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

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## 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)} = *P*^{n}*R*_{P} ∩ *R* be the symbolic power of *P* and *R*_{s}(*P*) = ⊕_{i≥ 0}*P*^{(n)}*t*^{n} 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.

## Keywords

Primary decomposition Symbolic power Numerical semigroup ring## Mathematics Subject Classification (2010)

Primary 13A30 Secondary 13H10## 1 Introduction

If *I* is an ideal in a Noetherian ring *R*, then the subring *R*(*I*) = *R*[*I**t*] 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 *P*^{n} 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*[*P**t*,*P*^{(2)}*t*^{2},*P*^{(3)}*t*^{3},…] 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*[*t*^{25},*t*^{72},*t*^{29}] = *k*[*x*,*y*,*z*]/(*x*^{11} − *y**z*^{7},*y*^{3} − *x*^{4}*z*^{4},*z*^{11} − *x*^{7}*y*^{2}). Hochster [4] has shown that *P*^{(n)} = *P*^{n} for all *n* if *k*[*x*_{1},…,*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.

## 2 Numerical Semigroup Rings

If \(R=k[t^{n_{1}},\ldots ,t^{n_{s}}]\), we map *k*[*x*_{1},…,*x*_{s}] to *k*[*t*], by \(x_{i}\mapsto t^{n_{i}}\). Then *R* ≃ *k*[*x*_{1},…,*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)} = *P*^{n} or *P*^{(n)} ∩ *Q*, where *Q* is (*x*_{1},…,*x*_{s})-primary. If *P*^{(n)} = *P*^{n} for all *n*, i.e., if the semigroup ring is a complete intersection, then *R*_{s}(*P*) = *R*(*P*) is Noetherian and *P*^{n} 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 3-Generated Numerical Semigroups

*P*is generated by the three 2 × 2-minors of a matrix (the relation matrix)

*R*

_{s}(

*P*) =

*R*[

*P*

*t*,

*P*

^{(2)}

*t*

^{2}] if and only if

*a*

_{1}=

*a*

_{2},

*b*

_{1}≤

*b*

_{2},

*c*

_{1}≥

*c*

_{2}(or a permutation). Huneke [5] has shown that if

*P*is a 2-dimensional prime in a 3-dimensional ring, then

*P*

^{(2)}/

*P*

^{2}is generated by one element Δ. Schenzel [9] has, in the case of a 3-generated semigroup, determined Δ. The result is, if

*a*

_{1}≤

*a*

_{2},

*b*

_{1}≥

*b*

_{2},

*c*

_{1}≥

*c*

_{2}, in particular this holds if

*R*

_{s}(

*P*) =

*R*[

*P*

*t*,

*P*

^{(2)}

*t*

^{2}] according to the result of Herzog and Ulrich,

*a*

_{1}≥

*a*

_{2},

*b*

_{1}≥

*b*

_{2},

*c*

_{1}≥

*c*

_{2}, there is a similar result:

*R*

_{s}(

*P*)≠

*R*[

*P*

*t*,

*P*

^{(2)}

*t*

^{2}], we can only determine the primary decomposition of

*P*

^{2}, but if

*R*

_{s}(

*P*) =

*R*[

*P*

*t*,

*P*

^{(2)}

*t*

^{2}], we will construct the primary decomposition of

*P*

^{n}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 that**R* = *k*[*t*^{a},*t*^{b},*t*^{c}] = *k*[*x*,*y*,*z*]/*P**is not a complete intersection. Then*\(P^{2} = (({\Delta }) + P^{2}) \cap ((z^{c_{2}}) + P^{2})\)*is**a primary decomposition. If furthermore**R*_{s}(*P*) = *R*[*P**t*,*P*^{(2)}*t*^{2}]*,**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) + *P*^{2} 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*[*P**t*,*P*^{(2)}*t*^{2}], then \(P^{(2n)} ={\sum }_{i = 0}^{n}(P^{(2)})^{i}P^{2n-2i} = (P^{(2)})^{n}\) since *P*^{2} ⊆ *P*^{(2)}. In the same way, we see that *P*^{(2n+ 1)} = *P**P*^{(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*[*P**t*,*P*^{(2)}*t*^{2}]. □

### 3.1 Arithmetic Sequences

*m*,

*m*+

*d*,

*m*+ 2

*d*, \(\gcd (m,m+d,m + 2d)= 1\). The semigroup is symmetric (so the semigroup ring is a complete intersection) if

*m*is even and

*d*odd. Otherwise, the relation matrix is

*R*

_{s}(

*P*) =

*R*[

*P*

*t*,

*P*

^{(2)}

*t*

^{2}]. We use this result also below to see when

*R*

_{s}(

*P*) =

*R*[

*P*

*t*,

*P*

^{(2)}

*t*

^{2}].

