How Does Morphism Affect Prime Spectra and Module Support?

[Euge]

Here is this week's POTW:

-----
Given commutative rings with unity $R$ and $S$, let $\phi : R \to S$ be a morphism of rings. It induces a morphism $\phi^* : \operatorname{Spec}(S) \to \operatorname{Spec}(R)$ of prime spectra such that $\phi^*(\mathfrak{q}) = \phi^{-1}(\mathfrak{q})$ for all $\mathfrak{q}\in \operatorname{Spec}(S)$. Show that if $X$ is a finitely generated $R$-module, the support of $S\otimes_R X$ is the inverse image of the support of $X$ under the induced map $\phi^*$.

-----

Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Euge]

I'm going to give one extra week for members to attempt a solution.

 

[Euge]

No one answered this week's problem. You can read my solution below.

Spoiler

If $\mathfrak{a}$ is an ideal of $R$, let $\mathfrak{a}^e$ denote the extension of $\mathfrak{a}$ in $S$. Let $x_1,\ldots, x_n$ be generators of $M$. If $\mathfrak{a}_i := \operatorname{Ann}(x_i)$, then $$S\otimes_R X \approx \sum S\otimes_R Rx_i \approx \sum S\otimes_R R/\mathfrak{a}_i \approx \sum S/\mathfrak{a}_i^e$$ Thus $$\operatorname{Supp}(S\otimes_R X) = \bigcup \operatorname{Supp}(S/\mathfrak{a}_i^e) = \bigcup V(\mathcal{a}_i^e) = \bigcup \phi^{*-1}(V(\mathfrak{a}_i)) = \phi^{*-1}(\cup V(\mathfrak{a}_i)) = \phi^{*-1}(\operatorname{Ann}(X)) = \phi^{*-1}(\operatorname{Supp}(X))$$