Proposition 10.156.15 (Nagata). Let $R$ be a ring. The following are equivalent:

$R$ is a Nagata ring,

any finite type $R$-algebra is Nagata, and

$R$ is universally Japanese and Noetherian.

Proposition 10.156.15 (Nagata). Let $R$ be a ring. The following are equivalent:

$R$ is a Nagata ring,

any finite type $R$-algebra is Nagata, and

$R$ is universally Japanese and Noetherian.

**Proof.**
It is clear that a Noetherian universally Japanese ring is universally Nagata (i.e., condition (2) holds). Let $R$ be a Nagata ring. We will show that any finitely generated $R$-algebra $S$ is Nagata. This will prove the proposition.

Step 1. There exists a sequence of ring maps $R = R_0 \to R_1 \to R_2 \to \ldots \to R_ n = S$ such that each $R_ i \to R_{i + 1}$ is generated by a single element. Hence by induction it suffices to prove $S$ is Nagata if $S \cong R[x]/I$.

Step 2. Let $\mathfrak q \subset S$ be a prime of $S$, and let $\mathfrak p \subset R$ be the corresponding prime of $R$. We have to show that $S/\mathfrak q$ is N-2. Hence we have reduced to the proving the following: (*) Given a Nagata domain $R$ and a monogenic extension $R \subset S$ of domains then $S$ is N-2.

Step 3. Let $R$ be a Nagata domain and $R \subset S$ a monogenic extension of domains. Let $R \subset R'$ be the integral closure of $R$ in its fraction field. Let $S'$ be the subring of the fraction field of $S$ generated by $R'$ and $S$. As $R'$ is finite over $R$ (by the Nagata property) also $S'$ is finite over $S$. Since $S$ is Noetherian it suffices to prove that $S'$ is N-2 (Lemma 10.155.7). Hence we have reduced to proving the following: (**) Given a normal Nagata domain $R$ and a monogenic extension $R \subset S$ of domains then $S$ is N-2.

Step 4: Let $R$ be a normal Nagata domain and let $R \subset S$ be a monogenic extension of domains. Suppose the induced extension of fraction fields of $R$ and $S$ is purely transcendental. In this case $S = R[x]$. By Lemma 10.155.13 we see that $S$ is N-2. Hence we have reduced to proving the following: (**) Given a normal Nagata domain $R$ and a monogenic extension $R \subset S$ of domains inducing a finite extension of fraction fields then $S$ is N-2.

Step 5. Let $R$ be a normal Nagata domain and let $R \subset S$ be a monogenic extension of domains inducing a finite extension of fraction fields $L/K$. Choose an element $x \in S$ which generates $S$ as an $R$-algebra. Let $L \subset M$ be a finite extension of fields. Let $R'$ be the integral closure of $R$ in $M$. Then the integral closure $S'$ of $S$ in $M$ is equal to the integral closure of $R'[x]$ in $M$. Also the fraction field of $R'$ is $M$ and $R \subset R'$ is finite (by the Nagata property of $R$). This implies that $R'$ is a Nagata ring (Lemma 10.156.5). To show that $S'$ is finite over $S$ is the same as showing that $S'$ is finite over $R'[x]$. Replace $R$ by $R'$ and $S$ by $R'[x]$ to reduce to the following statement: (***) Given a normal Nagata domain $R$ with fraction field $K$, and $x \in K$, the ring $S \subset K$ generated by $R$ and $x$ is N-1.

Step 6. Let $R$ be a normal Nagata domain with fraction field $K$. Let $x = b/a \in K$. We have to show that the ring $S \subset K$ generated by $R$ and $x$ is N-1. Note that $S_ a \cong R_ a$ is normal. Hence by Lemma 10.155.15 it suffices to show that $S_{\mathfrak m}$ is N-1 for every maximal ideal $\mathfrak m$ of $S$.

