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I thought I understood the proof last time, but now that I look at it again, there seems to be some puzzling part. I am reading the Gortz, Wedhorn, Algebraic Geometry, 2nd edition, proof of the Lemma 7.43 and struggling with some step ( see below ). The questions I summarized while trying to understand the steps were as follows. ( Originally this summarized question was posted separately, however it is recognized as a duplicate, so it is moved here. )

We want to show next : Let $A$ be a commutative ring and $M$ is a $A$-module. Assume that $\mathcal{E}:=\tilde{M}$ is finite locally free $\mathcal{O}_{\operatorname{Spec}A}$-module of constant rank $r$. Let $\{ \mathfrak{m}_1 ,\dots , \mathfrak{m}_n\} \subset A$ be a finite set of maximal ideals. Let $S$ be the complement of $\bigcup_i \mathfrak{m}_i$, which is a multiplicative subset of $A$. We set $A':= S^{-1}A$ and $M'=S^{-1}M$. Then $A'$ is a semi-local ring. ( True? Why? Can we exhibit explicit maximal ideals? )

Q.1. Then, first question is, can we show that $M'$ is free $A'$-module of rank $r$?

I am trying to apply next statement :

Let $R$ be a commutative ring with finitely many maximal ideals $\mathfrak m_1,\ldots,\mathfrak m_n$, and let $M$ be a finitely generated projective module such that $M_{\mathfrak m_i}$ has the same rank for every $i$, then $M$ is free. ( C.f. wxu's answer in How can I find an element $x\not\in\mathfrak mM_{\mathfrak m}$ for every maximal ideal $\mathfrak m$ )

Can we apply this statement? ; i.e., can we show that $M'$ is finitely generated projective module over $A'$ and its localizations with respect to the maximal ideals of $A'$ have same rank of $r$ ? I don't know how to connect the finite locally freeness of $\tilde{M}$ to property of the localization $M'=S^{-1}M$.

( Continuing argument ) Now, let $\tau$ denotes the jacobson radical of $A'$. Then $A'/\tau$ is a product of fields ( $\because$ The chinese remainder theorem )

Q.2. Then, second question is, $M'/\tau M'$ is free $A'/\tau$-module of rank $r$?

I am trying to apply next argument ( C.f. Qiaochu Yuan's answer for About modules over finite direct product of fields ( finite direct product of local rings is local ? ) : Let $\{\mathfrak{m}'_1, \dots , \mathfrak{m}'_l \} \subseteq A'$ be the finitely many maximal ideals. Then $A'/\tau \cong A'/\mathfrak{m}'_1 \times \cdot \times A'/\mathfrak{m}'_l$. Then $M'/\tau M'$ can be decomposible into the form $V_1 \oplus \cdots \oplus V_{l}$, where each $V_i$ is a $A'/\mathfrak{m}'_i$-vector space ( Can we explicitly exhibit them ? ). These modules are always projective, but they are free iff each $V_i$ has the same dimension. But I'm lost at this point. I think I can handle this problem by knowing explicit form of $\mathfrak{m}'_i$ and $V_i$.

Can anyone help?

This question originaes from following proof of the Lemma 7.43 of the Gortz, Wedhorn's Algebraic Geometry book :

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Why the underlined statements ( the freeness of $M'/\tau M'$ of rank $r$ and the freeness of $M'$ of rank $r$ ; I guess that we can even show the freeness of $M'$, instead of local freeness. True? ; If so, then since surjective homomorphism between finite free modules of same rank is isomorphism, the $A'^r \to M'$ in the proof is an isomorphism ) are true? I want to understand this stuff desperately.

Plantation
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1 Answers1

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For the first question appearing in the post: the prime ideals of $S^{-1}A$ identify with the prime ideals of $A$ not meeting $S$, ie the primes of $A$ contained in $\bigcup_{i=1}^r{\mathfrak{m}_i}$. By the prime avoidance theorem, a prime of $A$ is contained in $\bigcup_{i=1}^r{\mathfrak{m}_i}$ iff it is contained in some $\mathfrak{m}_i$.

