Abdallah Assi
email: assi@univ-angers.fr

Pedro A García-Sánchez

Vincenzo Micale
email: vmicale@dmi.unict.it

P A Garc Ía

Bases of subalgebras of K[[x]] and K[x]

come

Introduction

Let K be a field and let K x be the ring of formal power series over K. Let f 1 (x), . . . , f s (x) be s elements of K x and let R = K f 1 , . . . , f s be the subalgebra of K x generated by f 1 , . . . , f s . Given f ∈ R, let o(f ) the order of f . The set o(R) = {o(f ) | f ∈ R} is a submonoid of N, and the knowledge of a system of generators of this monoid is important for the understanding of the subalgebra R. When furthermore K x is an R-module of finite length, then o(R) is a numerical semigroup.

A similar construction can be made in the ring of polynomials K[x]. More precisely let f 1 (x), . . . , f s (x) be s elements of K[x] and let A = K[f 1 , . . . , f s ] be the subalgebra of K[x] generated by f 1 , . . . , f s . Given f ∈ A, let d(f ) the degree of f . The set d(A) = {d(f ) | f ∈ A} is a submonoid d(A) of N, and the knowledge of a system of generators of this monoid is important for the understanding of the subalgebra A. When furthermore K[x] is an A-module of finite length, then d(A) is a numerical semigroup.

A numerical semigroup S is a submonoid of the set of nonnegative integers under addition such that the N \ S is finite, or equivalently, gcd(S) = 1 (the greatest common divisor of the elements of S), see for instance [START_REF] Rosales | Numerical semigroups[END_REF]. In this case, there exists a minimum c ∈ S such that c + N ⊆ S. We call this element the conductor of S, and denote it by c(S) (the motivation of this name and others coming from Algebraic Geometry is explained in [START_REF] Barucci | Properties in numerical semigroups and applications to one-dimensional analytically irreducible local domains[END_REF][START_REF] Fröberg | On numerical semigroups[END_REF]).

Assume that f i is a monomial x a i for every i ∈ {1, . . . , s}. Then o(R) (respectively d(A)) is generated by a 1 , . . . , a s . In this case, R K X 1 , . . . , X s /T (respectively A K[X 1 , . . . , X s ]/T ), where T is a prime binomial ideal and, thus, V(T ) is a toric variety.

Given a subalgebra R = K[[f 1 , . . . , f s ]] (respectively A = K[f 1 , . . . , f s ]), the main objective of this paper is to describe an algorithm that calculates a generating system of o(R) (respectively d(A)). The algorithm we present here allows us, by using the technique of homogenization, to construct a flat K umodule (respectively K[u]-module) which is a deformation of R (respectively A) to a binomial ideal. This technique is well known when R = K f 1 , f 2 and K is algebraically closed field of characteristic zero (see [START_REF] Goldin | Resolving singularities of plane analytic branches with one toric morphism[END_REF] and [START_REF] Teissier | in Le probléme des pour les branches planes[END_REF]). It turns out that the same holds wherever we can associate a semigroup to the local subalgebra, and also that the same technique can be adapted to the global setting. As a particular case we prove that a plane polynomial curve has a deformation into a complete intersection monomial space curve.

The paper is organized as follows. In Section 2 we focus on the local case, namely the case of a subalgebra R of K x . We introduce the notion of basis of R and we show how to construct such a basis. We also show that if o(R) is a numerical semigroup, then every element of a reduced basis is a polynomial. In Section 3 we show how to construct a deformation from R to a toric ideal (or a formal toric variety) by using the technique of homogenization. In Section 4 we focus on the case when R = K[[f (x), g(x)]] and K is an algebraically closed field of characteristic zero. The existence in this case of the theory of Newton-Puiseux allows us to precise the results of Sections 2 and 3. The difference with the procedure The first author is partially supported by the project GDR CNRS 2945 and a GENIL-SSV 2014 grant. The second author is supported by the projects MTM2014-55367-P, FQM-343, FQM-5849, and FEDER funds.

1 presented in Section 2 is that it does not rely in the computation of successive kernels. Then in Sections 5 and 6 we adapt the local results to the case of a subalgebra A of K[x]. When A = K[f (x), g(x)] and K is algebraically closed field of characteristic zero, a basis of A can be obtained by using the theory of approximate roots of the resultant of X -f (x), Y -g(x), which is a polynomial with one place at infinity.

The procedures presented here have been implemented in GAP ( [START_REF]GAP -Groups, Algorithms, and Programming[END_REF]) and will be part of the forthcoming stable release of the package numericalsgps ( [START_REF] Delgado | NumericalSgps", A GAP package for numerical semigroups[END_REF]).

1.1. Some notation. We denote by A the monoid generated by A, A ⊆ N, that is, the set {n

1 x 1 + • • • + n m x m | m ∈ N, n i ∈ N, x i ∈ A for all i ∈ {1, . . . , m}}.
Associated to each numerical semigroup S we can define a natural partial ordering ≤ S , where for two elements s and r in S we have s ≤ S r if there exists u ∈ S such that r = s + u. The set g i of minimal elements in S \ {0} with this ordering is called a minimal set of generators for S. The set of minimal generators is finite since for any s ∈ S \ {0}, we have x ≡ y (mod s) if x = y are minimal elements with respect to ≤ S . The cardinality of the minimal generating system is known as the embedding dimension of S.

Semigroup of a formal space curve

Let K be a field. In this section we will consider rings R that are subalgebras of K x and such that, if we denote the integral closure of R in its quotient field by R, then R = K x and λ R ( R/R) < ∞ where λ R (•) is the length as R-module. Part of the results of this section are inspired in [START_REF] Micale | Order bases of subalgebras of k X , Commutative rings[END_REF] and in [START_REF] Micale | Order bases of subalgebras of power series rings[END_REF]. An alternative procedure (implemented in Maple) is provided in [START_REF] Castellanos | Algorithm for the Semigroup of a Space Curve Singularity[END_REF]. The main difference with our approach is that we do not rely on the multiplicity sequence, and thus we do not need to perform blow-ups. Also we take intrinsic advantage in our implementation of the GAP package numericalsgps ( [START_REF]GAP -Groups, Algorithms, and Programming[END_REF][START_REF] Delgado | NumericalSgps", A GAP package for numerical semigroups[END_REF]).

Let

f = i c i x i ∈ R * = R \ {0}. Define supp(f ) = {i, c i = 0}.
We call min supp(f ) the order of f and we denote it by o(f ). We also set

M o (f ) = c o(f ) x o(f ) . If M o (f ) = x o(f )
, then we shall say, by abuse of notation, that f is monic. We set o(0) = +∞.

We denote by o(R) the set of orders of elements in

R * = R \ {0}, that is, o(R) = {o(f ) | f ∈ R * }. We finally set M o (R) = K[M o (f ) | f ∈ R * ]. Proposition 2.1. Let f 1 , f 2 be elements of R * and let a = min{o(f 1 ), o(f 2 )}. (i) a ≤ o(f 1 + f 2 ). (ii) If o(f 1 ) = o(f 2 ) then a = o(f 1 + f 2 ). (iii) o(f 1 f 2 ) = o(f 1 ) + o(f 2 ).
Proof. This follows easily from the definition of order.

Proposition 2.2. [13, Lemma 3, p.486] Let R 1 and R 2 be rings of our type such that

R 1 ⊆ R 2 and o(R 1 ) = o(R 2 ). Then R 1 = R 2 . Proposition 2.3. [13, Proposition 1, p.488] Let R be a ring of our type. Then λ R ( R/R) = |N \ o(R)|.
The following two results appear in [START_REF] Micale | Order bases of subalgebras of k X , Commutative rings[END_REF], and since this paper is very hard to find, we include the proofs for sake of completeness. Proposition 2.4. [START_REF] Micale | Order bases of subalgebras of k X , Commutative rings[END_REF] Let R be a ring of our type. Then o(R) is a numerical semigroup.

Proof. Since R is a ring and by (iii) of Proposition 2.1, we have that o(R) is a subsemigroup of N. By λ R ( R/R) < ∞ and Proposition 2.3, we have the proof. Proposition 2.5. [START_REF] Micale | Order bases of subalgebras of k X , Commutative rings[END_REF] Let R be a ring of our type. Then R contains every element

f ∈ K x of order o(f ) ≥ c(o(R)). Proof. Use Proposition 2.2 with R 1 = R and R 2 = R + x c(o(R)) K x .
This later result allows to work with polynomials instead of series. Let f 1 , . . . , f s be in R * . Let R = K f 1 , . . . , f s be a subalgebra of K x as above, that is, the integral closure of R in its quotient field is R = K x and λ R ( R/R) < ∞.

