Lemma 10.144.2. Let $R \to S$ be a ring map. Let $\mathfrak q \subset S$ be a prime lying over the prime $\mathfrak p \subset R$. Assume $R \to S$ finite type and quasi-finite at $\mathfrak q$. Then there exists

1. an étale ring map $R \to R'$,

2. a prime $\mathfrak p' \subset R'$ lying over $\mathfrak p$,

3. a product decomposition

$R' \otimes _ R S = A \times B$

with the following properties

1. $\kappa (\mathfrak p) = \kappa (\mathfrak p')$,

2. $R' \to A$ is finite,

3. $A$ has exactly one prime $\mathfrak r$ lying over $\mathfrak p'$, and

4. $\mathfrak r$ lies over $\mathfrak q$.

Proof. Let $S' \subset S$ be the integral closure of $R$ in $S$. Let $\mathfrak q' = S' \cap \mathfrak q$. By Zariski's Main Theorem 10.122.12 there exists a $g \in S'$, $g \not\in \mathfrak q'$ such that $S'_ g \cong S_ g$. Consider the fibre rings $F = S \otimes _ R \kappa (\mathfrak p)$ and $F' = S' \otimes _ R \kappa (\mathfrak p)$. Denote $\overline{\mathfrak q}'$ the prime of $F'$ corresponding to $\mathfrak q'$. Since $F'$ is integral over $\kappa (\mathfrak p)$ we see that $\overline{\mathfrak q}'$ is a closed point of $\mathop{\mathrm{Spec}}(F')$, see Lemma 10.35.19. Note that $\mathfrak q$ defines an isolated closed point $\overline{\mathfrak q}$ of $\mathop{\mathrm{Spec}}(F)$ (see Definition 10.121.3). Since $S'_ g \cong S_ g$ we have $F'_ g \cong F_ g$, so $\overline{\mathfrak q}$ and $\overline{\mathfrak q}'$ have isomorphic open neighbourhoods in $\mathop{\mathrm{Spec}}(F)$ and $\mathop{\mathrm{Spec}}(F')$. We conclude the set $\{ \overline{\mathfrak q}'\} \subset \mathop{\mathrm{Spec}}(F')$ is open. Combined with $\mathfrak q'$ being closed (shown above) we conclude that $\overline{\mathfrak q}'$ defines an isolated closed point of $\mathop{\mathrm{Spec}}(F')$ as well.

An additional small remark is that under the map $\mathop{\mathrm{Spec}}(F) \to \mathop{\mathrm{Spec}}(F')$ the point $\overline{\mathfrak q}$ is the only point mapping to $\overline{\mathfrak q}'$. This follows from the discussion above.

By Lemma 10.23.3 we may write $F' = F'_1 \times F'_2$ with $\mathop{\mathrm{Spec}}(F'_1) = \{ \overline{\mathfrak q}'\}$. Since $F' = S' \otimes _ R \kappa (\mathfrak p)$, there exists an $s' \in S'$ which maps to the element $(r, 0) \in F'_1 \times F'_2 = F'$ for some $r \in R$, $r \not\in \mathfrak p$. In fact, what we will use about $s'$ is that it is an element of $S'$, not contained in $\mathfrak q'$, and contained in any other prime lying over $\mathfrak p$.

Let $f(x) \in R[x]$ be a monic polynomial such that $f(s') = 0$. Denote $\overline{f} \in \kappa (\mathfrak p)[x]$ the image. We can factor it as $\overline{f} = x^ e \overline{h}$ where $\overline{h}(0) \not= 0$. By Lemma 10.142.13 we can find an étale ring extension $R \to R'$, a prime $\mathfrak p'$ lying over $\mathfrak p$, and a factorization $f = h i$ in $R'[x]$ such that $\kappa (\mathfrak p) = \kappa (\mathfrak p')$, $\overline{h} = h \bmod \mathfrak p'$, $x^ e = i \bmod \mathfrak p'$, and we can write $a h + b i = 1$ in $R'[x]$ (for suitable $a, b$).

