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  • https://doi.org/10.5556/j.tkjm.31.2000.405Copy DOI Icon

Note on integral closures of semigroup rings

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Abstract

Let $S$ be a subsemigroup which contains 0 of a torsion-free abelian (additive) group. Then $S$ is called a grading monoid (or a $g$-monoid). The group $ \{s-s'|s,s'\in S\}$ is called the quotient group of $S$, and is denored by $q(S)$. Let $R$ be a commutative ring. The total quotient ring of $R$ is denoted by $q(R)$. Throught the paper, we assume that a $g$-monoid properly contains $ \{0\}$. A commutative ring is called a ring, and a non-zero-divisor of a ring is called a regular element of the ring. We consider integral elements over the semigroup ring $ R[X;S]$ of $S$ over $R$. Let $S$ be a $g$-monoid with quotient group $G$. If $ n\alpha\in S$ for an element $ \alpha$ of $G$ and a natural number $n$ implies $ \alpha\in S$, then $S$ is called an integrally closed semigroup. We know the following fact: ${\bf Theorem~1}$ ([G2, Corollary 12.11]). Let $D$ be an integral domain and $S$ a $g$-monoid. Then $D[X;S]$ is integrally closed if and only if $D$ is an integrally closed domain and $S$ is an integrally closed semigroup. Let $R$ be a ring. In this paper, we show that conditions for $R[X;S]$ to be integrally closed reduce to conditions for the polynomial ring of an indeterminate over a reduced total quotient ring to be integrally closed (Theorem 15). Clearly the quotient field of an integral domain is a von Neumann regular ring. Assume that $q(R)$ is a von Neumann regular ring. We show that $R[X;S]$ is integrally closed if and only if $R$ is integrally closed and $S$ is integrally closed (Theorem 20). Let $G$ be a $g$-monoid which is a group. If $R$ is a subring of the ring $T$ which is integrally closed in $T$, we show that $R[X;G]$ is integrally closed in $T[X;S]$ (Theorem 13). Finally, let $S$ be sub-$g$-monoid of a totally ordered abelian group. Let $R$ be a subring of the ring $T$ which is integrally closed in $T$. If $g$ and $h$ are elements of $T[X;S]$ with $h$ monic and $gh\in R[X;S]$, we show that $g\in R[X;S]$ (Theorem 24).

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