Under these hypotheses on R, we have that o(R) is a numerical semigroup (Proposition 2.4). We say that the set {f 1 , . . . ,

f s } ⊂ R * is a basis of R if o(R) = o(f 1 ), . . . , o(f s ) . The set {f 1 , . . . , f s } is a basis of R if and only if M o (R) = K[M o (f 1 ), . . . , M o (f s )].
Proposition 2.6. Let the notations be as above. Given f (x) ∈ K x , there exist g(x) ∈ R and r(x) ∈ K x such that the following conditions hold.

(i)

f (x) = g(x) + r(x) = α c α f α 1 1 • • • f αs s + r(x). (ii) If g(x) = 0 (respectively r(x) = 0), then o(g) ≥ o(f ) (respectively o(r) ≥ o(f )). (iii) Either r(x) = 0 or supp(r(x)) ⊆ N \ o(f 1 ), . . . , o(f s ) . Proof. The assertion is clear if f ∈ K. Suppose that f / ∈ K and let f (x) = i≥p c i x i with p = o(f ) ≥ 0.
(1) If p / ∈ o(f 1 ), . . . , o(f s ) , then we set g 1 = 0, r 1 = c p x p and f 1 = f -c p x p .

(

) If p ∈ o(f 1 ), . . . , o(f s ) , then c p x p = c θ M 0 (f 1 ) θ 1 • • • M 0 (f s ) θs . We set g 1 = c θ f θ 1 1 1 • • • f θ 1 s s , r 1 = 0 and f 1 = f -g 1 . 2
In such a way that f = f 1 + g 1 + r 1 , g 1 ∈ R, either r 1 = 0 or supp(r 1 ) ⊆ N \ o(f 1 ), . . . , o(f s ) , and if f 1 = 0, then o(f 1 ) > o(f ) = p. Then we restart with f 1 . We construct in this way sequences

(f k ) k≥1 , (g k ) k≥1 , (r k ) k≥1 such that for all k ≥ 1, f = f k + k i=1 g i + k i=1 r i , and o(f ) < o(f 1 ) < • • • < o(f k ), k i=1 g i ∈ R, supp( k i=1 r i ) ∈ N\ o(f 1 ), . . . , o(f s
) and for all i < j ≤ k, if g i = 0 = g j (respectively

r i = 0 = r j ), then o(f ) ≤ o(g i ) < o(g j ) (respectively o(f ) ≤ o(r i ) < o(r j )). Clearly lim k-→+∞ f k = 0.
Hence, if g = lim k-→+∞ k i=1 g i and r = lim k-→+∞ k i=1 r i , then f = g + r and g, r satisfy the conditions above.

We denote the series r(x) of the proposition above by R o (f, {f 1 , . . . , f s }). This series depend strongly on step (2) of the proof of Proposition 2.6. Let for example 11 . We shall see that r(x) becomes unique if f 1 , . . . , f s is a basis of R (see Proposition 2.8).

f 1 = x 6 , f 2 = x 4 + x 5 , f 3 = x 2 + x 5 , and let f = x 4 . We have f = f 2 -x 5 = f 2 3 -f 1 f 2 -2x 7 + x
Proposition 2.7. The set {f 1 , . . . , f s } is a basis of R if and only if R o (f, {f 1 , . . . , f s }) = 0 for all f ∈ R.

Proof. Suppose that {f 1 , . . . , f s } is a basis of R and let f ∈ R. Let r(x) = R o (f, {f 1 , . . . , f s }). Then r(x) ∈ R. If r = 0, then o(r) ∈ o(f 1 ), . . . , o(f s ) , which is a contradiction.

Conversely, suppose that {f 1 , . . . , f s } is not a basis of R and let 0 = f ∈ R such that o(f ) / ∈ o(f 1 ), . . . , o(f s ) . We have R o (f, {f 1 , . . . , f s }) = 0, which contradicts the hypothesis.

Proposition 2.8. If {f 1 , . . . , f s } is a basis of R then for all f ∈ K x , R o (f, {f 1 , . . . , f s }) is unique.
Proof. Suppose that f = g 1 (x) + r 1 (x) = g 2 (x) + r 2 (x) where g 1 (x), g 2 (x), r 1 (x), r 2 (x) satisfy conditions (i), (ii), (iii) of Proposition 2.6. We have

r 1 (x) -r 2 (x) = g 2 (x) -g 1 (x) ∈ R. If r 1 (x) -r 2 (x) = 0 then o(r 1 (x) -r 2 (x)) /
∈ o(f 1 ), . . . , o(f s ) . This contradicts the hypothesis.

Let, as above, R = K f 1 , . . . , f s . We shall suppose that f i is monic for all 1 ≤ i ≤ s. Define

φ : K[X 1 , . . . , X s ] -→ K[x], φ(X i ) = M o (f i ) for all i ∈ {1, . . . , s}.
Let {F 1 , . . . , F r } be a generating system of the kernel of φ. We can choose all of them to be binomials.

If F i = X α i 1 1 • • • X α i s s -X β i 1 1 • • • X β i s s , we set S i = f α i 1 1 • • • f α i s s -f β i 1 1 • • • f β i s s . Note that if p = s k=1 α i k o(f k ) = s k=1 β i k o(f k ), then o(S i ) > p.
Theorem 2.9. The system {f 1 , . . . , f s } is a basis of R if and only if R o (S i , {f 1 , . . . , f s }) = 0 for all i ∈ {1, . . . , r}.

Proof. Suppose that {f 1 , . . . , f s } is a basis of R. Since S i ∈ R for all i ∈ {1, . . . , r}, then, by Proposition 2.7, R o (S i , {f 1 , . . . , f s }) = 0.

For the sufficiency assume to the contrary that {f 1 , . . . , f s } is not a basis of R. Then there exists

f ∈ R such that o(f ) ∈ o(f 1 ), . . . , o(f s ) . Write f = θ c θ f θ 1 1 • • • f θs s .
For all θ, if c θ = 0, we set

p θ = s i=1 θ i o(f i ) = o(f θ 1 1 • • • f θs s ). Let p = min{p θ | c θ = 0}
and let {θ 1 , . . . , θ l } be such that p = o(f Algorithm 2.10. Let the notations be as above.

1. If R o (S k , {f 1 , . . . , f s }) = 0 for all k ∈ {1, . . . , r}, then {f 1 , . . . , f s } is a basis of R. 2. If r(x) = R o (S k , {f 1 , . . . , f s }) = 0
for some k ∈ {1, . . . , r}, and if M o (r(x)) = ax q , then we set f s+1 = 1 a r(x), and we restart with {f 1 , . . . , f s+1 }. Note that in this case,

o(f 1 ), . . . , o(f s ) o(f 1 ), . . . , o(f s ), o(f s+1 ) ⊆ o(R).
This process will stop, giving a basis of R, because the complement of o(R) in N is finite. Observe that r(x) is not in general a polynomial. So we must use a trick to compute it, or at least the relevant part of it. This is accomplished by using Proposition 2.5. If in the current step of the algorithm o(f 1 ), . . . , o(f s ) is a numerical semigroup, then we compute its conductor, say c. Then c ≥ c(o(R)). To compute R o (f, {f 1 , . . . , f s }) we do the following. Let p = o(f ). 1. If p ≥ c, then return 0. We implicitly assume that x a is in our generating set for a ∈ c + N (though we do not store them).

2. If p ∈ o(f 1 ), . . . , o(f s ) , then M o (f ) = θ c θ M o (f 1 ) θ 1 • • • M o (f s ) θs . Set f = f -θ c θ f θ 1 1 • • • f θs s
, and call recursively R o (f, {f 1 , . . . , f s }) (the process will stop because the order of the new f is larger, and eventually will become bigger than c after a finite number of steps).

3. If p ∈ o(f 1 ), . . . , o(f s ) , then return f . If o(f 1 ), . . . , o(f s ) is not a numerical semigroup, let d be its greatest common divisor. Set c = dc( o(f 1 ), . . . , o(f s ) /d).
In this case we proceed as follows. 1. If p ≥ c, then return f . We cannot ensure here that f will be reduced to zero, so we add it just in case.

If

p ∈ o(f 1 ), . . . , o(f s ) , then M o (f ) = θ c θ M o (f 1 ) θ 1 • • • M o (f s ) θs . Set f = f -θ c θ f θ 1 1 • • • f θs s , and call recursively R o (f, {f 1 , . . . , f s }). 3. If p ∈ o(f 1 ), . . . , o(f s ) , then return f . One might check first if d does not divide p, because in this case for sure p ∈ o(f 1 ), . . . , o(f s ) .
Observe that by adding the conditions p ≥ c, we are avoiding entering in an eventual infinite loop.