Consider the elements $h(s'), i(s') \in R' \otimes _ R S'$. By construction we have $h(s')i(s') = f(s') = 0$. On the other hand they generate the unit ideal since $a(s')h(s') + b(s')i(s') = 1$. Thus we see that $R' \otimes _ R S'$ is the product of the localizations at these elements:

$R' \otimes _ R S' = (R' \otimes _ R S')_{i(s')} \times (R' \otimes _ R S')_{h(s')} = S'_1 \times S'_2$

Moreover this product decomposition is compatible with the product decomposition we found for the fibre ring $F'$; this comes from our choices of $s', i, h$ which guarantee that $\overline{\mathfrak q}'$ is the only prime of $F'$ which does not contain the image of $i(s')$ in $F'$. Here we use that the fibre ring of $R'\otimes _ R S'$ over $R'$ at $\mathfrak p'$ is the same as $F'$ due to the fact that $\kappa (\mathfrak p) = \kappa (\mathfrak p')$. It follows that $S'_1$ has exactly one prime, say $\mathfrak r'$, lying over $\mathfrak p'$ and that this prime lies over $\mathfrak q$. Hence the element $g \in S'$ maps to an element of $S'_1$ not contained in $\mathfrak r'$.

The base change $R'\otimes _ R S$ inherits a similar product decomposition

$R' \otimes _ R S = (R' \otimes _ R S)_{i(s')} \times (R' \otimes _ R S)_{h(s')} = S_1 \times S_2$

It follows from the above that $S_1$ has exactly one prime, say $\mathfrak r$, lying over $\mathfrak p'$ (consider the fibre ring as above), and that this prime lies over $\mathfrak q$.

Now we may apply Lemma 10.144.1 to the ring maps $R' \to S'_1 \to S_1$, the prime $\mathfrak p'$ and the element $g$ to see that after replacing $R'$ by a principal localization we can assume that $S_1$ is finite over $R'$ as desired. $\square$

Comment #2353 by Eric Ahlqvist on

"x^e = h \bmod \mathfrak p'" should be "\overline{h} = h \bmod \mathfrak p'"?

Comment #4955 by awllower on

I am confused why $\overline{\mathfrak q}'$ is the only prime of $F'$ which does not contain the image of $h(s')$ in $F'$. I thought $h(s')i(s')=0$ implies that, since $\overline{\mathfrak q}'$ does not contain the image of $i(s')$ (as $i(s')\equiv (s')^e\pmod{\mathfrak p'}$ and $\overline{\mathfrak q}'$ does not contain $s'$ by construction), we shall have $h(s')\in\overline{\mathfrak q}'$. Here we may assume $e\geq1$ since we can multiply $f$ by $x$ if necessary. Also, for any other prime $\mathfrak q''$ of $F'$ than $\overline{\mathfrak q}'$, since $\overline h(0)\ne0$ we see that the constant term of $h$ is not in $\mathfrak q''$ while all other terms of $h(s')$ are in $\mathfrak q''$. Hence it seems we arrive at the conslusion that $\overline{q}'$ is the only prime of $F'$ which contains the image of $h(s')$. Maybe $\overline{\mathfrak q}'$ is the only prime of $F'$ which does not contain the image of $i(s')$ in $F'$ instead? I must have missed something elementary. Thanks for any help.

Comment #5210 by on

Thanks very much for pointing out this mistake. The fix is to replace $h$ by $i$ as you point out. Of course, the only thing we are using is that the product decomposition of $R' \otimes_R S'$ agrees with the given one on $F'$. So really all this is saying is that the images of $h$ and $i$ under the map $R' \otimes_R S' \to F'$ described in the proof (after the erroneous sentence) give the product decomposition $F' = F'_1 \times F'_2$. This you can just verify by hand using that $i(s')$ maps to $(r^e, 0)$. Thanks again! The fix is here.

Comment #5836 by Yicheng on

The very original version of the proof might have been right...since with the actual choice of $h$ and $i$, $S'_1$ and $S'_2$, it seems to me that $i(s')$ maps to $(0,r^e)$ rather than $(r^e,0)$.

Comment #5837 by on

@#5836 OK, I think what you mean is that in the displayed formula we should switch the factors so that what is now $S'_1$ becomes $S'_2$? Everything else seems correct to me. Thanks!

Comment #5838 by Yicheng on

Yes exactly!

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