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⁾ = PⁿR_P ∩ R be the symbolic power of P and R_s(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 [1] asked if the symbolic Rees algebra R_s(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 [2] showed that for n ≥ 4 then k[t^7n− 3,t^(5n− 2)n,t^8n− 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]/(x¹¹ − yz⁷,y³ − x⁴z⁴,z¹¹ − x⁷y²). Hochster [4] has shown that P⁽ⁿ⁾ = Pⁿ for all n if k[x₁,…,x_s]/P is a complete intersection. Huneke [5] 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,…,x_s] to k[t], by $x_{i}\mapsto t^{n_{i}}$. Then R ≃ k[x₁,…,x_s]/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,…,x_s)-primary. If P⁽ⁿ⁾ = Pⁿ for all n, i.e., if the semigroup ring is a complete intersection, then R_s(P) = R(P) is Noetherian and Pⁿ is P-primary. Kunz [6] 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 n ∈ S if and only if D − n∉S.

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 [3] have shown that R_s(P) = R[Pt,P⁽²⁾t²] if and only if a₁ = a₂,b₁ ≤ b₂,c₁ ≥ c₂ (or a permutation). Huneke [5] 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 [9] has, in the case of a 3-generated semigroup, determined Δ. The result is, if a₁ ≤ a₂,b₁ ≥ b₂,c₁ ≥ c₂, in particular this holds if R_s(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, a₁ ≥ a₂, b₁ ≥ b₂, c₁ ≥ c₂, 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 R_s(P)≠R[Pt,P⁽²⁾t²], we can only determine the primary decomposition of P², but if R_s(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 c − a ≤ 4, or if the multiplicity is ≤ 4.

Theorem 1

Suppose thatR = k[t^a,t^b,t^c] = 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 furthermoreR_s(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 R_s(P) = R[Pt,P⁽²⁾t²], then $P^{(2n)} ={\sum }_{i = 0}^{n}(P^{(2)})^{i}P^{2n-2i} = (P^{(2)})^{n}$ since P² ⊆ P⁽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 R_s(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 R_s(P) = R[Pt,P⁽²⁾t²]. We use this result also below to see when R_s(P) = R[Pt,P⁽²⁾t²].

Theorem 2

LetR = k[t^m,t^m+d,t^m+ 2d] = k[x,y,z]/Pbe nonsymmetric. Then the primary decomposition isP²ⁿ = ((Δ) + P²)ⁿ ∩ ((zⁿ) + P²)ⁿandP^2n+ 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 R_s(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 R_s(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 R_s(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 R_s(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 R_s(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 R_s(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 R_s(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 R_s(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 R_s(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 R_s(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 R_s(P) = R[Pt,P⁽²⁾t²].

Theorem 3

If the semigroup is generated bya < b < c,c − a ≤ 4,not symmetric, anda,b,cnot an arithmetic sequence, thenR_s(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 [5] 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 R_s(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 R_s(P) = R[Pt,P⁽²⁾t²] if and only if 5l − 3k + 3 ≤ 0. If k ≤ l, the relation matrix is

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

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

Theorem 4

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