## 10.162 Nagata rings

Here is the definition.

Definition 10.162.1. Let $R$ be a ring.

1. We say $R$ is universally Japanese if for any finite type ring map $R \to S$ with $S$ a domain we have that $S$ is N-2 (i.e., Japanese).

2. We say that $R$ is a Nagata ring if $R$ is Noetherian and for every prime ideal $\mathfrak p$ the ring $R/\mathfrak p$ is N-2.

It is clear that a Noetherian universally Japanese ring is a Nagata ring. It is our goal to show that a Nagata ring is universally Japanese. This is not obvious at all, and requires some work. But first, here is a useful lemma.

Lemma 10.162.2. Let $R$ be a Nagata ring. Let $R \to S$ be essentially of finite type with $S$ reduced. Then the integral closure of $R$ in $S$ is finite over $R$.

Proof. As $S$ is essentially of finite type over $R$ it is Noetherian and has finitely many minimal primes $\mathfrak q_1, \ldots , \mathfrak q_ m$, see Lemma 10.31.6. Since $S$ is reduced we have $S \subset \prod S_{\mathfrak q_ i}$ and each $S_{\mathfrak q_ i} = K_ i$ is a field, see Lemmas 10.25.4 and 10.25.1. It suffices to show that the integral closure $A_ i'$ of $R$ in each $K_ i$ is finite over $R$. This is true because $R$ is Noetherian and $A \subset \prod A_ i'$. Let $\mathfrak p_ i \subset R$ be the prime of $R$ corresponding to $\mathfrak q_ i$. As $S$ is essentially of finite type over $R$ we see that $K_ i = S_{\mathfrak q_ i} = \kappa (\mathfrak q_ i)$ is a finitely generated field extension of $\kappa (\mathfrak p_ i)$. Hence the algebraic closure $L_ i$ of $\kappa (\mathfrak p_ i)$ in $K_ i$ is finite over $\kappa (\mathfrak p_ i)$, see Fields, Lemma 9.26.11. It is clear that $A_ i'$ is the integral closure of $R/\mathfrak p_ i$ in $L_ i$, and hence we win by definition of a Nagata ring. $\square$

Lemma 10.162.3. Let $R$ be a ring. To check that $R$ is universally Japanese it suffices to show: If $R \to S$ is of finite type, and $S$ a domain then $S$ is N-1.

Proof. Namely, assume the condition of the lemma. Let $R \to S$ be a finite type ring map with $S$ a domain. Let $L$ be a finite extension of the fraction field of $S$. Then there exists a finite ring extension $S \subset S' \subset L$ such that $L$ is the fraction field of $S'$. By assumption $S'$ is N-1, and hence the integral closure $S''$ of $S'$ in $L$ is finite over $S'$. Thus $S''$ is finite over $S$ (Lemma 10.7.3) and $S''$ is the integral closure of $S$ in $L$ (Lemma 10.36.16). We conclude that $R$ is universally Japanese. $\square$

Lemma 10.162.4. If $R$ is universally Japanese then any algebra essentially of finite type over $R$ is universally Japanese.

Proof. The case of an algebra of finite type over $R$ is immediate from the definition. The general case follows on applying Lemma 10.161.3. $\square$

Lemma 10.162.5. Let $R$ be a Nagata ring. If $R \to S$ is a quasi-finite ring map (for example finite) then $S$ is a Nagata ring also.

Proof. First note that $S$ is Noetherian as $R$ is Noetherian and a quasi-finite ring map is of finite type. Let $\mathfrak q \subset S$ be a prime ideal, and set $\mathfrak p = R \cap \mathfrak q$. Then $R/\mathfrak p \subset S/\mathfrak q$ is quasi-finite and hence we conclude that $S/\mathfrak q$ is N-2 by Lemma 10.161.5 as desired. $\square$

Proof. Clear from Lemma 10.161.3. $\square$

Lemma 10.162.7. Let $R$ be a ring. Let $f_1, \ldots , f_ n \in R$ generate the unit ideal.

1. If each $R_{f_ i}$ is universally Japanese then so is $R$.

2. If each $R_{f_ i}$ is Nagata then so is $R$.

Proof. Let $\varphi : R \to S$ be a finite type ring map so that $S$ is a domain. Then $\varphi (f_1), \ldots , \varphi (f_ n)$ generate the unit ideal in $S$. Hence if each $S_{f_ i} = S_{\varphi (f_ i)}$ is N-1 then so is $S$, see Lemma 10.161.4. This proves (1).