Suppose that {f 1 , . . . , f s } is a basis of R. Also suppose that for all i ∈ {1, . . . , s}, f i is monic. We say that {f 1 , . . . ,

f s } is a minimal basis of R if o(f 1 ), . . . , o(f s ) generate minimally the semigroup o(R). We say that {f 1 , . . . , f s } is a reduced basis of R if supp(f i (x) -M 0 (f i )) ⊆ N \ o(R). Let i ∈ {1, . . . , s}. If o(f i ) ∈ o(f 1 ), . . . , o(f i-1 ), o(f i+1 ), . . . , o(f s ) , then {f 1 , . . . , f i-1 , f i+1 , . . . , f s } is also a basis of R. Furthermore,
by applying the division process of Proposition 2.6 to f i -M o (f i ), we can always construct a reduced basis of R.

Corollary 2.11. The algebra R has a unique minimal reduced basis.

Proof. Let {f 1 , . . . , f s } and {g 1 , . . . , g s } be two minimal reduced bases of R. Hence s is the embedding dimension of o(R), and the same holds for s ; whence they are equal. Let i = 1. There exists j 1 such that o(f 1 ) = o(g j 1 ), because minimal generating systems of numerical semigroups are unique. If

f 1 -g j 1 = 0, then o(f 1 -g j 1 ) / ∈ o(R) (the basis is reduced), which is a contradiction because f 1 -g j 1 ∈ R. The same argument shows that {f 1 , . . . , f s } = {g 1 , . . . , g s } Remark 2.12. Let R = K f 1 , . . . , f s and assume that f i is monic for all 1 ≤ i ≤ s. Also assume that o(f 1 ) ≤ o(f 2 ) ≤ . . . ≤ o(f s ). Set n = o(f 1 ) and let f 1 = x n + i>n c 1 i x i
. By an analytic change of variables, we may assume that f 1 = x n , hence, up to an analytic isomorphism, we may assume that R = K x n , f 2 , . . . , f s . In particular, we may assume that R has a minimal reduced basis of the form x n , g 2 (x), . . . , g s (x).

Example 2.13. Let R = K x 4 + x 5 , x 6 , x 15 + x 16 + n≥20 x n , with K a field of characteristic zero. Then R is a one-dimensional ring. Since the conductor of 4, 6, 15 is 18, we have that λ R (K X /R) < ∞ and we know by Propsition 2.5 that R = K x 4 + x 5 , x 6 , x 15 + x 16 . Let us denote x 4 + x 5 by f 1 , x 6 by f 2 and

x 15 + x 16 by f 3 . The kernel of φ : K[X 1 , X 2 , X 3 ] -→ K[x], φ(X 1 ) = x 4 , φ(X 2 ) = x 6 and φ(X 3 ) = x 15 is generated by X 3 1 -X 2 2 , X 2 3 -X 5 2
, hence we get S 1 = x 13 +x 14 + 1 3 x 15 and S 2 = x 31 + 1 2 x 32 . As 13 ∈ 4, 6, 15 , we add it as f 4 = S 1 . We do not care about S 2 , because the conductor of 4, 6, 15 is 18. Now, the conductor of 4, 6, 13, 15 is 12. If we compute a system of generators of the kernel of 15 and φ(X 4 ) = x 13 , then all the elements S i have orders greater than 12, and so the algorithm ends. We conclude that o(R) = 4, 6, 13, 15 .

ϕ : K[X 1 , X 2 , X 3 , X 4 ] -→ K[x], φ(X 1 ) = x 4 , φ(X 2 ) = x 6 , φ(X 3 ) = x
We have implemented this algorithm in the numericalsgps ( [START_REF] Delgado | NumericalSgps", A GAP package for numerical semigroups[END_REF]) GAP ( [START_REF]GAP -Groups, Algorithms, and Programming[END_REF]) package. Next we illustrate how to compute this semigroup with the functions we have implemented (that will be available in the next release of the package). Proposition 1]). The integer λ R ( R/R) is the degree of singularity of R, and this is why the genus of o(R) is called by some authors the degree of singularity of the semigroup (see [START_REF] Barucci | Properties in numerical semigroups and applications to one-dimensional analytically irreducible local domains[END_REF][START_REF] Matsuoka | On the degree of singularity of one-dimensional analytically irreducible noetherian local rings[END_REF]). In the setting of Weirstrass numerical semigroups the genus coincides with the geometrical genus of the curve used to define the semigroup ( [START_REF] Del Centina | Weierstrass points and their impact in the study of algebraic curves: a historical account from the "Lückensatz" to the 1970s[END_REF]).

(R) is 12. The conductor of R in R is precisely (x c(o(R)) ) ([3]). We have that λ R ( R/R) = |[0, c(o(R)) -1] ∩ (N \ o(R))| = 7 counts the number of gaps of o(R) ([13,
The number of sporadic elements of o Example 2.15. Let R = K x 4 , x 6 + x 7 , x 13 + a 14 x 14 + a 15 x 15 + . . . with K a field. Using the same argument as in the Example 2.13, we find that if char K = 2, we have that if a 15 -a 14 + 1/2 = 0, then {x 4 , x 6 + x 7 , x 13 } is the reduced basis of R. Furthermore, since 4, 6, 13 is a symmetric numerical semigroup (the number of nonnegative integers not in the semigroup equals the conductor divided by two), then, by [START_REF] Kunz | The value-semigroup of a one-dimensional Gorenstein ring[END_REF], R is Gorenstein. Finally λ R ( R/R) = 8. Otherwise if a 15 -a 14 + 1/2 = 0, then {x 4 , x 6 + x 7 , x 13 , x 15 } is the reduced basis of R with R a non Gorenstein ring. Furthermore λ R ( R/R) = 7.

(R) is λ R (R/(R : R)) = |[0, c(o(R)) -1] ∩ o(R)| = 5
Otherwise, if char K = 2, then the reduced basis of R is {x 4 , x 6 + x 7 , x Remark 2.18. We do not know a priori if R = K[[x]] and we do not have a general procedure to check it. If the algorithm is called with such an R, it will eventually not stop.

Deformation to a toric ideal

Let the notations be as in Section 2. Given

f (x) = i≥p c i x i ∈ K x , we set H f (u, x) = i≥p c i u i-p x i . In particular, if we consider the linear form L : N 2 -→ N, L(a, b) = b -a, then H f is L-homogeneous of order p, that is, L(i -p, i) = p for all i ∈ supp(f ). We set H R = K u, H f | f ∈ R .
With these notations we have the following.

Proposition 3.1. The set {f 1 , . . . , f s } is a basis of R if and only if H R = K u, H f 1 , . . . , H fs . Proof. Note first that if H g ∈ H R for some g ∈ K x , then g ∈ R. Suppose that {f 1 , . . . , f s } is a basis of R and let f (x) ∈ R. Write H f (u, x) = i≥p c i u i-p x i . We have M o (f ) = c p x p = c p s i=1 M o (f i ) p k i , hence H f -c p s i=1 H p k i f i = u q H f 1
with f 1 ∈ R and either f 1 = 0, or o(f 1 ) > p. In the second case we restart with f 1 . A similar argument as in Proposition 2.6 proves our assertion. Conversely, suppose that H R = K u, H f 1 , . . . , H fs and let f ∈ R. Let P (X 0 , X 1 , . . . , X s ) ∈ K X 0 , X 1 , . . . , X s such that H f = P (u, H f 1 , . . . , H fs ). Setting u = 0, we get that M o (f

) = P (M o (f 1 ), . . . , M o (f s )) ∈ K M o (f 1 ), . . . , M o (f s ) . Hence M o (f ) ∈ K[M o (f 1 ), . . . , M o (f s )]. Suppose that {f 1 , . . . , f s } is a basis of R. Then T = K u H f 1 , . . . , H fs is a K u -module. When u = 1 (respectively u = 0), we get T | u=1 = R (respectively T | u=0 = K M (f 1 ), . . . , M (f s ) ). Hence we get a deformation from R to K M o (f 1 ), . . . , M o (f s ) . More precisely let ψ : K X 1 , . . . , X s -→ R = K f 1 , . . . , f s and H ψ : K u X 1 , . . . , X s -→ T = K u H f 1 , .
. . , H fs be the morphisms of rings such that H ψ (u) = u, ψ(X i ) = f i and H ψ (X i ) = H f i for all i ∈ {1, . . . , s}. Let, as in Section 2,