With assumptions as in the preceding paragraph, pick such a maximal ideal and set $\mathfrak n = R \cap \mathfrak m$. The residue field extension $\kappa (\mathfrak n) \subset \kappa (\mathfrak m)$ is finite (Theorem 10.33.1) and generated by the image of $x$. Hence there exists a monic polynomial $f(X) = X^ d + \sum _{i = 1, \ldots , d} a_ iX^{d -i}$ with $f(x) \in \mathfrak m$. Let $K \subset K''$ be a finite extension of fields such that $f(X)$ splits completely in $K''[X]$. Let $R'$ be the integral closure of $R$ in $K''$. Let $S' \subset K'$ be the subring generated by $R'$ and $x$. As $R$ is Nagata we see $R'$ is finite over $R$ and Nagata (Lemma 10.156.5). Moreover, $S'$ is finite over $S$. If for every maximal ideal $\mathfrak m'$ of $S'$ the local ring $S'_{\mathfrak m'}$ is N-1, then $S'_{\mathfrak m}$ is N-1 by Lemma 10.155.15, which in turn implies that $S_{\mathfrak m}$ is N-1 by Lemma 10.155.7. After replacing $R$ by $R'$ and $S$ by $S'$, and $\mathfrak m$ by any of the maximal ideals $\mathfrak m'$ lying over $\mathfrak m$ we reach the situation where the polynomial $f$ above split completely: $f(X) = \prod _{i = 1, \ldots , d} (X - a_ i)$ with $a_ i \in R$. Since $f(x) \in \mathfrak m$ we see that $x - a_ i \in \mathfrak m$ for some $i$. Finally, after replacing $x$ by $x - a_ i$ we may assume that $x \in \mathfrak m$.

To recapitulate: $R$ is a normal Nagata domain with fraction field $K$, $x \in K$ and $S$ is the subring of $K$ generated by $x$ and $R$, finally $\mathfrak m \subset S$ is a maximal ideal with $x \in \mathfrak m$. We have to show $S_{\mathfrak m}$ is N-1.

We will show that Lemma 10.156.12 applies to the local ring $S_{\mathfrak m}$ and the element $x$. This will imply that $S_{\mathfrak m}$ is analytically unramified, whereupon we see that it is N-1 by Lemma 10.156.10.

We have to check properties (1), (2), (3)(a) and (3)(b). Property (1) is trivial. Let $I = \mathop{\mathrm{Ker}}(R[X] \to S)$ where $X \mapsto x$. We claim that $I$ is generated by all linear forms $aX + b$ such that $ax = b$ in $K$. Clearly all these linear forms are in $I$. If $g = a_ d X^ d + \ldots a_1 X + a_0 \in I$, then we see that $a_ dx$ is integral over $R$ (Lemma 10.122.1) and hence $b := a_ dx \in R$ as $R$ is normal. Then $g - (a_ dX - b)X^{d - 1} \in I$ and we win by induction on the degree. As a consequence we see that

\[ S/xS = R[X]/(X, I) = R/J \]

where

\[ J = \{ b \in R \mid ax = b \text{ for some }a \in R\} = xR \cap R \]

By Lemma 10.151.6 we see that $S/xS = R/J$ has no embedded primes as an $R$-module, hence as an $R/J$-module, hence as an $S/xS$-module, hence as an $S$-module. This proves property (2). Take such an associated prime $\mathfrak q \subset S$ with the property $\mathfrak q \subset \mathfrak m$ (so that it is an associated prime of $S_{\mathfrak m}/xS_{\mathfrak m}$ – it does not matter for the arguments). Then $\mathfrak q$ is minimal over $xS$ and hence has height $1$. By the sequence of equalities above we see that $\mathfrak p = R \cap \mathfrak q$ is an associated prime of $R/J$, and so has height $1$ (see Lemma 10.151.6). Thus $R_{\mathfrak p}$ is a discrete valuation ring and therefore $R_{\mathfrak p} \subset S_{\mathfrak q}$ is an equality. This shows that $S_{\mathfrak q}$ is regular. This proves property (3)(a). Finally, $(S/\mathfrak q)_{\mathfrak m}$ is a localization of $S/\mathfrak q$, which is a quotient of $S/xS = R/J$. Hence $(S/\mathfrak q)_{\mathfrak m}$ is a localization of a quotient of the Nagata ring $R$, hence Nagata (Lemmas 10.156.5 and 10.156.6) and hence analytically unramified (Lemma 10.156.13). This shows (3)(b) holds and we are done. $\square$

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