The Stacks project

Lemma 10.145.2. 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 10.123.12 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 10.36.19. Note that $\mathfrak q$ defines an isolated closed point $\overline{\mathfrak q}$ of $\mathop{\mathrm{Spec}}(F)$ (see Definition 10.122.3). 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 10.24.3 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 $\overline{h}(0) \not= 0$. After replacing $f$ by $x f$ if necessary, we may assume $e \geq 1$. By Lemma 10.143.13 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 10.145.1 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. $\square$

Comments (10)

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 is the only prime of which does not contain the image of in . I thought implies that, since does not contain the image of (as and does not contain by construction), we shall have . Here we may assume since we can multiply by if necessary. Also, for any other prime of than , since we see that the constant term of is not in while all other terms of are in . Hence it seems we arrive at the conslusion that is the only prime of which contains the image of . Maybe is the only prime of which does not contain the image of in 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 by as you point out. Of course, the only thing we are using is that the product decomposition of agrees with the given one on . So really all this is saying is that the images of and under the map described in the proof (after the erroneous sentence) give the product decomposition . This you can just verify by hand using that maps to . 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 and , and , it seems to me that maps to rather than .

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 becomes ? Everything else seems correct to me. Thanks!

Comment #5838 by Yicheng on

Yes exactly!

Comment #7188 by nkym on

implicitly being assumed to be positive, as mentioned in earlier comments, is not included in the proof. Also, the last of the third last paragraph might be in my guess.

Comment #7317 by on

Thanks! There must be a curse on this lemma. Changes are here.

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 00UJ. Beware of the difference between the letter 'O' and the digit '0'.