S i = f α i 1 1 • • • f α i s s -f β i 1 1 • • • f β i s s = θ i c i θ i f θ i 1 1 • • • f θ i s s , 1 ≤ i ≤ r with o(f θ i 1 1 • • • f θ i s s ) = D i θ i > s k=1 α i k o(f k ) = s k=1 β i k o(f k ) = p i . Let I (respectively J) be the ideal generated by (G i = X α i 1 1 • • • X α i s s -X β i 1 1 • • • X β i s s -θ i c i θ i X θ i 1 1 • • • X θ i s s ) 1≤i≤r (respectively (H i = X α i 1 1 • • • X α i s s - X β i 1 1 • • • X β i s s -θ i u D i θ i -p i c i θ i X θ i 1 1 • • • X θ i s s ) 1≤i≤r ) in K X 1 , . . . , X s (respectively K u X 1 , . . . , X s ). Well shall consider on N s (respectively, N s+1 ) the linear form O(θ 1 , . . . , θ s ) = s i=1 θ i o(f i ) (respectively O h (θ 0 , θ 1 , . . . , θ s ) = -θ 0 + s i=1 θ i o(f i )). Given a monomial X θ 1 1 • • • X θs s (respectively u θ 0 X θ 1 1 • • • X θs s ), we set O(X θ 1 1 • • • X θs s ) = O(θ 1 , . . . , θ s ) (respectively O h (u θ 0 X θ 1 1 • • • X θs s ) = O h (θ 0 , θ 1 , . . . , θ s )). For G = θ c θ X θ 1 1 • • • X θs s (respectively H = θ c θ u θ 0 X θ 1 1 • • • X θs s ), we say that G (respectively H) is O-homogeneous of order a (respectively O h -homogeneous of order b) if O(θ 1 , . . . , θ s ) = a (respectively O h (θ 0 , θ 1 , . . . , θ s ) = b
) for all (θ 1 , . . . , θ s ) (respectively (θ 0 , θ 1 , . . . , θ s )) such that c θ = 0. More generally let G = k≥0 G p k where p 0 < p 1 < • • • and G p k is O-homogeneous of order p k . We set O(G) = p 0 . We also set in(G) = G p 0 and we call it the initial form of G.

We finally set O(0) = +∞, and we recall that O(G) = +∞ if and only if G = 0.

Lemma 3.2. With the standing notations and hypothesis, the kernel of ψ is generated by I.

Proof. Let, as in Section 2, F 1 , . . . , F r be a generating system of the kernel of the morphism

φ : K X 1 , . . . , X s -→ K x , φ(X i ) = M o (f i )
for all i ∈ {1, . . . , s}. In particular F i is O-homogeneous of order o(f i ) for all i ∈ {1, . . . , s}, and

K X 1 , . . . , X s /(F 1 , . . . , F r ) K M o (f 1 ), . . . , M o (f s ) .
For all i ∈ {1, . . . , r}, ψ(G i ) = 0. Hence I ⊆ ker(ψ).

For the other inclusion, let

G = θ c θ X θ 1 1 • • • X θs s ∈ ker(ψ). Write G = k≥0 c θ k X k,O(θ k )=O(θ 0 ) c θ k f θ k 1 1 • • • f θ k s s = r i=1 λ 0 i (f 1 , . . . , f s )S i . Let G 1 = G -r i=1 λ 0 i G i . We have G 1 ∈ ker(ψ). If G 1 = 0, then O(G) < O(G 1
). Then we restart with G 1 . We construct in the same way G 2 , and λ 1 1 , . . . , λ 1 r such that

G 1 = G 2 + r j=1 λ 1 j G j with O(G) < O(G 1 ) < O(G 2 ), λ 1 i O-homogeneous and O(λ 0 i ) < O(λ 1 i
) for all i ∈ {1, . . . , r}. If we continue in this way, we get that for all k ≥ 0,

G = G k+1 + r i=1 (λ 0 i + λ 1 i + . . . + λ k i )G i , with O(G) < O(G 1 ) < . . . < O(G k+1 ), λ j i O-homogeneous, and O(λ 0 i ) < O(λ 1 i ) < • • • < O(λ k i
) for all i ∈ {1, . . . , r} and for all j ∈ {1, . . . , k}. If G k = 0 for some k, then we are done. Otherwise, let λ i = ∞ k=0 λ k i , and let Ḡ = lim k→+∞ G k . We have Ḡ = 0 and G = r i=1 λ i G i . This proves our assertion.

Let the notations be as above. Let

G = θ c θ X θ 1 1 • • • X θs s ∈ K X 1 , . . . , X s and write G = i≥0 G p i with p 0 < p 1 < • • • and G p i O-homogeneous. We set H G = i≥0 u p i -p 0 G p i , in such a way that H G is O h -homogeneous of order p 0 . Given an ideal S of K X 1 , . . . , X s , we set in(S) = (in(G) | G ∈ S \ {0}).
We also denote by

H S = (H G | G ∈ S \ {0})K u, X 1 , . . . , X s .
With these notations we have in(S i ) = F i and H G i = H i for all i ∈ {1, . . . , r}. Lemma 3.3. Let the notations be as above. We have in(I) = (F 1 , . . . , F r ), and

H I = (H 1 , . . . , H r ) = J.
Proof. The first assertion follows from the proof of Lemma 3.2. To prove the second assertion, let H ∈ H I and assume that H is O h -homogeneous. We have

G = H(1, X 1 , . . . , H s ) ∈ I. Furthermore, H = u e H G for some e ≥ 0. Write H G = in(G) + H 1 where H 1 (0, X 1 , . . . , X s ) = 0. We have in(G) = r i=1 λ i F i where λ i is O-homogeneous of order p 0 -O(F i ). Let H G 1 = H G -r i=1 λ i H G i = H G -r i=1 λ i H i . Then H G 1 ∈ H I is O h -homogeneous and O h (H G ) < O h (H G 1 ). Now we restart with H G 1 . We prove in this way that H ∈ (H 1 , . . . , H r ). Let H = θ c θ u θ X θ 1 1 • • • X θs s ∈ ker(H ψ ). Write H = k H k , where H k is O h -homogeneous. For all k, we have H ψ (H k ) = 0. Setting G k = H k (1, X 1 , .
. . , X s ), we have ψ(G k ) = 0. This implies that G k ∈ I. Hence H G k ∈ (H 1 , . . . , H r ) by Lemma 3.3. But H k = u e k H G k for some e k ∈ N. Consequently H k ∈ (H 1 , . . . , H r ). Finally H ∈ (H 1 , . . . , H r ), which proves that ker(H ψ ) ⊆ J. As the inclusion J ⊆ ker(H ψ ) is obvious, we conclude that J = ker(H ψ ). Now the morphism K u -→ K u X 1 , . . . , X s /J is flat because u is not a zero divisor. Hence we get a family of formal space curves parametrized by u that gives us a deformation from K X 1 , . . . , X r /I to K X 1 , . . . , X r /(F 1 , . . . , F r ).

In particular we get the following.

Theorem 3.4. Every formal space curve of K l , parametrized by Y 1 = g 1 (x), . . . , Y l = g l (x) has a deformation into a formal monomial curve of K r for some positive integer r.

Basis of K f (x), g(x)

In this section we study the particular case of a subalgebra R of K[[x]] generated by two elements, and see that a different approach can be considered to study o(R), with some interesting applications.

Let f (x) = i≥n a i x i and g(x) = j≥m b j x j be two elements of K x and suppose, without loss of generality, that the following conditions hold:

(1)

a n = b m = 1, (2) n ≤ m, (3)
the greatest common divisor of supp(f (x)) ∪ supp(g(x)) is equal to 1 (in particular for all d > 1, f (x), g(x) / ∈ K x d ). Let the notations be as in Section 2, in particular R = K f, g . By the analytic change of variables f (x) = xn , we may assume that R = K x n , g(x) . Let F (X, Y ) be the x-resultant of X -x n , Y -g(x), that is, F (X, Y ) is the generator of the kernel of the map ρ : K X, Y -→ K x , ρ(X) = x n and ρ(Y ) = g(x).

Since K f, g = K f, g -f k for all k ≥ 1, then we shall assume that n < m and also that n does not divide m. Given a nonzero element G(X, Y ) / ∈ (F (X, Y ))K X, Y , we set int(F, G) = o(G(f (x), g(x))). Condition (3) implies that the set of int(F, G), G(X, Y ) / ∈ (F (X, Y ))K X, Y , is a numerical semigroup. We denote it by Γ(F ). We have the following.

Proposition 4.1. o(R) = Γ(F ).

Proof. We have a ∈ Γ(F ) if and only if a = o(G(f (x), g(x))) for some G(X, Y ) ∈ K X, Y if and only if a ∈ o(R).

Suppose that K is algebraically closed with characteristic zero, and let

d 1 = n, m 1 = inf{i ∈ supp(g) | d 1 i}, that is, m 1 = m,
≤ k ≤ h, r k = r k-1 e k-1 + m k -m k-1 .
With these notations we have the following (see [START_REF] Abhyankar | On the semigroup of a meromorphic plane curve, Part I[END_REF]).

(1) Γ(F ) = o(R) is generated by {r 0 , r 1 , . . . , r h }.

(2) r k d k < r k+1 d k+1 for all k ∈ {1, . . . , h -1}.