Thus the primes of $S^{-1}A$ identify with the primes of $A$ contained in some $\mathfrak{m}_i$. As such, it is clear that the maximal ideals of $S^{-1}A$ are the $\mathfrak{m}_i(S^{-1}A)$ (which can also be denoted $S^{-1}\mathfrak{m}_i$) for each $i$, so $S^{-1}A$ is semi-local.

For Q1, the assumption implies that $M$ is a finitely presented flat $A$-module, so it is projective (somewhere in Stacks). Thus $M’$ is a projective $A’$-module. The localizations of $M’$ at the maximal ideals of $A’$ are the same as the localizations of $M$ at each of the $\mathfrak{m}_i$ (see Localizing a localization at a prime ideal), so are free of rank $r$ by the assumption.

For Q2, yes, this is simply a base change of $M’$ as a $A’$-module.

Aphelli
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  • O.K. Thank you. Q.2. We can view $M'/\tau M' \cong M' \otimes_{A'} A'/\tau$. From the freeness of $M'$, how can we deduce the freeness of $M'/\tau M'$? Associated theorem? – Plantation Jan 28 '25 at 10:38
  • Note that $M'/\tau M' \cong (M' \otimes_{A'} (A'/\mathfrak{m}1')) \oplus \cdots \oplus (M' \otimes{A'} (A'/\mathfrak{m}_n')) $, where $\mathfrak{m}_i'$ are maximal ideals of $A'$. Each component has dimension $r$ over $A'/\mathfrak{m}_i'$? Then we may apply the argument below Q.2.in my question. – Plantation Jan 28 '25 at 10:53
  • O.K. $(M' \otimes_{A'} (A'/\mathfrak{m}1')) = A'^{r} \otimes{A'} ( A'/\mathfrak{m}1') = (A'\oplus \cdots \oplus A') \otimes{A'}(A'/\mathfrak{m}_1') = (A'/\mathfrak{m'}_1)^{r}$. Thank you. I learned a lot. – Plantation Jan 28 '25 at 11:01
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    Your first comment is somewhat puzzling. If the answer is not clear to you, then you really should consolidate your commutative algebra rather than learn schemes for the time being… – Aphelli Jan 28 '25 at 12:17
  • The first comment leads to the second comment, since $A'/\tau \cong A'/\mathfrak{m}'_1 \times \cdot \times A'/\mathfrak{m}'_n$ ( Chinese remainder theorem ) as in the post. I think I understand the question Q.2. Thank you :) – Plantation Jan 28 '25 at 12:32
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    No, the answer to the question in the first comment has nothing to do with the CRT. If $M’$ is a free $A’$-module of rank $r$, then $M’ \simeq (A’)^{\oplus r}$. Then $M’/\tau M’ \simeq (A’/\tau A’)^{\oplus r}$, and $M’/\tau M’$ is thus free of rank $r$ over $A’/\tau A’$. – Aphelli Jan 28 '25 at 13:20
  • Aha.. O.K. Such a simple route. Thanks for pointing out. – Plantation Jan 28 '25 at 13:24
  • P.s. In the proof of the Gortz's Lemma 7.43 above, to show that $M'/\tau M'$ is free of rank $r$ over $A'/\tau$, where the property that "$A'/\tau$ is finite product of fields " is used? Redundant? – Plantation Feb 01 '25 at 12:02
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    No, it’s not redundant because they’re not using the freeness criterion for modules over semi-local rings (rather, they reprove it). Again, this and your adding in the linked MSE question in the edit strike me as adding details to things you should master inside out by now. Do some more commutative algebra, the scheme theory will only be getting harder! – Aphelli Feb 01 '25 at 17:16
  • I understood your comment as, in the proof of the Gortz's Lemma 7.43, "$A'/\tau$ is finite product of fields" is used for proof of the statement "As the rank of $\mathcal{E}$ is constant , $M'/\tau M'$ is free (of rank $r$)" 'implicitly', but in our approach we bypassed this argument by using the freeness criterion for modules over semi-local rings ( so that we deduce $M' \cong A'^r \Rightarrow M'/\tau M' \cong (A'/\tau)^{r} $ ). Right? – Plantation Feb 02 '25 at 00:37