Lemma 10.126.6 (00QR)—The Stacks project

Lemma 10.126.6. Let $\varphi : R \to S$ be a ring map. Let $\mathfrak q \subset S$ be a prime lying over $\mathfrak p \subset R$. Assume

$S$ is of finite presentation over $R$,
$\varphi $ induces an isomorphism $R_\mathfrak p \cong S_\mathfrak q$.

Then there exist $f \in R$, $f \not\in \mathfrak p$ and an $R_ f$-algebra $C$ such that $S_ f \cong R_ f \times C$ as $R_ f$-algebras.

Proof. Write $S = R[x_1, \ldots , x_ n]/(g_1, \ldots , g_ m)$. Let $a_ i \in R_\mathfrak p$ be an element mapping to the image of $x_ i$ in $S_\mathfrak q$. Write $a_ i = b_ i/f$ for some $f \in R$, $f \not\in \mathfrak p$. After replacing $R$ by $R_ f$ and $x_ i$ by $x_ i - a_ i$ we may assume that $S = R[x_1, \ldots , x_ n]/(g_1, \ldots , g_ m)$ such that $x_ i$ maps to zero in $S_\mathfrak q$. Then if $c_ j$ denotes the constant term of $g_ j$ we conclude that $c_ j$ maps to zero in $R_\mathfrak p$. After another replacement of $R$ we may assume that the constant coefficients $c_ j$ of the $g_ j$ are zero. Thus we obtain an $R$-algebra map $S \to R$, $x_ i \mapsto 0$ whose kernel is the ideal $(x_1, \ldots , x_ n)$.

Note that $\mathfrak q = \mathfrak pS + (x_1, \ldots , x_ n)$. Write $g_ j = \sum a_{ji}x_ i + h.o.t.$. Since $S_\mathfrak q = R_\mathfrak p$ we have $\mathfrak p \otimes \kappa (\mathfrak p) = \mathfrak q \otimes \kappa (\mathfrak q)$. It follows that $m \times n$ matrix $A = (a_{ji})$ defines a surjective map $\kappa (\mathfrak p)^{\oplus m} \to \kappa (\mathfrak p)^{\oplus n}$. Thus after inverting some element of $R$ not in $\mathfrak p$ we may assume there are $b_{ij} \in R$ such that $\sum b_{ij} g_ j = x_ i + h.o.t.$. We conclude that $(x_1, \ldots , x_ n) = (x_1, \ldots , x_ n)^2$ in $S$. It follows from Lemma 10.21.5 that $(x_1, \ldots , x_ n)$ is generated by an idempotent $e$. Setting $C = eS$ finishes the proof. $\square$

Comments (5)

Comment #251 by Roland Paulin on July 25, 2013 at 12:35

Hi!

I think this Lemma (Tag 00QR) is false. The proof says we can assume that $\varphi$ is surjective. This seems to be a mistake. Here is a counterexample: Let $A$ be any local ring with maximal ideal $\mathfrak{m}$ . Let $R=S'=A$ and $\mathfrak{q}' = \mathfrak{m}$ . Let $S = A \times A$ and $\mathfrak{q} = \mathfrak{m} \times A$ . Then $S' \to S$ is of finite presentation, $\varphi^{-1}(\mathfrak{\mathfrak{q}}) = \mathfrak{q}'$ , and $A = S'_{\mathfrak{q}'} \to S_{\mathfrak{q}} = A$ is an isomorphism. However if $g \in S' \setminus \mathfrak{q'}$ , then $g \in (S')^{\times}$ , hence the map $S'_g \to S_{\varphi(g)}$ is just $\varphi$ , which is not an isomorphism (it is injective, but not surjective!).

The statement of the lemma is a bit peculiar, because it seems that $R$ could be omitted: we could just assume that $S' \to S$ is of finite presentation. I wonder, maybe the conditions were intended to say finite $R$ -module / finitely presented $R$ -module instead of $R$ -algebra of finite type / of finite presentation?

Roland

Comment #252 by Johan on July 25, 2013 at 19:31

Woops! Yes, that is completely wrong! And indeed the formulation was suspiciously confusing. I've now changed the statement (and the proof) to something more true (hopefully). This will be online above in about 5 minutes. For those of you who want to see the wrong statement and wrong proof take a look at the red text in the commit.

This also means you win the last Stacks project T-shirt!

Comment #3252 by Dario Weißmann on April 11, 2018 at 15:42

Typos: 'Then if $c_j$ denotes the constant term of $g_i$ we conclude that $c_i$ ...' The $i$ should be $j$ .

$A=(a_{ij})$ should be $A=(a_{ji})$ .

Comment #3348 by Johan on May 19, 2018 at 16:27

Thanks, fixed here.

Comment #8315 by Et on April 12, 2023 at 12:27

Could more details be added to the proof? I've spent some time on the second paragraph and I cannot understand most of the justifications there. Particularly, I don't see why the two asserted tensor products are equal, why the map between vector spaces is surjective, and why the expressions you find for the x_i imply the ideal (x_1,...x_n) is idempotent.

Comments (5)

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