# Primary Decomposition of Powers of the Prime Ideal of a Numerical Semigroup Ring

## Abstract

Let $$R=k[t^{n_{1}},\ldots ,t^{n_{s}}]=k[x_{1},\ldots ,x_{s}]/P$$ be a numerical semigroup ring and let P(n) = PnRPR be the symbolic power of P and Rs(P) = ⊕i≥ 0P(n)tn the symbolic Rees ring of P. It is hard to determine symbolic powers of P; there are even non-Noetherian symbolic Rees rings for 3-generated semigroups. We determine the primary decomposition of powers of P for some classes of 3-generated numerical semigroups.

## Introduction

If I is an ideal in a Noetherian ring R, then the subring R(I) = R[It] of R[t] is called the Rees ring of I. This was introduced by Rees, in his proof of the Artin-Rees lemma. If P is a prime ideal, then the primary decomposition of Pn always contains a P-primary component, it is $$R=k[t^{n_1},\ldots ,t^{n_s}]=k[x_1,\ldots ,x_n]/P$$ and is called the symbolic n th power of P. Cowsik  asked if the symbolic Rees algebra Rs(P) = R[Pt,P(2)t2,P(3)t3,…] is always Noetherian. This was shown not to be true by Roberts [7, 8]. There are even counterexamples when $$R=k[t^{n_{1}},t^{n_{2}},t^{n_{3}}]=k[x,y,z]/P$$. Goto-Nishida-Watanabe  showed that for n ≥ 4 then k[t7n− 3,t(5n− 2)n,t8n− 3] does not have a finitely generated symbolic Rees algebra if char(k) = 0. Their smallest counterexample is k[t25,t72,t29] = k[x,y,z]/(x11yz7,y3x4z4,z11x7y2). Hochster  has shown that P(n) = Pn for all n if k[x1,…,xs]/P is a complete intersection. Huneke  has shown that if P is a prime ideal of height 2 in a 3-dimensional ring, then the opposite holds.

## Numerical Semigroup Rings

If $$R=k[t^{n_{1}},\ldots ,t^{n_{s}}]$$, we map k[x1,…,xs] to k[t], by $$x_{i}\mapsto t^{n_{i}}$$. Then Rk[x1,…,xs]/P, and P is a prime ideal since R is a domain. In the case of numerical semigroup rings, R is 1-dimensional, so P(n) = Pn or P(n)Q, where Q is (x1,…,xs)-primary. If P(n) = Pn for all n, i.e., if the semigroup ring is a complete intersection, then Rs(P) = R(P) is Noetherian and Pn is P-primary. Kunz  has shown that the semigroup ring of S is Gorenstein if and only if the semigroup S is symmetric, i.e., if for some D, we have nS if and only if DnS.

## 3-Generated Numerical Semigroups

In the sequel, we mean numerical semigroup when we write semigroup. A 3-generated semigroup is Gorenstein if and only if it is a complete intersection. This is so, since if we factor with a generator, we get a ring of embedding dimension 2. The factor ring is Gorenstein (complete intersection) if and only if the semigroup ring is Gorenstein (complete intersection). For a ring of embedding dimension 2, the concepts are equivalent. If the semigroup is generated by 3 elements, and is not a complete intersection, then $$R=k[t^{n_{1}},t^{n_{2}},t^{n_{3}}]\simeq k[x,y,z]/P$$ where P is generated by the three 2 × 2-minors of a matrix (the relation matrix)

$$\left( \begin{array}{ccc} x^{a_{1}}&y^{b_{1}}&z^{c_{1}}\\ z^{c_{2}}&x^{a_{2}}&y^{b_{2}} \end{array}\right).$$

Herzog and Ulrich  have shown that Rs(P) = R[Pt,P(2)t2] if and only if a1 = a2,b1b2,c1c2 (or a permutation). Huneke  has shown that if P is a 2-dimensional prime in a 3-dimensional ring, then P(2)/P2 is generated by one element Δ. Schenzel  has, in the case of a 3-generated semigroup, determined Δ. The result is, if a1a2,b1b2,c1c2, in particular this holds if Rs(P) = R[Pt,P(2)t2] according to the result of Herzog and Ulrich,

$${\Delta}=\left|\begin{array}{ccc} x^{a_{1}}&y^{b_{1}}&z^{c_{1}}\\ z^{c_{2}}&x^{a_{2}}&y^{b_{2}}\\ y^{b_{1}}&x^{a_{2}-a_{1}}y^{b_{1}-b_{2}}z^{c_{2}}&y^{a_{1}}z^{c_{1}-c_{2}} \end{array}\right|.$$

He also showed that $$(x^{a_{1}},y^{b_{2}},z^{c_{2}}){\Delta }\in P^{2}$$.

