The Stacks project

Proof. By assumption the image $T$ of $V(g) \subset \mathop{\mathrm{Spec}}(S')$ under the morphism $\mathop{\mathrm{Spec}}(S') \to \mathop{\mathrm{Spec}}(R)$ does not contain $\mathfrak p$. By Section 10.41 especially, Lemma 10.41.6 we see $T$ is closed. Pick $f \in R$, $f \not\in \mathfrak p$ such that $T \cap D(f) = \emptyset $. Then we see that $g$ becomes invertible in $S'_ f$. Hence $S'_ f \cong S_ f$. Thus $S_ f$ is both of finite type and integral over $R_ f$, hence finite. $\square$

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:

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

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. $\square$

Proof. Denote $F = S \otimes _ R \kappa (\mathfrak p)$ the fibre ring of $S/R$ at the prime $\mathfrak p$. As $F$ is of finite type over $\kappa (\mathfrak p)$ it is Noetherian and hence $\mathop{\mathrm{Spec}}(F)$ has finitely many isolated closed points. If there are no isolated closed points, i.e., no primes $\mathfrak q$ of $S$ over $\mathfrak p$ such that $S/R$ is quasi-finite at $\mathfrak q$, then the lemma holds. If there exists at least one such prime $\mathfrak q$, then we may apply Lemma 10.145.2. This gives a diagram

as in said lemma. Since the residue fields at $\mathfrak p$ and $\mathfrak p'$ are the same, the fibre rings of $S/R$ and $(A_1 \times B')/R'$ are the same. Hence, by induction on the number of isolated closed points of the fibre we may assume that the lemma holds for $R' \to B'$ and $\mathfrak p'$. Thus we get an étale ring map $R' \to R''$, a prime $\mathfrak p'' \subset R''$ and a decomposition

We omit the verification that the ring map $R \to R''$, the prime $\mathfrak p''$ and the resulting decomposition

is a solution to the problem posed in the lemma. $\square$

Proof. The strategy of the proof is to make two étale ring extensions: first we control the residue fields, then we apply Lemma 10.145.3.

Denote $F = S \otimes _ R \kappa (\mathfrak p)$ the fibre ring of $S/R$ at the prime $\mathfrak p$. As in the proof of Lemma 10.145.3 there are finitely may primes, say $\mathfrak q_1, \ldots , \mathfrak q_ n$ of $S$ lying over $R$ at which the ring map $R \to S$ is quasi-finite. Let $\kappa (\mathfrak p) \subset L_ i \subset \kappa (\mathfrak q_ i)$ be the subfield such that $\kappa (\mathfrak p) \subset L_ i$ is separable, and the field extension $\kappa (\mathfrak q_ i)/L_ i$ is purely inseparable. Let $L/\kappa (\mathfrak p)$ be a finite Galois extension into which $L_ i$ embeds for $i = 1, \ldots , n$. By Lemma 10.144.3 we can find an étale ring extension $R \to R'$ together with a prime $\mathfrak p'$ lying over $\mathfrak p$ such that the field extension $\kappa (\mathfrak p')/\kappa (\mathfrak p)$ is isomorphic to $\kappa (\mathfrak p) \subset L$. Thus the fibre ring of $R' \otimes _ R S$ at $\mathfrak p'$ is isomorphic to $F \otimes _{\kappa (\mathfrak p)} L$. The primes lying over $\mathfrak q_ i$ correspond to primes of $\kappa (\mathfrak q_ i) \otimes _{\kappa (\mathfrak p)} L$ which is a product of fields purely inseparable over $L$ by our choice of $L$ and elementary field theory. These are also the only primes over $\mathfrak p'$ at which $R' \to R' \otimes _ R S$ is quasi-finite, by Lemma 10.122.8. Hence after replacing $R$ by $R'$, $\mathfrak p$ by $\mathfrak p'$, and $S$ by $R' \otimes _ R S$ we may assume that for all primes $\mathfrak q$ lying over $\mathfrak p$ for which $S/R$ is quasi-finite the field extensions $\kappa (\mathfrak q)/\kappa (\mathfrak p)$ are purely inseparable.

Next apply Lemma 10.145.3. The result is what we want since the field extensions do not change under this étale ring extension. $\square$