" in that list, and they all redirect to an article using lower case. It's the list that's inconsistent, not the article titles. Celia Homeford (talk) 12:34, 1 December 2016 (UTC)
 * I mean what I wrote, and the first ten of a long list is not a representative sample. See Mangrai Dynasty in the same country, and Phagmodrupa Dynasty in the section above, plus lots more.  What do we do about "House", "Empire" and "Kingdom"?   D b f i r s   12:58, 1 December 2016 (UTC)

 * Support. Chicago Manual of Style, the copy editor's Bible, says to lower case "dynasty." Section 8.71 gives "Shang dynasty" as an example. Dynasties are considered descriptive designations of periods rather than political divisions. So it's Shang Kingdom but Shang dynasty. Pandas and people (talk) 06:48, 8 December 2016 (UTC)
 * We normally follow what reliable sources do rather than obey one particular giver of advice, but reliable sources seem to be split here. (We need some in the article.)  How important is consistency between articles?  Should this be widened to be discussed at WP:Manual of Style ( which doesn't seem to follow its own advice )?    D b f i r s   08:44, 8 December 2016 (UTC)
 * The Manual of Style appears to disagree. It recommends CMOS in two different places. This is a style issue, so the style manuals are the relevant authorities. I don't think there is any style manual that tells you to capitalize. It's just sloppy editing. Pandas and people (talk) 11:14, 8 December 2016 (UTC)
 * We must be reading different articles. I see no recommendation, only a partial disagreement.    D b f i r s   18:04, 8 December 2016 (UTC)
 * At the bottom of the article, it says, "Wikipedians are encouraged to familiarize themselves with other guides to style and usage, which may cover details not included in this Manual of Style. Among these are:" Then a bunch of style guides listed. Being one item on a long list may not seem like much of a recommendation. But most of the books listed aren't relevant here. They are specifically for British English, Australian English, or something like that. Pandas and people (talk) 02:08, 12 December 2016 (UTC)
 * Please see my comments in the section above regarding the different meanings of dynasty, i.e. historical period vs ruling house. Does the CMOS address the latter at all? --Paul_012 (talk) 14:37, 8 December 2016 (UTC)
 * I didn't see anything like that. The manual online at Amazon, so you can check for yourself. It's page 418, paragraph 8.71. Pandas and people (talk) 23:21, 8 December 2016 (UTC)
 * Chicago 's reasoning is obviously correct, and MoS uses similar proper name versus description reasoning throughout. Anyway, a few points, in two different directions: Familiarizing yourself with other style guidelines is a process, and MoS pointing out some major ones does not in any way mean "MoS says do exactly what these style guides say". MoS its is own style guide. It addresses only matters of perennial dispute on Wikipedia itself, and leaves the rest up to editorial consensus, which obviously should be informed by a comparison of major style guides. However, the idea that "We normally follow what reliable sources do" is also frequently misinterpreted to mean "do what the specialist sources do", when what WP really does is what the general-audience sources do (see WP:SSF for why).  — SMcCandlish ☺ ☏ ¢ ≽ʌⱷ҅ᴥⱷʌ≼  14:47, 15 December 2016 (UTC)

 * Support per Google Ngram. Timmyshin (talk) 07:44, 11 December 2016 (UTC)
 * It's interesting to see that it was more often published with a capital "D" in the early 1980s.   D b f i r s   08:02, 11 December 2016 (UTC)
 * If you like Timmyshin (talk) 08:06, 11 December 2016 (UTC)
 * Yes, I agree that it was only transient and there are not many publications, but the frequencies are comparable. The only reason I was inclined to oppose was that the only links in the article all had a capital "D".    D b f i r s   08:24, 11 December 2016 (UTC)

 * Support per MOS:CAPS and for WP:CONSISTENCY with the other "Foo dynasty" articles. The general rule of MOSCAPS is do not capitalize if in doubt (i.e., if reliable sources are not consistent on the matter), and they are not about "dynasty" and the like, including in this specific case. Proof: Google Books results; weeding out titles Capitalized In Title Case, the usage is about a 60/40 split, favoring lower case, and we would still use lower case even if it were a 30/70 split favoring capitalization. We've already had the same basic discussion many times previously, for "dynasty", "family", "regime", yadda yadda, and the result is always the same: the descriptor goes in lower case.  See, e.g., two RMs at Talk:Kim dynasty (North Korea), among many others (I'm having connectivity issues right now, so I can't diff-dig for a bunch more of these).  There are a few other "Dynasty" stragglers that should also be moved, aside from a handful where the proper name of the entity actually includes that word (or a cognate in another language we're translating). PS: Sometimes dynasty articles use a plural of the family name, without "dynasty", if it's more familiar that way, which I doubt is the case with Chakris, though that should exist as a redirect.  — SMcCandlish ☺ ☏ ¢ ≽ʌⱷ҅ᴥⱷʌ≼  14:47, 15 December 2016 (UTC)
 * The Kim move discussion was really about including "Communist", but I'll take your word that the matter has been discussed many times before because you make a very good argument. I would have been happier to support in the first place if the article had even one reference to a lower-case usage instead of all links being to "Chakri Dynasty".  What are we going to do about all the other capitalisations of Dynasty in other articles?  How do we decide whether there are some cases of "proper name" as  suggests above?  What about "House", "Empire" and "Kingdom"?    D b f i r s   19:20, 15 December 2016 (UTC)

 * Support per what has been said above. Other examples can also be found here on Wikipedia. Dynasty isn't a proper noun here, thus I think there's no need for a capital letter. Keivan.f  Talk 22:10, 20 December 2016 (UTC)

 * The above discussion is preserved as an archive of a requested move. Please do not modify it. Subsequent comments should be made in a new section on this talk page or in a move review. No further edits should be made to this section.