In the other case, a1a2, b1b2, c1c2, there is a similar result:

$${\Delta}=\left|\begin{array}{ccc} x^{a_{1}}&y^{b_{1}}&z^{c_{1}}\\ z^{c_{2}}&x^{a_{2}}&y^{b_{2}}\\ x^{a_{1}-a_{2}}&y^{b_{1}-b_{2}}z^{c_{1}}&x^{a_{1}}z^{c_{1}-c_{2}} \end{array}\right|$$

and $$(x^{a_{2}},y^{b_{2}},z^{c_{2}}){\Delta }\in P^{2}$$. If Rs(P)≠R[Pt,P(2)t2], we can only determine the primary decomposition of P2, but if Rs(P) = R[Pt,P(2)t2], we will construct the primary decomposition of Pn for all n in the following cases: if the semigroup is generated by an arithmetic sequence, or if it is generated by a < b < c with ca ≤ 4, or if the multiplicity is ≤ 4.

### Theorem 1

Suppose thatR = k[ta,tb,tc] = k[x,y,z]/Pis not a complete intersection. Then$$P^{2} = (({\Delta }) + P^{2}) \cap ((z^{c_{2}}) + P^{2})$$isa primary decomposition. If furthermoreRs(P) = R[Pt,P(2)t2],then$$P^{2n} = (P^{(2)})^{n}\cap ((z^{nc_{2}}) + P^{2n})$$and$$P^{2n + 1} = P(P^{(2)})^{n}\cap ((z^{nc_{2}}) + P^{2n})$$.

### Proof

Since P(2) = (Δ) + P2 and since $$(z^{c_{2}}) + P^{2}$$ is (x,y,z)-primary, it suffices to note that $$(z^{c_{2}})\cap ({\Delta })=z^{c_{2}} {\Delta }\subseteq P^{2}$$ ([9, Theorem 10.3]) to see the first statement. If Rs(P) = R[Pt,P(2)t2], then $$P^{(2n)} ={\sum }_{i = 0}^{n}(P^{(2)})^{i}P^{2n-2i} = (P^{(2)})^{n}$$ since P2P(2). In the same way, we see that P(2n+ 1) = PP(2n). Finally, $$(z^{nc_{2}})\cap P^{(2n)} =(z^{nc_{2}})\cap (({\Delta })+(P^{2})^{n}) \subseteq P^{2n}$$ since $$z^{c_{2}}{\Delta }\subseteq P^{2}$$.

The remaining part is a search for examples when Rs(P) = R[Pt,P(2)t2]. □

### Arithmetic Sequences

Now suppose that the semigroup is generated by m,m + d,m + 2d, $$\gcd (m,m+d,m + 2d)= 1$$. The semigroup is symmetric (so the semigroup ring is a complete intersection) if m is even and d odd. Otherwise, the relation matrix is

$$\left( \begin{array}{ccc} x^{k+d}&y&z\\ z^{k}&x&y \end{array}\right).$$

Thus, according to the theorem by Herzog and Ulrich Rs(P) = R[Pt,P(2)t2]. We use this result also below to see when Rs(P) = R[Pt,P(2)t2].

### Theorem 2

LetR = k[tm,tm+d,tm+ 2d] = k[x,y,z]/Pbe nonsymmetric. Then the primary decomposition isP2n = ((Δ) + P2)n ∩ ((zn) + P2)nandP2n+ 1 = (P(Δ) + P2)n ∩ ((zn) + P2)n.

### Semigroups Generated by a < b < c, c − a ≤ 4

If the semigroup is not generated by an arithmetic sequence, the generators are m,m + 1,m + 3 or m,m + 1,m + 4 or m,m + 2,m + 3 or m,m + 3,m + 4.

If the semigroup is generated by m,m + 1,m + 3, it is symmetric if m ≡ 0 (mod 3).

If m = 3k + 1 the relation matrix is

$$\left( \begin{array}{ccc} x^{k}&y&z\\ z^{k}&x^{2}&y^{2} \end{array}\right)$$

and Rs(P)≠R[Pt,P(2)t2] for all k.

If m = 3k + 2 the relation matrix is

$$\left( \begin{array}{ccc} x^{k + 1}&y^{2}&z\\ z^{k}&x^{2}&y \end{array}\right)$$

and Rs(P)≠R[Pt,P(2)t2] unless k = 1.

If the semigroup is generated by m,m + 1,m + 4, it is symmetric if m ≡ 0 (mod 4) (and if m = 5).

If m = 4k + 1, k ≥ 2, the relation matrix is

$$\left( \begin{array}{ccc} x^{k-1}&y&z\\ z^{k}&x^{3}&y^{3} \end{array}\right)$$

and Rs(P)≠R[Pt,P(2)t2].

