The Stacks project

Proof. In the statement and in the proof below, we take the depth of $M$ as an $R$-module, the depth of $M \otimes _ R N$ as an $S$-module, and the depth of $N/\mathfrak m_ RN$ as an $S/\mathfrak m_ RS$-module. Denote $n$ the right hand side. First assume that $n$ is zero. Then both $\text{depth}(M) = 0$ and $\text{depth}(N/\mathfrak m_ RN) = 0$. This means there is a $z \in M$ whose annihilator is $\mathfrak m_ R$ and a $\overline{y} \in N/\mathfrak m_ RN$ whose annihilator is $\mathfrak m_ S/\mathfrak m_ RS$. Let $y \in N$ be a lift of $\overline{y}$. Since $N$ is flat over $R$ the map $z : R/\mathfrak m_ R \to M$ produces an injective map $N/\mathfrak m_ RN \to M \otimes _ R N$. Hence the annihilator of $z \otimes y$ is $\mathfrak m_ S$. Thus $\text{depth}(M \otimes _ R N) = 0$ as well.

Assume $n > 0$. If $\text{depth}(N/\mathfrak m_ RN) > 0$, then we may choose $f \in \mathfrak m_ S$ mapping to $\overline{f} \in S/\mathfrak m_ RS$ which is a nonzerodivisor on $N/\mathfrak m_ RN$. Then $\text{depth}(N/\mathfrak m_ RN) = \text{depth}(N/(f, \mathfrak m_ R)N) + 1$ by Lemma 10.72.7. According to Lemma 10.99.1 the element $f \in S$ is a nonzerodivisor on $N$ and $N/fN$ is flat over $R$. Hence by induction on $n$ we have

Because $N/fN$ is flat over $R$ the sequence

is exact where the first map is multiplication by $f$ (Lemma 10.39.12). Hence by Lemma 10.72.7 we find that $\text{depth}(M \otimes _ R N) = \text{depth}(M \otimes _ R N/fN) + 1$ and we conclude that equality holds in the formula of the lemma.

If $n > 0$, but $\text{depth}(N/\mathfrak m_ RN) = 0$, then we can choose $f \in \mathfrak m_ R$ which is a nonzerodivisor on $M$. As $N$ is flat over $R$ it is also the case that $f$ is a nonzerodivisor on $M \otimes _ R N$. By induction on $n$ again we have

In this case $\text{depth}(M \otimes _ R N) = \text{depth}(M/fM \otimes _ R N) + 1$ and $\text{depth}(M) = \text{depth}(M/fM) + 1$ by Lemma 10.72.7 and we conclude that equality holds in the formula of the lemma. $\square$

Proof. This is a special case of Lemma 10.163.1. $\square$

Proof. Follows from the definitions and Lemmas 10.163.2 and 10.112.7. $\square$

Proof. Let $\mathfrak q$ be a prime of $S$ lying over a prime $\mathfrak p$ of $R$. By Lemma 10.163.2 we have

On the other hand, we have

by Lemma 10.112.6. (Actually equality holds, by Lemma 10.112.7 but strictly speaking we do not need this.) Finally, as the fibre rings of the map are assumed $(S_ k)$ we see that $\text{depth}(S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}) \geq \min (k, \dim (S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}))$. Thus the lemma follows by the following string of inequalities

as desired. $\square$

Proof. Let $\mathfrak q$ be a prime of $S$ lying over a prime $\mathfrak p$ of $R$. Assume that $\dim (S_{\mathfrak q}) \leq k$. Since $\dim (S_{\mathfrak q}) = \dim (R_{\mathfrak p}) + \dim (S_{\mathfrak q}/\mathfrak pS_{\mathfrak q})$ by Lemma 10.112.7 we see that $\dim (R_{\mathfrak p}) \leq k$ and $\dim (S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}) \leq k$. Hence $R_{\mathfrak p}$ and $S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}$ are regular by assumption. It follows that $S_{\mathfrak q}$ is regular by Lemma 10.112.8. $\square$

Proof. For Noetherian rings reduced is the same as having properties $(S_1)$ and $(R_0)$, see Lemma 10.157.3. Thus we know $R$ and the fibre rings have these properties. Hence we may apply Lemmas 10.163.4 and 10.163.5 and we see that $S$ is $(S_1)$ and $(R_0)$, in other words reduced by Lemma 10.157.3 again. $\square$

Proof. Observe that $R \to S$ is flat with regular fibres (see the list of results on smooth ring maps in Section 10.142). In particular, the fibres are reduced. Thus if $R$ is Noetherian, then $S$ is Noetherian and we get the result from Lemma 10.163.6.

In the general case we may find a finitely generated $\mathbf{Z}$-subalgebra $R_0 \subset R$ and a smooth ring map $R_0 \to S_0$ such that $S \cong R \otimes _{R_0} S_0$, see remark (10) in Section 10.142. Now, if $x \in S$ is an element with $x^2 = 0$, then we can enlarge $R_0$ and assume that $x$ comes from an element $x_0 \in S_0$. After enlarging $R_0$ once more we may assume that $x_0^2 = 0$ in $S_0$. However, since $R_0 \subset R$ is reduced we see that $S_0$ is reduced and hence $x_0 = 0$ as desired. $\square$

Proof. For a Noetherian ring being normal is the same as having properties $(S_2)$ and $(R_1)$, see Lemma 10.157.4. Thus we know $R$ and the fibre rings have these properties. Hence we may apply Lemmas 10.163.4 and 10.163.5 and we see that $S$ is $(S_2)$ and $(R_1)$, in other words normal by Lemma 10.157.4 again. $\square$

Proof. Observe that $R \to S$ is flat with regular fibres (see the list of results on smooth ring maps in Section 10.142). In particular, the fibres are normal. Thus if $R$ is Noetherian, then $S$ is Noetherian and we get the result from Lemma 10.163.8.

The general case. First note that $R$ is reduced and hence $S$ is reduced by Lemma 10.163.7. Let $\mathfrak q$ be a prime of $S$ and let $\mathfrak p$ be the corresponding prime of $R$. Note that $R_{\mathfrak p}$ is a normal domain. We have to show that $S_{\mathfrak q}$ is a normal domain. To do this we may replace $R$ by $R_{\mathfrak p}$ and $S$ by $S_{\mathfrak p}$. Hence we may assume that $R$ is a normal domain.

Assume $R \to S$ smooth, and $R$ a normal domain. We may find a finitely generated $\mathbf{Z}$-subalgebra $R_0 \subset R$ and a smooth ring map $R_0 \to S_0$ such that $S \cong R \otimes _{R_0} S_0$, see remark (10) in Section 10.142. As $R_0$ is a Nagata domain (see Proposition 10.162.16) we see that its integral closure $R_0'$ is finite over $R_0$. Moreover, as $R$ is a normal domain it is clear that $R_0' \subset R$. Hence we may replace $R_0$ by $R_0'$ and $S_0$ by $R_0' \otimes _{R_0} S_0$ and assume that $R_0$ is a normal Noetherian domain. By the first paragraph of the proof we conclude that $S_0$ is a normal ring (it need not be a domain of course). In this way we see that $R = \bigcup R_\lambda $ is the union of normal Noetherian domains and correspondingly $S = \mathop{\mathrm{colim}}\nolimits R_\lambda \otimes _{R_0} S_0$ is the colimit of normal rings. This implies that $S$ is a normal ring. Some details omitted. $\square$

Proof. This follows from Lemma 10.163.5 applied for all $(R_ k)$ using Lemma 10.140.3 to see that the hypotheses are satisfied. $\square$