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

1. $S$ is of finite presentation over $R$,

2. $\varphi$ 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. $\square$

Comment #251 by Roland Paulin on

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 on

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

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 $A=(a_{ji})$.

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