If m = 4k + 2, k ≥ 2, the relation matrix is

$$\left( \begin{array}{ccc} x^{k}&y^{2}&z\\ z^{k}&x^{3}&y^{2} \end{array}\right)$$

and Rs(P) = R[Pt,P(2)t2] if and only if k ≥ 3 and k = 1.

If m = 4k + 3, the relation matrix is

$$\left( \begin{array}{ccc} x^{k + 1}&y^{3}&z\\ z^{k}&x^{3}&y \end{array}\right)$$

and Rs(P) = R[Pt,P(2)t2] only if k = 1 or k = 2.

If the semigroup is generated by m,m + 2,m + 3, it is symmetric if m ≡ 0 (mod 3) (and if m = 4).

If m = 3k + 1, the relation matrix is

$$\left( \begin{array}{ccc} x^{k + 1}&y^{2}&z^{2}\\ z^{k-1}&x&y \end{array}\right)$$

and Rs(P)≠R[Pt,P(2)t2] for all k.

If m = 3k + 2 the relation matrix is

$$\left( \begin{array}{ccc} x^{k}&y^{2}&z^{2}\\ z^{k}&x&y \end{array}\right)$$

and Rs(P)≠R[Pt,P(2)t2] for all k.

If the semigroup is generated by m,m + 3,m + 4, it is symmetric if m ≡ 0 (mod 4) (and if m = 6 or m = 9).

If m = 4k + 1, k ≥ 2, the relation matrix is

$$\left( \begin{array}{ccc} x^{k + 1}&y^{3}&z^{3}\\ z^{k-1}&x&y \end{array}\right)$$

and Rs(P)≠R[Pt,P(2)t2].

If m = 4k + 2, k ≥ 2, the relation matrix is

$$\left( \begin{array}{ccc} x^{k + 1}&y^{2}&z^{3}\\ z^{k-1}&x&y^{2} \end{array}\right)$$

and Rs(P) = R[Pt,P(2)t2] if and only if k ≥ 4.

If m = 4k + 3 the relation matrix is

$$\left( \begin{array}{ccc} x^{k + 1}&y&z^{3}\\ z^{k}&x&y^{3} \end{array}\right)$$

and Rs(P) = R[Pt,P(2)t2] only if k = 3.

For m = 5 the relation matrix is

$$\left( \begin{array}{ccc} x^{2}&y^{2}&z\\ z&x^{3}&y \end{array}\right)$$

and Rs(P) = R[Pt,P(2)t2].

### Theorem 3

If the semigroup is generated bya < b < c,ca ≤ 4,not symmetric, anda,b,cnot an arithmetic sequence, thenRs(P) = R[Pt,P(2)t2] if and only if the generators are 5,6,8 or 6,7,10 or 15,18,19 or 7,8,11 or 11,12,15 or 15,18,19 or 4k + 2,4k + 3,4k + 6,k ≥ 3 or 4k + 2,4k + 5,4k + 6,k ≥ 4.

### Semigroups of Multiplicity 3

We note that if the semigroup is of multiplicity ≤ 4, then Huneke  has shown that the symbolic Rees ring is Noetherian. Suppose that the semigroup is generated by 3,3k + 1,3l + 2. In order to have a 3-generated semigroup, we must have l ≤ 2k and k ≤ 2l + 1. The semigroup is never symmetric. The relation matrix is

$$\left( \begin{array}{ccc} x^{2l-k + 1}&y&z\\ z&x^{2k-l}&y \end{array}\right)$$

and Rs(P) = R[Pt,P(2)t2].

### Semigroups of Multiplicity 4

If a 3-generated semigroup has multiplicity 4 and is not symmeteric, it has generators 4,4k + 1,4l + 3. If k > l the relation matrix is

$$\left( \begin{array}{ccc} x^{3l-k + 2}&y&z^{2}\\ z&x^{2k-2l-1}&y \end{array}\right)$$

and Rs(P) = R[Pt,P(2)t2] if and only if 5l − 3k + 3 ≤ 0. If kl, the relation matrix is

$$\left( \begin{array}{ccc} x^{2l-2k + 1}&y&z\\ z&x^{3k-l}&y^{2} \end{array}\right)$$

and Rs(P) = R[Pt,P(2)t2] if and only if 3l − 5k + 1 ≥ 0.

### Theorem 4

If the semigroup has multiplicity 3,thenRs(P) = R[Pt,P(2)t2].If the multiplicity is 4 and not symmetric, it is generated by 4,4k + 1,4l + 3,andRs(P) = R[Pt,P(2)t2] if and only ifk > land 5l − 3k + 3 ≤ 0 or ifkland 3l − 5k + 1 ≥ 0.

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Correspondence to Ralf Fröberg.