Ex. 2.6 from Eisenbud: localization and infinite intersections

Question

Can someone please explain to me why, given an infinite field $k$, we have that $$\left(\bigcap_{a \in k}{(x-a)}\right) [U^{-1} ] \neq \bigcap_{a \in k}{\left((x-a)[U^{-1}] \right)},$$ where $U$ is the set of nonzero elements of $k[x]$?

Answer 1

Any polynomial in the intersection of all the ideals $(x-a)$ must have infinitely many roots, i.e. be $0$. So, the localization on the left-hand side is $0$ as well.

On the right hand, note that $x-a\in U$ for every $a$, so that we see $\frac{x-a}{x-a}=1$ is in $(x-a)[U^{-1}]$ for every $a$, and thus in the intersection. In fact, the localizations $(x-a)[U^{-1}]$ are isomorphic to $k[x][U^{-1}],$ that is to $k(x)$, as modules, via the map $\frac{p(x)}{q(x)}\to\frac{p(x)}{q(x)(x-a)}(x-a)$, so that the right-hand side is $k(x)$.

It's good to check that we would have gotten $k(x)$ the left-hand side as well in case $k$ was a finite field, since then $\prod_{a\in k} (x-a)$ is nonzero, and we can generate $k(x)$ in essentially the same way as we did for the ideals above generated by degree-1 polynomials.