Spring 2016 - Problem 4

ring theory

Let $A$ be a commutative ring, $S$ a multiplicatively closed subset of $A$ , $A \to A\br{S^{-1}}$ the localization.

Which elements of $A$ map to zero in $A\br{S^{-1}}$ ?
Let $\mathfrak{p}$ be a prime ideal in $A$ . Show that the ideal generated by the image of $\mathfrak{p}$ in $A\br{S^{-1}}$ is prime if and only if the intersection of $\mathfrak{p}$ with $S$ is empty.

Suppose $a$ maps to $0$ in $A\br{S^{-1}}$ ; that is, $\frac{a}{1} = 0$ . By definition, there exists $s \in S$ so that $as = 0$ in $A$ . Conversely, suppose $a \in A$ is annihilated by $S$ , i.e., $as = 0$ for some $s \in S$ . Then in $A\br{S^{-1}}$ , we have
$\frac{a}{1} = \frac{as}{s} = 0,$
so the elements in $A$ which map to zero in the localization are precisely the elements annihilated by $S$ .
" $\implies$ "

Assume the ideal $I = S^{-1}\mathfrak{p}$ is prime. If there exists $s \in \mathfrak{p} \cap S$ , then $I$ is all of $A\br{S^{-1}}$ , but this means $I$ is not prime. Thus, $\mathfrak{p} \cap S = \emptyset$ .

" $\impliedby$ "

Suppose $\mathfrak{p} \cap S = \emptyset$ , and assume $\frac{a}{s} \cdot \frac{b}{t} \in I$ , where $I$ is as above. Then $\frac{ab}{st} = \frac{p}{r}$ for some $p \in \mathfrak{p}$ . Thus, $rab = pst \in \mathfrak{p}$ , and by assumption, $r \notin \mathfrak{p}$ , so because $\mathfrak{p}$ is prime, we have $ab \in \mathfrak{p}$ . Thus, $a$ or $b$ is in $\mathfrak{p}$ , which implies $\frac{a}{s}$ or $\frac{b}{t}$ is in $I$ , so $I$ is prime.