### **Theorem 2**

*Let**R* = *k*[*t*^{m},*t*^{m+d},*t*^{m+ 2d}] = *k*[*x*,*y*,*z*]/*P**be nonsymmetric. Then the primary decomposition is**P*^{2n} = ((Δ) + *P*^{2})^{n} ∩ ((*z*^{n}) + *P*^{2})^{n}*and**P*^{2n+ 1} = (*P*(Δ) + *P*^{2})^{n} ∩ ((*z*^{n}) + *P*^{2})^{n}*.*

### 3.2 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).

*m*= 3

*k*+ 1 the relation matrix is

*R*

_{s}(

*P*)≠

*R*[

*P*

*t*,

*P*

^{(2)}

*t*

^{2}] for all

*k*.

*m*= 3

*k*+ 2 the relation matrix is

*R*

_{s}(

*P*)≠

*R*[

*P*

*t*,

*P*

^{(2)}

*t*

^{2}] 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).

*m*= 4

*k*+ 1,

*k*≥ 2, the relation matrix is

*R*

_{s}(

*P*)≠

*R*[

*P*

*t*,

*P*

^{(2)}

*t*

^{2}].

*m*= 4

*k*+ 2,

*k*≥ 2, the relation matrix is

*R*

_{s}(

*P*) =

*R*[

*P*

*t*,

*P*

^{(2)}

*t*

^{2}] if and only if

*k*≥ 3 and

*k*= 1.

*m*= 4

*k*+ 3, the relation matrix is

*R*

_{s}(

*P*) =

*R*[

*P*

*t*,

*P*

^{(2)}

*t*

^{2}] 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).

*m*= 3

*k*+ 1, the relation matrix is

*R*

_{s}(

*P*)≠

*R*[

*P*

*t*,

*P*

^{(2)}

*t*

^{2}] for all

*k*.

*m*= 3

*k*+ 2 the relation matrix is

*R*

_{s}(

*P*)≠

*R*[

*P*

*t*,

*P*

^{(2)}

*t*

^{2}] 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).

*m*= 4

*k*+ 1,

*k*≥ 2, the relation matrix is

*R*

_{s}(

*P*)≠

*R*[

*P*

*t*,

*P*

^{(2)}

*t*

^{2}].

*m*= 4

*k*+ 2,

*k*≥ 2, the relation matrix is

*R*

_{s}(

*P*) =

*R*[

*P*

*t*,

*P*

^{(2)}

*t*

^{2}] if and only if

*k*≥ 4.

*m*= 4

*k*+ 3 the relation matrix is

*R*

_{s}(

*P*) =

*R*[

*P*

*t*,

*P*

^{(2)}

*t*

^{2}] only if

*k*= 3.

*m*= 5 the relation matrix is

*R*

_{s}(

*P*) =

*R*[

*P*

*t*,

*P*

^{(2)}

*t*

^{2}].

### **Theorem 3**

*If the semigroup is generated by**a* < *b* < *c**,**c* − *a* ≤ 4*,**not symmetric, and**a*,*b*,*c**not an arithmetic sequence, then**R*_{s}(*P*) = *R*[*P**t*,*P*^{(2)}*t*^{2}] *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* 4*k* + 2,4*k* + 3,4*k* + 6*,**k* ≥ 3 *or* 4*k* + 2,4*k* + 5,4*k* + 6*,**k* ≥ 4*.*

### 3.3 Semigroups of Multiplicity 3

*k*+ 1,3

*l*+ 2. In order to have a 3-generated semigroup, we must have

*l*≤ 2

*k*and

*k*≤ 2

*l*+ 1. The semigroup is never symmetric. The relation matrix is

*R*

_{s}(

*P*) =

*R*[

*P*

*t*,

*P*

^{(2)}

*t*

^{2}].

### 3.4 Semigroups of Multiplicity 4

*k*+ 1,4

*l*+ 3. If

*k*>

*l*the relation matrix is

*R*

_{s}(

*P*) =

*R*[

*P*

*t*,

*P*

^{(2)}

*t*

^{2}] if and only if 5

*l*− 3

*k*+ 3 ≤ 0. If

*k*≤

*l*, the relation matrix is

*R*

_{s}(

*P*) =

*R*[

*P*

*t*,

*P*

^{(2)}

*t*

^{2}] if and only if 3

*l*− 5

*k*+ 1 ≥ 0.

### **Theorem 4**

*If the semigroup has multiplicity* 3*,**then**R*_{s}(*P*) = *R*[*P**t*,*P*^{(2)}*t*^{2}]*.**If the multiplicity is* 4 *and not symmetric, it is generated by* 4,4*k* + 1,4*l* + 3*,**and**R*_{s}(*P*) = *R*[*P**t*,*P*^{(2)}*t*^{2}] *if and only if**k* > *l**and* 5*l* − 3*k* + 3 ≤ 0 *or if**k* ≤ *l**and* 3*l* − 5*k* + 1 ≥ 0*.*

## Notes

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