(3) Γ(F ) = o(R) is free with respect to the arrangement (r 0 , . . . , r h ). More precisely, let

e k = d k d k+1 for all k ∈ {1, . . . , h}. Then e k r k ∈ r 0 , . . . , r k-1 . (4) C = h k=1 k -1)r k -n + 1 is the conductor of Γ(F ) = o(R). Example 4.2. Let f = x 7 and g = x 4 + x 2 . The above resultant is then F = y 7 -7x 2 y 3 -x 4 -14x 2 y 2 - 7x 2 y -x 2 . Then Γ(F ) = o(R) = 2,
k ≥ 2, let G k (X, Y ) ∈ K X, Y such that o(G k (x n , g(x))) = r k . It follows from [1] that deg Y G k = n d k . If g k (x) = G k (x n , g(x)
), then we have the following. Proposition 4.3. The set {x n , g, g 2 , . . . , g h } is a basis of R, that is, R = K x n , g, g 2 , . . . , g h and M o (R) = K x n , x m , x r 2 , . . . , x r h . Note that, by a similar argument as in Section 2, we may assume that f = x n , g = x m + i∈G(Γ(F )) c 1 i x i , and for all k ≥ 2,

g k = x r k + i∈G(Γ(F )) c k i x i , where G(Γ(F )) = {j ∈ N | j / ∈ Γ(F )} is the set of gaps of Γ(F ).
Let the notations be as in Section 3. The morphism

D : K u -→ T = K u H f , H g , H g 2 , . . . , H g h gives us a deformation of T | u=1 = R = K f (x), g(x), g 2 (x), . . . , g h (x) to T | u=0 = K x n , x m , x r 2 , . . . , x r h .
Note that, since n, m, r 2 , . . . , r h is free with respect to the given arrangement, then it is a complete intersection (see for instance [START_REF] Rosales | Numerical semigroups[END_REF]). For all k ∈ {1, . . . , h}, write

e k r k = k-1 i=0 θ k i r i with 0 ≤ θ k i < e i for all i ∈ {1, . . . , k -1}. If B is the ideal of K[X 0 , X 1 , . . . , X h ] generated by {X e 2 1 -X m d 2 0 , X e 2 2 -X θ 2 0 0 X θ 2 1 1 , . . . , X e h h -X θ h 0 0 X θ h 1 1 . . . X θ h h-1 h-1 } then K x n , x m , x r 2 , . . . , x r h K X 0 , X 1 , . . . , X h /B.
Let F (X, Y ) be the x-resultant of X -x n , y -g(x). By hypothesis, F (X, Y ) is a polynomial. Furthermore,

F (X, Y ) = Y n + a 1 (X)Y n-1 + . . . + a n (X) with o(a i (X)) > i for all 2 ≤ i ≤ n. Set G h+1 = F and for all k ≥ 1, let G k+1 = G e k k -X θ k 0 k-1 i=1 G θ k i i + α k c k α k X α k 0 G α k 1 1 • • • G α k k k ,
where the following conditions hold:

(1) for all i ∈ {1, . . . , k -1}, 0 ≤ θ k i < e i ; (2) for all α k , if c k α k = 0, then for all i ∈ {1, . . . , k}, 0

≤ α k i < e i ; (3) for all α k , if c k α k = 0, then α k 0 n + k i=1 α k i r i = D k i > e k r k = θ k 0 r 0 + k-1 i=1 θ k i r i . It follows from Section 3. that if I (respectively is the ideal generated by (X e k k -X θ k 0 0 k-1 i=1 X θ k i i + α k c k α k X α k 0 0 X α k 1 1 • • • X α k k k ) 1≤k≤h (respectively (X e k k -X θ k 0 0 k-1 i=1 X θ k i i + α k c k α k u D k i -e k r k X α k 0 0 X α k 1 1 • • • X α k k k ) 1≤k≤h ) in K X 0 , . . . , X h (respec- tively K u X 0 , . . . , X h ), then R = K x n , g(x), g 2 (x), . . . , g h (x) K X 0 , X 1 , . . . , X h /I and K u x n , H g , H g 2 , . . . , H g h K u X 0 , X 1 , . . . , X h /J. Furthermore, K u X 0 , X 1 , . . . , X h /J is a flat K u -module.
This gives us a family of formal space curves parametrized by u which is a deformation from K X 0 , X 1 , . . . , X h /I to the formal toric variety K X 0 , X 1 , . . . , X h /B. The later being a complete intersection, we get the following. Theorem 4.4. Every irreducible singularity of a plane curve X = f (x), Y = g(x) of K 2 has a deformation into a formal monomial complete intersection curve of K h+1 for some h ≥ 1.

Example 4.5. Let f (x) = x 4 , g(x) = x 6 + x 7 . The minimal polynomial of (f (x), g(x)) is given by:

F (X, Y ) = Y 4 -2X 3 Y 2 + X 6 -4X 5 Y -X 7 = (Y 2 -X 3 ) 2 -4X 5 Y -X 7 Let r 0 = 4 = d 1 , r 1 = 6 = m 1 and G 1 = Y . We have d 2 = gcd(6, 4) = 2, hence m 2 = 7. It follows that r 2 = 13. Note that if G 2 = Y 2 -X 3 , then g 2 (x) = G 2 (f (x), g(x)) = 2x 13 + x 14 . Hence Γ(F ) = o(R) = 4, 6, 13 and {f (x), g(x), g 2 (x)} is a basis of R. Let us double check it. gap> SemigroupOfValuesOfCurve_Local([x^4,x^6+x^7],"basis"); [ x^4, x^7+x^6, -1/2*x^15+x^13 ]
(Observe that the output is different, since this is a reduced basis: we change 2x 13 + x 4 with x 13 + 1 2 x 14 , and then using that 14 = 2 × 4 + 6, we replace this last polynomial with x 13 -1 2 x 15 .) Consequently, H R = K u, x 4 , x 6 + ux 7 , 2x 13 + ux 14 . With the notations above,

e 1 = 3, e 2 = 2, hence K x 4 , x 6 , x 13 T = K X 0 , X 1 , X 2 /(X 2 1 -X 3 0 , X 2 2 -X 5 0 X 1 ), and K u -→ K u X 0 , X 1 , X 2 /(X 2 1 -X 3 0 , X 2 2 -4X 5 0 X 1 -u 2 X 7 0 )
gives us a deformation from R to T (we can also change X 2 with 1 2 X 2 , and then

B = (X 2 1 -X 3 0 , X 2 2 - 4X 4 0 X 1 )).

Semigroup of a polynomial curve

Let K be a field and let f 1 (x), . . . , f s (x) be s polynomials of K

[x]. Let A = K[f 1 , . . . , f s ] be a subalgebra of K[x],
and assume, without loss of generality, that f i is monic for all i ∈ {1, . . . , s}. Given f (x) = p i=0 c i x i ∈ A, with c p = 0, we set d(f ) = p and M(f ) = c p x p , the degree and leading monomial, respectively. We also define supp

(f ) = {i | c i = 0}. The set d(A) = {d(f ) | f ∈ A} is a submonoid of N. We shall assume that λ A (K[x]/A) < ∞. In particular d(A) is a numerical semigroup. We say that {f 1 , . . . , f s } is a basis of A if {d(f 1 ), . . . , d(f s )} generates d(A). Clearly, {f 1 , . . . , f s } is a basis of A if and only if K[M (f ), f ∈ A] = K[M (f 1 ), . . . , M (f s )].
For several variables, these basis are known in the literature as SAGBI basis ( [START_REF] Robbiano | Subalgebra bases[END_REF][START_REF] Bravo | Some facts about canonical subalgebra bases[END_REF]). Since there are already algorithms in the literature to calculate a basis of A, we will not include the procedure here.

We would like just mention that if we follow a similar argument to the one used in Section 2, the sequences of degrees decrease, and thus the finiteness conditions are easier to deduce. In this setting a basis for A is unique up to constants.

Deformation to a toric ideal

Let the notations be as in Section 5. Given

f (x) = p i=0 c i x i ∈ K[x], we set h f (u, x) = p i=0 c i u p-i x i , in particular, if we consider the linear form L h : N 2 -→ N, L(a, b) = a + b, then h f is L h -homogeneous of degree p, that is, L h (i, p -i) = p for all i ∈ Supp(f ) . We set h A = K[u, h f | f ∈ A].
With these notations we have the following result, and its proof is similar to that of Proposition 3.1.