Consistency?
The consensus was clearly that the word "dynasty" is not part of a proper noun so I've changed Mangrai Dynasty to lower case, and Celia has changed Phagmodrupa Dynasty. There are lots more to do if we are serious about consistency. The article still has links only to Chakri Dynasty with a capital "D". How about adding some references which use a lower case "d"? Also, does this policy apply to "House", "Empire" and "Kingdom"?  D b f i r s   20:53, 21 December 2016 (UTC)
 * That's a question to ask at the talk page for MOS:CAPS, not here. I'd suggest requesting input there. However, the general guidelines with regard to capitalisation, MOS:CAPS for prose and WP:NCCAPS for article titles, make it clear that unnecessary capitalisation is avoided on Wikipedia.  RGloucester  — ☎ 21:21, 21 December 2016 (UTC)
 * Thanks for replying. It seems that we have to search Google for every individual article to determine whether the article title is always regarded as a proper name (in which case it's Dynasty, as in Dragon Dynasty) (or just usually, as in this decision on the Egyptian Dynasties), or whether some reliable sources treat the word dynasty as not part of the proper title (as here).  I still think that the first place to look should be the English references given in the article, and this article hasn't any!  All of its English links use a capital D.  Sorry to go on about this.  I'll shut up now!    D b f i r s   22:42, 21 December 2016 (UTC)

Start of Rama X's reign
The Thai gov't has determined that Vajiralongkorn's reign began upon his father's death, 13 October 2016. -- GoodDay (talk) 22:44, 1 December 2016 (UTC)

Line of Succession Tree
Why does the line of succession tree only begin with King Bhumibol Adulyadej who died in 2016, and not include the 7 previous kings? -- TadgStirkland401 (TadgTalk) 02:37, 22 August 2018 (UTC)
 * Just added! Sodacan (talk) 03:05, 22 August 2018 (UTC)

DNA.
Has any of the Chakri descendants done a Y-DNA test? --Yomal Sidoroff-Biarmskii (talk) 15:12, 30 April 2019 (UTC)

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title: Talk:Chakri dynasty


text:
Talk:Cassandra Trelawney

This article has been merged with Sybill Trelawney after this VFD debate. Actually, the suggested target was List of characters in the Harry Potter books, but I chose to merge with Sybill Trelawney instead since the list does not accomodate descriptions. If anybody disagrees with my decision, it can be undone. Sjakkalle (Check!)  08:13, 26 July 2005 (UTC)

title: Talk:Cassandra Trelawney


text: Category:Autobianchi

title: Category:Autobianchi


text:
Thomas Edward

Thomas Edward may refer to:

 * Thomas Edward (naturalist) (1814–1886), Scottish naturalist
 * Thomas Edward (MP), MP for Taunton

title: Thomas Edward


text:
Escudo

The escudo (Portuguese: 'shield') is a unit of currency which is used in Cape Verde, and which has been used by Portugal, Spain and their colonies. The original coin was worth 16 silver reais. The Cape Verdean escudo is, and the Portuguese escudo was, subdivided into 100 centavos. Its symbol is the Cifrão, a letter S with two vertical bars superimposed used between the units and the subdivision (for example, ).

In Spain and its colonies, the escudo refers to a gold coin worth sixteen reales de plata or forty reales de vellón.

Circulating

 * Cape Verdean escudo

Obsolete

 * Angolan escudo
 * Chilean escudo
 * French écu
 * Mozambican escudo
 * Portuguese escudo
 * Portuguese Guinean escudo
 * Portuguese Indian escudo
 * Portuguese Timorese escudo
 * São Tomé and Príncipe escudo
 * Spanish escudo

title: Escudo


text:
User talk:Diogenes zosimus

Please click here to leave me a new message.

Welcome to my Talk Page. Please use the box above, or manually enter new messages at the end of my page so I can find them easily. Thanks!'''

PS - I borrowed the above from Cecropia's discussion board - hope that's OK! I'm still learning how to use all the fancy commands of Wikiworld.

Duplicate images uploaded
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If you have uploaded other unlicensed media, please check whether they're used in any articles or not. You can find a list of 'image' pages you have edited by clicking on the "my contributions" link (it is located at the very top of any Wikipedia page when you are logged in), and then selecting "Image" from the dropdown box. Note that any non-free images not used in any articles will be deleted after seven days, as described on criteria for speedy deletion. Thank you. BetacommandBot 08:56, 21 June 2007 (UTC)

Disputed fair use rationale for Image:UGB.PNG
Thanks for uploading Image:UGB.PNG. However, there is a concern that the rationale you have provided for using this image under "fair use" may be invalid. Please read the instructions at Non-free content carefully, then go to the image description page and clarify why you think the image qualifies for fair use. Using one of the templates at Fair use rationale guideline is an easy way to ensure that your image is in compliance with Wikipedia policy, but remember that you must complete the template. Do not simply insert a blank template on an image page.

If it is determined that the image does not qualify under fair use, it will be deleted within a couple of days according to our criteria for speedy deletion. If you have any questions please ask them at the media copyright questions page. Thank you.BetacommandBot (talk) 21:50, 26 November 2007 (UTC)

ArbCom elections are now open!
MediaWiki message delivery (talk) 13:05, 23 November 2015 (UTC)

ArbCom elections are now open!
MediaWiki message delivery (talk) 13:32, 23 November 2015 (UTC)

title: User talk:Diogenes zosimus