I'm trying to prove that the localization $S^{-1}A$ is a principal ring. Here, $A$ is a principal ring and $S$ a multiplicative subset of $A$. Furthermore we assume $0 \notin S$.
The term principal ring is used in Lang's algebra, referring to a nontrivial commutative ring whose all ideals are principal. It allows zero divisors, which is the only difference form PID.
Anyway I start with an ideal $\mathfrak a'$ in $S^{-1}A$. Let $f:A \to S^{-1}A$ be the canonical map. Then $f^{-1}(\mathfrak a')$ is a principal ideal, say $aA$, in $A$.
Now $\mathfrak a'$ seems to be the principal ideal $(a/1)(S^{-1}A)$. Showing that this ideal is contained in $\mathfrak a'$ is easy, but I got stuck to show the converse.
Any braekthrough?
Best Answer
Take an ideal $\mathfrak a\leqslant S^{-1}A$ and consider the underlying ideal in $A$, i.e. $\mathfrak a\cap A$. By assumption, this ideal is principal and hence $\mathfrak a\cap A=aA$ for some $a\in A$. Let $x\in\mathfrak a$. Then there is some $s\in S$ such that $sx\in A$ by construction and hence $sx\in \mathfrak a\cap A=aA$. So $x\in a(S^{-1}A)$ lifting the principal ideal to the localization. This shows at once that $\mathfrak a$ is principal with generator $a$.
The map $\mathfrak a\mapsto\mathfrak a\cap A$ is a left inverse of your canonical map. One can show that composing these the other way around is the identity on prime ideals disjoint from $S$, which might be an interesting fact. See Milne's Algebraic Number Theory $ยง1$ for the details.