Proposition 6.1. The set {f 1 , . . . , f r } is a basis of A if and only if h A = K[u, h f 1 , . . . , h fs ].
Suppose that {f 1 , . . . , f s } is a basis of A. By the inclusion morphism of rings D :

K[u] -→ B = K[u, h f 1 , . . . , h fs ], B is a K[u]-module. When u = 1 (respectively u = 0), we get B | u=1 = A (respec- tively B | u=0 = K[M(f 1 ), . . . , M(f s )]). Hence we get a deformation from A to K[M(f 1 ), . . . , M(f s )]. More precisely let ψ : K[X 1 , . . . , X s ] -→ K[f 1 , . . . , f s ] and h ψ : K[u][X 1 , . . . , X s ] -→ K[u][h f 1 , . . . , h fs ]
be the morphisms of rings such that h ψ (u) = u, ψ(X i ) = f i and h ψ (X i ) = h f i for all i = 1, . . . , s. For all i = 1, . . . , r, let

S i = f α i 1 1 • • • f α i s s -f β i 1 1 • • • f β i s s = θ i c i θ i f θ i 1 1 • • • f θ i s s with d(f θ i 1 1 • • • f θ i s s ) = D i θ i > s k=1 α i k d(f k ) = s k=1 β i k d(f k ) = p i . Let I (respectively J) be the ideal generated by (G i = X α i 1 1 • • • X α i s s -X β i 1 1 • • • X β i s s -θ i c i θ i X θ i 1 1 • • • X θ i s s ) 1≤i≤r (respectively (H i = X α i 1 1 • • • X α i s s - X β i 1 1 • • • X β i s s -θ i u p i -D i θ i c i θ i X θ i 1 1 • • • X θ i s s ) 1≤i≤r ) in K[X 1 , . . . , X s ] (respectively K[u][X 1 , . . . , X s ]
). Well shall consider on N s (respectively, N s+1 ) the linear form

D(θ 1 , . . . , θ s ) = s i=1 θ i d(f i ) (respectively D h (θ 0 , θ 1 , . . . , θ s ) = θ 0 + s i=1 θ i d(f i )). Given a monomial X θ 1 1 • • • X θs s (respectively u θ 0 X θ 1 1 • • • X θs s ), we set D(X θ 1 1 • • • X θs s ) = D(θ 1 , . . . , θ s ) (respectively D h (u θ 0 X θ 1 1 • • • X θs s ) = D h (θ 0 , θ 1 , . . . , θ s )). For G = θ c θ X θ 1 1 • • • X θs s (respectively H = θ c θ u θ 0 X θ 1 1 • • • X θs s )
, we say that G (respectively H) is D-homogeneous of degree a (respectively D h -homogeneous of degree b) if D(θ 1 , . . . , θ s ) = a (respectively D h (θ 0 , θ 1 , . . . , θ s ) = b) for all (θ 1 , . . . , θ s ) (respectively (θ 0 , θ 1 , . . . , θ s )) such that c θ = 0. More generally let G = m k=0 G p k where p 0 > p 1 > • • • > p m and G p k is D-homogeneous of degree p k . We set D(G) = p 0 . We also set In(G) = G p 0 and we call it the initial form of G. Lemma 6.2. With the standing notations and hypothesis, the kernel of ψ is generated by I.

Proof. Let F 1 , . . . , F r be a generating system of the kernel of the morphism

φ : K[X 1 , . . . , X s ] -→ K[x], φ(X i ) = M(f i )
for all i ∈ {1, . . . , s}. In particular F i is D-homogeneous of degree d(f i ) for all i ∈ {1, . . . , s}, and

K[X 1 , . . . , X s ]/(F 1 , . . . , F r ) K[M(f 1 ), . . . , M(f s )].
For all i ∈ {1, . . . , r}, ψ(G i ) = 0. Hence I ⊆ ker(ψ). For the other inclusion, let G

= θ c θ X θ 1 1 • • • X θs s ∈ ker(ψ). Write G = m k=0 c θ k X Let G 1 = G -r i=1 λ 0 i G i . It follows that G 1 ∈ ker(ψ). If G 1 = 0, then D(G) > D(G 1
). Then we restart with G 1 . We construct in the same way G 2 , and λ 1 1 , . . . , λ 1 r such that

G 1 = G 2 + r j=1 λ 1 j G j with D(G) > D(G 1 ) > D(G 2 ), λ 1
i D-homogeneous and D(λ 0 i ) > D(λ 1 i ) for all i ∈ {1, . . . , r}. If we continue in this way, we get that for all k ≥ 0,

G = G k+1 + r i=1 (λ 0 i + λ 1 i + . . . + λ k i )G i , with D(G) > D(G 1 ) > • • • > D(G k+1 ), λ j i D-homogeneous, and D(λ 0 i ) > D(λ 1 i ) > • • • > D(λ k i
) for all i ∈ {1, . . . , r} and for all j ∈ {1, . . . , k}. Thus, there exists l such that G l+1 = 0. Hence G = r i=1 (λ 0 i + λ 1 i + . . . + λ l i )G i . This proves our assertion. Let the notations be as above.

Let G = θ c θ X θ 1 1 • • • X θs s ∈ K[X 1 , .
. . , X s and write G = m i=0 G p i with p 0 > p 1 > • • • > p m and G p i D-homogeneous. We set H G = m i=0 u p 0 -p i G p i , in such a way that H G is D h -homogeneous of degree p 0 . Given an ideal S of K[X 1 , . . . , X s ], we set In(S) = (In(G) | G ∈ S \ {0}). We also denote by

H S = (H G | G ∈ S \ {0})K[u, X 1 , . . . , X s ].
With these notations we have In(S i ) = F i and H G i = H i for all i ∈ {1, . . . , r}. Lemma 6.3. Let the notations be as above. We have In(I) = (F 1 , . . . , F r ), and H I = (H 1 , . . . , H r ) = J.

Proof. The first assertion follows from the proof of Lemma 6.2.

To prove the second assertion, let H ∈ H I and assume that H is D h -homogeneous. We have

G = H(1, X 1 , . . . , H s ) ∈ I. Furthermore, H = u e H G for some e ≥ 0. Write H G = In(G) + H 1 where H 1 (0, X 1 , . . . , X s ) = 0. We have In(G) = r i=1 λ i F i where λ i is D-homogeneous of degree D(G) -D(F i ). Let H G 1 = H G -r i=1 λ i H G i = H G -r i=1 λ i H i . Then H G 1 ∈ H I is D h -homogeneous and D h (H G ) > D h (H G 1
). Now we restart with H G 1 . In this way we show that H ∈ (H 1 , . . . , H r ).

Let H = θ c θ u θ X θ 1 1 • • • X θs s ∈ ker(h ψ ). Write H = n k=0 H k where H k is D h -homogeneous. For all k, we have h ψ (H k ) = 0. Setting G k = H k (1, X 1 , . . . , X s ), we have ψ(G k ) = 0. This implies that G k ∈ I, and thus H G k ∈ (H 1 , . . . , H r ) by Proposition 6.3. But H k = u e k H G k for some e k ∈ N, whence H k ∈ (H 1 , . . . , H r )
. Finally H ∈ (H 1 , . . . , H r ), which proves that ker(h ψ ) ⊆ J. The inclusion J ⊆ ker(h ψ ) is obvious, and we can conclude that J = ker(h ψ ). Now the morphism

K[u] -→ K[u][X 1 , .
. . , X r ]/J is flat (because p(u) is not a zero divisor for all p(u) ∈ K[u]). Hence we get a family of polynomial space curves parametrized by u which gives us a deformation from K[X 1 , . . . , X r ]/I to K[X 1 , . . . , X r ]/(F 1 , . . . , F r ).

In particular, we get the following analogue to Theorem 3.4, which can be seen as a a geometric reinterpretation of [START_REF] Sturmfels | Gröbner bases and convex polytopes[END_REF]Corollary 11.6] (also [4, Corollary 6.1]). Proof. If gcd(n, m) = 1, then Γ(F ) = d(A) = n, m . Hence {f (x), g(x)} is a basis of A. Lemma 7.5. Suppose that gcd(n, m) = p 1 • • • p l where p i is a positive prime number for all i ∈ {1, . . . , l} (and the p i 's are not necessarily distinct). The set {f (x), g(x)} is not a basis of A. Furthermore, if c is the cardinality of a basis of A, then 2 ≤ c ≤ l + 2. In particular, if gcd(n, m) is a prime number p > 1, then a basis of A has either two or three elements.

