Lemma 10.158.1. Let $R \to S$ be a ring map. Assume that

$R \to S$ is faithfully flat, and

$S$ is Noetherian.

Then $R$ is Noetherian.

In this section we start proving some algebraic facts concerning the “descent” of properties of rings. It turns out that it is often “easier” to descend properties than it is to ascend them. In other words, the assumption on the ring map $R \to S$ are often weaker than the assumptions in the corresponding lemma of the preceding section. However, we warn the reader that the results on descent are often useless unless the corresponding ascent can also be shown! Here is a typical result which illustrates this phenomenon.

Lemma 10.158.1. Let $R \to S$ be a ring map. Assume that

$R \to S$ is faithfully flat, and

$S$ is Noetherian.

Then $R$ is Noetherian.

**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.
$\square$

Lemma 10.158.2. Let $R \to S$ be a ring map. Assume that

$R \to S$ is faithfully flat, and

$S$ is reduced.

Then $R$ is reduced.

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

Lemma 10.158.3. Let $R \to S$ be a ring map. Assume that

$R \to S$ is faithfully flat, and

$S$ is a normal ring.

Then $R$ is a normal ring.

**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 10.38.17). 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.
$\square$

Lemma 10.158.4. Let $R \to S$ be a ring map. Assume that

$R \to S$ is faithfully flat, and

$S$ is a regular ring.

Then $R$ is a regular ring.

**Proof.**
We see that $R$ is Noetherian by Lemma 10.158.1. Let $\mathfrak p \subset R$ be a prime. Choose a prime $\mathfrak q \subset S$ lying over $\mathfrak p$. Then Lemma 10.109.9 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.
$\square$

Lemma 10.158.5. Let $R \to S$ be a ring map. Assume that

$R \to S$ is faithfully flat, and

$S$ is Noetherian and has property $(S_ k)$.

Then $R$ is Noetherian and has property $(S_ k)$.

**Proof.**
We have already seen that (1) and (2) imply that $R$ is Noetherian, see Lemma 10.158.1. 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 10.111.7) and $\text{depth}(A) = \text{depth}(B)$ (Lemma 10.157.2). Hence since $B$ has $(S_ k)$ we see that $A$ has $(S_ k)$.
$\square$

Lemma 10.158.6. Let $R \to S$ be a ring map. Assume that

$R \to S$ is faithfully flat, and

$S$ is Noetherian and has property $(R_ k)$.

Then $R$ is Noetherian and has property $(R_ k)$.

**Proof.**
We have already seen that (1) and (2) imply that $R$ is Noetherian, see Lemma 10.158.1. 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 10.111.7). As $S$ has $(R_ k)$ we conclude that $B$ is a regular local ring. By Lemma 10.109.9 we conclude that $A$ is regular.
$\square$

Lemma 10.158.7. Let $R \to S$ be a ring map. Assume that

$R \to S$ is smooth and surjective on spectra, and

$S$ is a Nagata ring.

Then $R$ is a Nagata ring.

**Proof.**
Recall that a Nagata ring is the same thing as a Noetherian universally Japanese ring (Proposition 10.156.15). We have already seen that $R$ is Noetherian in Lemma 10.158.1. Let $R \to A$ be a finite type ring map into a domain. According to Lemma 10.156.3 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 10.156.15). Since $A \to B$ is smooth (Lemma 10.135.4) we see that $B$ is reduced (Lemma 10.157.7). Since $B$ is Noetherian it has only a finite number of minimal primes $\mathfrak q_1, \ldots , \mathfrak q_ t$ (see Lemma 10.30.6). As $A \to B$ is flat each of these lies over $(0) \subset A$ (by going down, see Lemma 10.38.18) The total ring of fractions $Q(B)$ is the product of the $L_ i = \kappa (\mathfrak q_ i)$ (Lemmas 10.24.4 and 10.24.1). 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 10.36.16). 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

\[ Q(A) \otimes _ A B = (A \setminus \{ 0\} )^{-1}A \otimes _ A B = (A \setminus \{ 0\} )^{-1}B \to Q(B) \]

is injective (we used Lemma 10.11.15). Via this map $A'$ maps into $B'$. This induces a map

\[ A' \otimes _ A B \longrightarrow B' \]

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. $\square$

Remark 10.158.8. The property of being “universally catenary” does not descend; not even along étale ring maps. In Examples, Section 102.16 there is a construction of a finite ring map $A \to B$ with $A$ local Noetherian and not universally catenary, $B$ semi-local with two maximal ideals $\mathfrak m$, $\mathfrak n$ with $B_{\mathfrak m}$ and $B_{\mathfrak n}$ regular of dimension $2$ and $1$ respectively, and the same residue fields as that of $A$. Moreover, $\mathfrak m_ A$ generates the maximal ideal in both $B_{\mathfrak m}$ and $B_{\mathfrak n}$ (so $A \to B$ is unramified as well as finite). By Lemma 10.147.11 there exists a local étale ring map $A \to A'$ such that $B \otimes _ A A' = B_1 \times B_2$ decomposes with $A' \to B_ i$ surjective. This shows that $A'$ has two minimal primes $\mathfrak q_ i$ with $A'/\mathfrak q_ i \cong B_ i$. Since $B_ i$ is regular local (since it is étale over either $B_{\mathfrak m}$ or $B_{\mathfrak n}$) we conclude that $A'$ is universally catenary.

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