The Stacks project

Proof. Let $I_0 \subset I_1 \subset I_2 \subset \ldots$ be a growing sequence of ideals of $R$. By assumption we have $I_ nS = I_{n +1}S = I_{n + 2}S = \ldots$ for some $n$. Since $R \to S$ is flat we have $I_ kS = I_ k \otimes _ R S$. Hence, as $R \to S$ is faithfully flat we see that $I_ nS = I_{n +1}S = I_{n + 2}S = \ldots$ implies that $I_ n = I_{n +1} = I_{n + 2} = \ldots$ as desired. $\square$

Proof. This is clear as $R \to S$ is injective. $\square$

Proof. Since $S$ is reduced it follows that $R$ is reduced. Let $\mathfrak p$ be a prime of $R$. We have to show that $R_{\mathfrak p}$ is a normal domain. Since $S_{\mathfrak p}$ is faithfully over $R_{\mathfrak p}$ too we may assume that $R$ is local with maximal ideal $\mathfrak m$. Let $\mathfrak q$ be a prime of $S$ lying over $\mathfrak m$. Then we see that $R \to S_{\mathfrak q}$ is faithfully flat (Lemma 10.39.17). Hence we may assume $S$ is local as well. In particular $S$ is a normal domain. Since $R \to S$ is faithfully flat and $S$ is a normal domain we see that $R$ is a domain. Next, suppose that $a/b$ is integral over $R$ with $a, b \in R$. Then $a/b \in S$ as $S$ is normal. Hence $a \in bS$. This means that $a : R \to R/bR$ becomes the zero map after base change to $S$. By faithful flatness we see that $a \in bR$, so $a/b \in R$. Hence $R$ is normal. $\square$

Proof. We see that $R$ is Noetherian by Lemma 10.164.1. Let $\mathfrak p \subset R$ be a prime. Choose a prime $\mathfrak q \subset S$ lying over $\mathfrak p$. Then Lemma 10.110.9 applies to $R_\mathfrak p \to S_\mathfrak q$ and we conclude that $R_\mathfrak p$ is regular. Since $\mathfrak p$ was arbitrary we see $R$ is regular. $\square$

Proof. We have already seen that (1) and (2) imply that $R$ is Noetherian, see Lemma 10.164.1. Let $\mathfrak p \subset R$ be a prime ideal. Choose a prime $\mathfrak q \subset S$ lying over $\mathfrak p$ which corresponds to a minimal prime of the fibre ring $S \otimes _ R \kappa (\mathfrak p)$. Then $A = R_{\mathfrak p} \to S_{\mathfrak q} = B$ is a flat local ring homomorphism of Noetherian local rings with $\mathfrak m_ AB$ an ideal of definition of $B$. Hence $\dim (A) = \dim (B)$ (Lemma 10.112.7) and $\text{depth}(A) = \text{depth}(B)$ (Lemma 10.163.2). Hence since $B$ has $(S_ k)$ we see that $A$ has $(S_ k)$. $\square$

Proof. We have already seen that (1) and (2) imply that $R$ is Noetherian, see Lemma 10.164.1. Let $\mathfrak p \subset R$ be a prime ideal and assume $\dim (R_{\mathfrak p}) \leq k$. Choose a prime $\mathfrak q \subset S$ lying over $\mathfrak p$ which corresponds to a minimal prime of the fibre ring $S \otimes _ R \kappa (\mathfrak p)$. Then $A = R_{\mathfrak p} \to S_{\mathfrak q} = B$ is a flat local ring homomorphism of Noetherian local rings with $\mathfrak m_ AB$ an ideal of definition of $B$. Hence $\dim (A) = \dim (B)$ (Lemma 10.112.7). As $S$ has $(R_ k)$ we conclude that $B$ is a regular local ring. By Lemma 10.110.9 we conclude that $A$ is regular. $\square$

Proof. Recall that a Nagata ring is the same thing as a Noetherian universally Japanese ring (Proposition 10.162.15). We have already seen that $R$ is Noetherian in Lemma 10.164.1. Let $R \to A$ be a finite type ring map into a domain. According to Lemma 10.162.3 it suffices to check that $A$ is N-1. It is clear that $B = A \otimes _ R S$ is a finite type $S$-algebra and hence Nagata (Proposition 10.162.15). Since $A \to B$ is smooth (Lemma 10.137.4) we see that $B$ is reduced (Lemma 10.163.7). Since $B$ is Noetherian it has only a finite number of minimal primes $\mathfrak q_1, \ldots , \mathfrak q_ t$ (see Lemma 10.31.6). As $A \to B$ is flat each of these lies over $(0) \subset A$ (by going down, see Lemma 10.39.19) The total ring of fractions $Q(B)$ is the product of the $L_ i = \kappa (\mathfrak q_ i)$ (Lemmas 10.25.4 and 10.25.1). Moreover, the integral closure $B'$ of $B$ in $Q(B)$ is the product of the integral closures $B_ i'$ of the $B/\mathfrak q_ i$ in the factors $L_ i$ (compare with Lemma 10.37.16). Since $B$ is universally Japanese the ring extensions $B/\mathfrak q_ i \subset B_ i'$ are finite and we conclude that $B' = \prod B_ i'$ is finite over $B$. Since $A \to B$ is flat we see that any nonzerodivisor on $A$ maps to a nonzerodivisor on $B$. The corresponding map

is injective (we used Lemma 10.12.15). Via this map $A'$ maps into $B'$. This induces a map

which is injective (by the above and the flatness of $A \to B$). Since $B'$ is a finite $B$-module and $B$ is Noetherian we see that $A' \otimes _ A B$ is a finite $B$-module. Hence there exist finitely many elements $x_ i \in A'$ such that the elements $x_ i \otimes 1$ generate $A' \otimes _ A B$ as a $B$-module. Finally, by faithful flatness of $A \to B$ we conclude that the $x_ i$ also generated $A'$ as an $A$-module, and we win. $\square$