Let $R\subset S$ be an extension of commutative rings. The integral closure of $R$ in $S$ is $\textrm{IC}(R)=\{s\in S\mid \exists f(x)\in\ R[x]\text{ monic polynomial such that }f(s)=0\}$, which is a ring.
How about "non-monic" integral closure, namely, $$\textrm{NMIC}(R)=\{s\in S\mid \exists f(x)\in\ R[x]\text{ (possiboly non-monic) nonzero polynomial such that }f(s)=0\}?$$ It is obvious that $R\subset\textrm{IC}(R)\subset\textrm{NMIC}(R)\subset S$. Is $\textrm{NMIC}(R)$ a ring, or an abelian group, or multiplicatively closed? I read several books and searched in web, but I could not find any clue to this question.