d 1 > d 2 > . . . > d h+1 = 1; (2) r k d k > r k-1 d k-1 for all k ∈ {1, . . . , h}; (3) e k r k ∈< r 0 , . . . , r k-1 > for all k = 1, . . . , h.
Such a sequence is called a δ-sequence and it is well known (see [START_REF] Abhyankar | On the semigroup of a meromorphic plane curve, Part I[END_REF]) that there exists a polynomial F (X, Y ) with one place at infinity such that the semigroup {rank K K[X, Y ]/( F , G), G / ∈ (F )} is generated by r. It follows from Theorem 7.1. that a polynomial curve has one place at infinity. The converse is not true in general. Abhyankar asked whether every semigroup generated by a δ-sequence (hence the semigroup of a curve with one place at infinity) is the semigroup of a polynomial curve (for example, the δ-sequence [START_REF]GAP -Groups, Algorithms, and Programming[END_REF][START_REF] Bravo | Some facts about canonical subalgebra bases[END_REF][START_REF] Castellanos | Algorithm for the Semigroup of a Space Curve Singularity[END_REF] generates the semigroup 4, 5 which is the semigroup of the polynomial curve A = K[x 4 , x 5 ]). It has been proved recently that the answer is no ( [START_REF] Fujimoto | On polynomial curves in the affine plane[END_REF]). It would be nice to see which supplementary conditions a δ-sequence should satisfy in order to generate the semigroup of a polynomial curve. Remark 7.7. Let f (x) and g(x) be as above, and let A = K[f (x), g(x)]. Let also F (X, Y ) be the xresultant of X -f (x) and Y -g(x). Let r 0 = n, r 1 = m, r 2 , . . . , r h be the generators of Γ(F ) calculated as above. Let 1 ≤ k ≤ h and let G k (X, Y ) = App(F, d k ). We have d(G k (f (x), g(x)) = r k , but G k is not the unique polynomial with this condition (for example, d((G k + λ)(f (x), g(x))) = r k for all λ = 0). Hence it is natural to ask the following: is there a polynomial G(X, Y ) (of degree < n in Y ) such that G is parametrized by polynomials in x such that d(G(f (x), g(x)) = r k ? Such a polynomial, if it exists, should be of degree n d k and should have the contact with F at a characteristic exponent of F (see [START_REF] Abhyankar | On the semigroup of a meromorphic plane curve, Part I[END_REF] for the definition of the characteristic exponents of a curve with one place at infinity and the notion of contact). Hence the existence of such a polynomial implies that a polynomial curve can be approximated by polynomial curves.

Let the notations be as above, in particular

F (X, Y ) = Y n + c 1 (X)Y n-1 + • • • + c n (X) is the x- resultant of (X -f (x), Y -g(x)). Let G 1 = Y, G 2 , .
. . , G h be the set of approximate roots of F (X, Y ) constructed algorithmically as above. In particular r 0 = n, r 1 = m, r 2 = int(F, G 2 ), . . . , r h = int(F, G h ) generate d(A). For all k = 2, . . . , h, let g k (x) = G k (f (x), g(x)) and let M(g k ) = b r k x r k . We have A = K[f (x), g(x), g 2 (x), . . . , g h (x)]. Furthermore, the map

D : K[u] -→ B = K[u][h f , h g , h g 2 , . . . , h g h ]
introduced in Section 6. gives us a deformation of the polynomial curve B | u=1 = A into B | u=0 = K[t n , t m , t r 2 , . . . , t r h ]. Note that, since n, m, r 2 , . . . , r h is free with respect to the given arrangement, then it is a complete intersection. For all k ∈ {1, . . . , h}, write e k r k = k-1 i=0 θ k i r i with 0 ≤ θ k i < e i for every i ∈ {1, . . . , k -1}. With the notations above, if T is the ideal of K[X 0 , X 1 , . . . , X h ] generated by Set G h+1 = F and for all k ≥ 1, let

G k+1 = G e k k -X θ k 0 k-1 i=1 G θ k i i + α k c k α k X α k 0 .G α k 1 1 • • • G α k k k ,
where the following conditions hold:

(1) for all i ∈ {1, . . . , k -1}, 0 ≤ θ k i < e i ; (2) for all α k , if c k α k = 0, then for all i ∈ {1, . . . , k}, 0 ≤ α k i < e i , (3) for all α k , if c k α k = 0, then α k 0 n + k i=1 α k i r i = D k i < e k r k = θ k 0 r 0 + k-1 i=1 θ k i r i .

It follows from Section 6 that if I (respectively J) is the ideal generated by (X e k k -X

θ k 0 0 k-1 i=1 X θ k i i + α k c k α k X α k 0 0 .X α k 1 1 • • • X α k k k ) 1≤k≤h (respectively (X e k k -X θ k 0 0 k-1 i=1 X θ k i i + α k c k α k u e k r k -D k i X α k 0 0 .X α k 1 1 • • • X α k k k ) 1≤k≤h
) in K[X 0 , . . . , X h ] (respectively K[u][X 0 , . . . , X h ]), then A = K[x n , g(x), g 2 (x), . . . , g h (x)] K[X 0 , X 1 , . . . , X h ]/I and K[u][x n , h g(x) , h g 2 (x) , . . . , h g h (x) ] K[u][X 0 , X 1 , . . . , X h ]/J. Furthermore, K[u][X 0 , X 1 , . . . , X h ]/J is a flat K[u]-module. This gives us a family of space curves parametrized by u which is a deformation from K[X 0 , X 1 , . . . , X h ]/I to the toric variety K[X 0 , X 1 , . . . , X h ]/T . The later being a complete intersection, we get the following result.

Theorem 7.8. Every polynomial curve X = f (x), Y = g(x) of K 2 has a deformation into a monomial complete intersection curve of K h+1 for some positive integer h. Example 7.9. Let f (x) = x 6 + x 3 , g(X) = x 4 . The minimal polynomial of (f (x), g(x)) is given by:

F (X, Y ) = Y 6 -2X 2 Y 3 -4XY 3 -Y 3 + X 4 .
Let r 0 = 6 = d 1 , r 1 = 4 and G 1 = Y . We have d 2 = gcd(6, 4) = 2, and G 2 = App(F, 2) = Y 3 -X 2 -2X -1 2 . Since g 2 (x) = G 2 (f (x), g(x)) = -2x 9 -3x 6 -2x 3 -1 2 , then r 2 = 9 and d 3 = 1, hence Γ(F ) = d(A) = 6, 4, 9 and {f (x), g(x), -g 2 (x) =} is a basis of A. Consequently, h A = K[u, x 6 + u 3 x 3 , x 4 , 2x 9 + 3u 3 x 6 + 2u 6 x 3 + 1 2 u 9 ]. Note that, with the notations above, e 1 = 3, e 2 = 2, hence K[x 6 , x 4 , 2x 9 ] K[X 0 , X 1 , X 2 ]/(X 3 1 -X 2 0 , X 2 2 -4X 3 0 ) = K[X 0 , X 1 , X 2 ]/T , K[x 6 + x 3 , x 4 , 2x 9 -3x 6 -2x 3 -

1 2 ] K[X 0 , X 1 , X 2 ]/(X 3 1 -X 2 0 -2X 0 -1 2 , X 2 2 -4X 3 0 -5X 2 0 -2X 0 -1 4
), and

K[u] -→ K[u][X 0 , X 1 , X 2 ]/(X 3 1 -X 2 0 -2u 6 X 0 - 1 2 u 9 , X 2 2 -4X 3 0 -5u 6 X 2 0 -2u 12 X 0 - 1 4 u 18 )
gives us a deformation from A to K[X 0 , X 1 , X 2 ]/T . The computation of the approximate roots and of Γ(F ) can be performed with the algorithm presented in [START_REF] Assi | Algorithms for curves with one place at infinity[END_REF]. gap> f:=y^6-2*x^2*y^3-4*x*y^3-y^3+x^4;; gap> SemigroupOfValuesOfPlaneCurveWithSinglePlaceAtInfinity(f); <Numerical semigroup with 3 generators> gap> MinimalGeneratingSystem(last); [START_REF] Bravo | Some facts about canonical subalgebra bases[END_REF][START_REF] Del Centina | Weierstrass points and their impact in the study of algebraic curves: a historical account from the "Lückensatz" to the 1970s[END_REF][START_REF] Fujimoto | On polynomial curves in the affine plane[END_REF] gap> SemigroupOfValuesOfPlaneCurveWithSinglePlaceAtInfinity(f); [ [START_REF] Del Centina | Weierstrass points and their impact in the study of algebraic curves: a historical account from the "Lückensatz" to the 1970s[END_REF][START_REF] Bravo | Some facts about canonical subalgebra bases[END_REF][START_REF] Fujimoto | On polynomial curves in the affine plane[END_REF], [ y, y^3-x^2-2*x-1/2 ] ] Example 7.10. Let f (x) = x 6 + x, g(x) = x 4 . The minimal polynomial of (f (x), g(x)) is given by:

F (X, Y ) = Y 6 -2X 2 Y 3 -4XY 2 -Y + X 4 .
Let r 0 = 6 = d 1 , r 1 = 4 and G 1 = Y . We have d 2 = gcd(6, 4) = 2, and G 2 = App(F, 2) = Y 3 -X 2 . Since g 2 (x) = G 2 (f (x), g(x)) = -2x 7 -x 2 , then r 2 = 7 and d 3 = 1, hence Γ(F ) = d(A) = 6, 4, 7 and {f (x), g(x), -g 2 (x)} is a basis of A. Consequently, h A = K[u, x 6 + u 5 x, x 4 , 2x 7 + u 5 x 2 ]. Note that,

  gap> x:=X(Rationals,"x");; gap> l:=[x^4+x^5,x^6,x^15+x^16];; gap> s:=SemigroupOfValuesOfCurve_Local(l);; gap> MinimalGeneratingSystem(s); [ 4, 6, 13, 15 ] gap> SemigroupOfValuesOfCurve_Local(l,13); x^13 Remark 2.14. It is known (cf. [3, Section II.1]) that there exist relations between algebraic characteristics and invariants of the semigroup o(R) and the ring R. Hence, in the Example 2.13, from o(R) = 4, 6, 13, 15 = {0, 4, 6, 8, 10, 12, -→}, we deduce that the conductor of o

  . The ring R is Cohen-Macaulay and its type is less than or equal to #T(o(R)) = 3, where for a numerical semigroup Γ, T(Γ) = {x ∈ Z \ Γ | x + Γ * ⊆ Γ} ([3, Proposition II.1.16]). According also to [3, Proposition II.1.16], equality holds if and only if o(R : m) = T(o(R)). However T(o(R)) = {2, 9, 11} and 2 ∈ o(R : m). So in our example we get an strict inequality.

and d 2

 2 = gcd(n, m 1 ). For all k ≥ 2 we set m k = inf{i ∈ supp(g) | d k i} and d k+1 = gcd(d k , m k ). It follows that there exists h ≥ 1 such that d h+1 = 1. The set {m 1 , . . . , m h } is called the set of Newton-Puiseux exponents of F (X, Y ). Let e k = d k d k+1 for all 1 ≤ k ≤ h and define the sequence (r k ) 0≤k≤h as follows: r 0 = n, r 1 = m, and for all 2

7 .

 7 gap> Resultant(x-t^7, y-t^4-t^2,t); y^7-7*x^2*y^3-x^4-14*x^2*y^2-7*x^2*y-x^2 gap> s:=SemigroupOfValuesOfCurve_Local([t^7,t^4+t^2]); <Modular numerical semigroup satisfying 7x mod 14 <= x > gap> MinimalGeneratingSystem(last); [ 2, 7 ] gap> IsFreeNumericalSemigroup(s); true Let the notations be as above. For all

Lemma 7 . 4 .

 74 [START_REF] Torstensson | Canonical bases for subalgebras on two generators in the univariate polynomial ring[END_REF] Theorem 2] If gcd(n, m) = 1, then {f (x), g(x)} is a basis of A.

Proof.Remark 7 . 6 .

 76 Since gcd(n, m) > 1, then the first assertion is clear. On the other hand, sinced 2 = gcd(n, m) = p 1 • • • p r , we have A = K[x], and h ≤ l + 1. Hence Γ(F ) = d(A) has at most l + 2 generators. The result now follows. Let r = (r 0 = n, r 1 = m, r 2 , . . . , r h ) be a sequence of integers and for all k ≥ 1, letd k = gcd(r 0 , • • • , r k-1) and e k = d k d k+1 . Assume that the following conditions hold:[START_REF] Abhyankar | On the semigroup of a meromorphic plane curve, Part I[END_REF]

  13 , x15 } and R is not a Gorenstein ring. Here, λ R ( R/R) = 7.Example 2.16. Let R = K x 8 , x 12 + x 14 + x 15 , with K a field of characteristic zero. Using the same argument as in the Example 2.13, we have that {x 8 , x 12 + x 14 + x 15 , x 26 + x 27 + x 29 -Example 2.17. The following battery of examples was provided by Lance Bryant as a test for our algorithm.

	1 2	x 31 ,
	x 53 +	1 2	x 55 -	1 2	x 57 -	1 8	x 63 +	25 8	x 67 -	95 32	x 71 -	15 16	x 75 -	135 32	x 83 }
	is the reduced basis of the Gorenstein ring R. Furthermore, we have λ R ( R/R) = 42.
	gap> l:=[x^8,x^12+x^14+x^15];;
	gap> SemigroupOfValuesOfCurve_Local(l,"basis");
	[ x^8, x^15+x^14+x^12, -1/2*x^31+x^29+x^27+x^26,
	-135/32*x^83-15/16*x^75-95/32*x^71+25/8*x^67-1/8*x^63-1/2*x^57+1/2*x^55+x^53 ]
	gap> l:=[ [ x^6,x^8+x^9,x^19], [x^7,x^9+x^10,x^19,x^31], [x^7,x^21+x^28+x^33],
	[x^4,x^6+x^7,x^13], [x^6,x^8+x^11,x^10+2*x^13,x^21], [x^5,-x^18-x^21,-x^23,-x^26],
	[x^5,-x^18-x^21,-x^26], [x^5,-x^18-x^21,x^23-x^26], [x^6,x^9+x^10,x^19],
	[x^7,x^9+x^10,x^19], [x^8,x^9+x^10,x^19], [x^7,x^9+x^10,x^17,x^19] ] ;;
	gap> List(l, i->MinimalGeneratingSystem(SemigroupOfValuesOfCurve_Local(i)));
	[ [ 6, 8, 19, 29 ], [ 7, 9, 19, 29, 31 ], [ 7, 33 ], [ 4, 6, 13, 15 ],
	[ 6, 8, 10, 21, 23, 25 ], [ 5, 18, 26, 39, 47 ], [ 5, 18, 26, 39, 47 ],
	[ 5, 18, 26, 39, 47 ], [ 6, 9, 19, 20 ], [ 7, 9, 19, 29 ], [ 8, 9, 19, 30 ],
	[ 7, 9, 17, 19, 29 ] ]

Acknowledgements

The authors would also like to thank Lance Bryant for providing plenty of examples for testing our algorithms, as well as suggesting us to implement new functionalities, that are now given as an extra argument of the function SemigroupOfValuesOfCurve Local.

A communication with the contents of this manuscript was presented by the third author at MEGA'2015. We would like to thank the anonymous referees assigned by the organization of this meeting for their comments and suggestions. All of them were taken into account in the current version of this manuscript.

Theorem 6.4. Every polynomial space curve of K l , parametrized by Y 1 = g 1 (x), . . . , Y l = g l (x) has a deformation into a monomial curve of K r for some positive integer r.

Basis of K[f (x), g(x)]

In [START_REF] Torstensson | Canonical bases for subalgebras on two generators in the univariate polynomial ring[END_REF], the case when a subalgebra A of K[x] has a (SAGBI) basis with two elements is treated. Here we study subalgebras generated by two elements of K[x], and see how a basis can be obtained by using a different approach to that of the general case, as we already did for K [[x]] in Section 4.

Let f (x) = n i=1 a i x i and g(x) = m j=1 b j x j be two polynomials of K[x] and suppose, without loss of generality, that the following conditions hold:

(1)

the greatest common divisor of supp(f (x)) ∪ supp(g(x)) is equal to 1 (in particular for all d > 1,

). Let the notations be as in Section 5, in particular

). Assume that K is algebraically closed with characteristic zero. Let d be a divisor of n, and let G be a monic polynomial in K

We say that G is a dth approximate root of F if α 1 = 0. There is a unique dth approximate root of F . We denote it by App(F, d). The following results can be found in [START_REF] Abhyankar | On the semigroup of a meromorphic plane curve, Part I[END_REF].

Theorem 7.1. Under the standing hypothesis.

(1) F (X, Y ) has one place at infinity, that is the affine curve F (X, Y ) = 0 has one point at infinity, and the projective closure of this curve in P 2 K is analytically irreducible at this point. Proof. In fact, h(x) ∈ A if and only if h(x) = P (f (x), g(x)) for some

be as above, and assume, after a possible change of variables X = X, Y = Y + c 1 n , that c 1 (X, Y ) = 0 (note that this does not change A). In particular App(F, n) = Y . A system of generators of Γ(F ) can be found algorithmically in the following way.

Let

, and d 2 = gcd(r 0 , r 1 ). We set

), and d 3 = gcd(r 3 , d 2 ), and so on. . . With these notations we have the following:

(1) d 1 > d 2 > . . . and there exists h ≥ 1 such that d h+1 = 1;

(2) Γ(F ) = d(A) is generated by {r 0 , r 1 , . . . , r h };

(3) with the notations above, e 1 = 3, e 2 = 2, hence K[x 6 , x 4 , x 7 ] K[X 0 , X 1 , X 2 ]/(X 3 1 -X 2 0 , X 2 2 -X 0 X 2 1 ) = K[X 0 , X 1 , X 2 ]/T , and

1 -X 2 0 , X 2 2 -4X 0 X 2 1 -u 10 X 1 ) gives us a deformation from A to K[X 0 , X 1 , X 2 ]/T (we can also change X 2 with 1 2 X 2 , and then we get (X 3 1 -X 2 0 , X 2 2 -X 0 X 2 1 -1 4 u 10 X 1 ) instead).