Let $R$ be a commutative, local ring and let $f$ be a monic polynomial in $R[x]$. How can I show that $R[x]/(f)$ is semilocal, respectively artinian?
Thank you for your help!
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@YACP Thanks for the link! One more question: Are the maximal ideals in $R[x]/(f)$ lying over the maximal ideal $\mathfrak{m}$ in $R$ the only maximal ideal in $R[x]/(f)$? Can't there exist a maximal ideal $\mathfrak{M}$ in $R[x]/(f)$ such that $\mathfrak{M}\cap R\subset \mathfrak{m}$? – Heffalump Oct 11 '13 at 09:15
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As you already know, the ring extension $R\subset R[x]/(f)$ is integral and finitely generated (actually generated by one element, that is, $\hat x$, the residue class of $x$ modulo $(f)$), so it's a finite extension. In this topic it's proved that there are only finitely many primes lying over the maximal ideal of $R$. Why are these maximal? Why are these the only maximal ideals in $S$? From the well known fact that for an integral extension $A\subset B$ of integral domains, $A$ is a field iff $B$ is a field it can be shown that for a integral extension the preimage of a maximal ideal is maximal.
I would not say that in general $R[x]/(f)$ is artinian since $\dim R[x]/(f)=\dim R$, but an artinian ring has dimension $0$.
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Sorry, I wasn't aware of that "tool". But since you ask, I think that the last part of the argumetation could be more precize. For example one could say, that from the "well known fact that A is field iff B is a field" it can be shown that for a integral extension the preimage of a maximal ideal is maximal.But anyway, thanks again for your help – Heffalump Oct 12 '13 at 09:36