I've been working on the following exercise from Eisenbud's book on Commutative Algebra.
Exercise 10.7: Show that if $R$ is an integral domain contained in the local ring $(S,Q)$, then there is a minimal prime of $S$ contracting to $0$ in $R$. (All rings in this chapter are Noetherian.)
Here's my attempt at a solution: Let $P$ be a minimal prime of $S$. Consider the contraction of the same, $P\cap R$ in $R$. If $a\in P\cap R$ and $a$ is nonzero, then since $a$ is an element of a minimal prime of $S$, there exists another $a'\in S$ such that $aa' = 0$. Now $(a')\cap R=(0)$ as $R$ is a domain. Now, if $(a')$ contains a prime ideal, we are done. Let's consider the ideals containing $(a')$ not containing any nonzero element of $R$. Since $S$ is Noetherian, there's an ideal maximal among these, say $I$. I would like to show that $I$ is prime. Suppose not. Then there can be two possibilities: first, there's an element $r\in R$ such that $(I:r)\supsetneq I$ and second, there's no such element - in other words, there's some element $s\in S\backslash R$ such that $(I:s)\supsetneq I$ and $(I:s)$ does not contain any nonzero element of $R$. The second case easily provides a contradiction to the maximality of $I$. In the first case, since $R$ is closed under multiplication, $(I:r)$ can't contain any nonzero element of $R$. Thus, we have a contradiction and $I$ must be prime as claimed, and this $I$ contracts to the zero ideal in $R$.
Now there are a couple of reasons why I'm not so sure if this proof is correct:
- I don't seem to be using the fact that $S$ is a local ring anywhere in my above ``proof".
- Is the Noetherian hypothesis really necessary? The step where I'm using it, as far as I can see, one can replace that by Zorn's lemma, and the proof thereby works for non-Noetherian rings too.
I would be really glad if someone can point out the flaws in my proof or provide some explanation for the above doubts.
$T=R\setminus\{0\}$ is a multiplicative subset of $S$. In the ring of fractions $T^{-1}S$ there is a prime ideal $P=T^{-1}\mathfrak p$, with $\mathfrak p\subset S$ a prime ideal such that $\mathfrak p\cap T=\emptyset$, that is, $\mathfrak p\cap R=(0)$. If $\mathfrak p$ is minimal we are done. Otherwise, there is $\mathfrak p'\subset\mathfrak p$ a minimal prime, and obviously $\mathfrak p'\cap R=(0)$.