If each $R_{f_ i}$ is Nagata, then each $R_{f_ i}$ is Noetherian and hence $R$ is Noetherian, see Lemma 10.23.2. And if $\mathfrak p \subset R$ is a prime, then we see each $R_{f_ i}/\mathfrak pR_{f_ i} = (R/\mathfrak p)_{f_ i}$ is N-2 and hence we conclude $R/\mathfrak p$ is N-2 by Lemma 10.161.4. This proves (2). $\square$

Proof. Let $R$ be a complete local Noetherian ring. Let $\mathfrak p \subset R$ be a prime. Then $R/\mathfrak p$ is also a complete local Noetherian ring, see Lemma 10.160.2. Hence it suffices to show that a Noetherian complete local domain $R$ is N-2. By Lemmas 10.161.5 and 10.160.11 we reduce to the case $R = k[[X_1, \ldots , X_ d]]$ where $k$ is a field or $R = \Lambda [[X_1, \ldots , X_ d]]$ where $\Lambda$ is a Cohen ring.

In the case $k[[X_1, \ldots , X_ d]]$ we reduce to the statement that a field is N-2 by Lemma 10.161.17. This is clear. In the case $\Lambda [[X_1, \ldots , X_ d]]$ we reduce to the statement that a Cohen ring $\Lambda$ is N-2. Applying Lemma 10.161.16 once more with $x = p \in \Lambda$ we reduce yet again to the case of a field. Thus we win. $\square$

Definition 10.162.9. Let $(R, \mathfrak m)$ be a Noetherian local ring. We say $R$ is analytically unramified if its completion $R^\wedge = \mathop{\mathrm{lim}}\nolimits _ n R/\mathfrak m^ n$ is reduced. A prime ideal $\mathfrak p \subset R$ is said to be analytically unramified if $R/\mathfrak p$ is analytically unramified.

At this point we know the following are true for any Noetherian local ring $R$: The map $R \to R^\wedge$ is a faithfully flat local ring homomorphism (Lemma 10.97.3). The completion $R^\wedge$ is Noetherian (Lemma 10.97.5) and complete (Lemma 10.97.4). Hence the completion $R^\wedge$ is a Nagata ring (Lemma 10.162.8). Moreover, we have seen in Section 10.160 that $R^\wedge$ is a quotient of a regular local ring (Theorem 10.160.8), and hence universally catenary (Remark 10.160.9).

Lemma 10.162.10. Let $(R, \mathfrak m)$ be a Noetherian local ring.

1. If $R$ is analytically unramified, then $R$ is reduced.

2. If $R$ is analytically unramified, then each minimal prime of $R$ is analytically unramified.

3. If $R$ is reduced with minimal primes $\mathfrak q_1, \ldots , \mathfrak q_ t$, and each $\mathfrak q_ i$ is analytically unramified, then $R$ is analytically unramified.

4. If $R$ is analytically unramified, then the integral closure of $R$ in its total ring of fractions $Q(R)$ is finite over $R$.

5. If $R$ is a domain and analytically unramified, then $R$ is N-1.

Proof. In this proof we will use the remarks immediately following Definition 10.162.9. As $R \to R^\wedge$ is a faithfully flat local ring homomorphism it is injective and (1) follows.

Let $\mathfrak q$ be a minimal prime of $R$, and assume $R$ is analytically unramified. Then $\mathfrak q$ is an associated prime of $R$ (see Proposition 10.63.6). Hence there exists an $f \in R$ such that $\{ x \in R \mid fx = 0\} = \mathfrak q$. Note that $(R/\mathfrak q)^\wedge = R^\wedge /\mathfrak q^\wedge$, and that $\{ x \in R^\wedge \mid fx = 0\} = \mathfrak q^\wedge$, because completion is exact (Lemma 10.97.2). If $x \in R^\wedge$ is such that $x^2 \in \mathfrak q^\wedge$, then $fx^2 = 0$ hence $(fx)^2 = 0$ hence $fx = 0$ hence $x \in \mathfrak q^\wedge$. Thus $\mathfrak q$ is analytically unramified and (2) holds.

Assume $R$ is reduced with minimal primes $\mathfrak q_1, \ldots , \mathfrak q_ t$, and each $\mathfrak q_ i$ is analytically unramified. Then $R \to R/\mathfrak q_1 \times \ldots \times R/\mathfrak q_ t$ is injective. Since completion is exact (see Lemma 10.97.2) we see that $R^\wedge \subset (R/\mathfrak q_1)^\wedge \times \ldots \times (R/\mathfrak q_ t)^\wedge$. Hence (3) is clear.

Assume $R$ is analytically unramified. Let $\mathfrak p_1, \ldots , \mathfrak p_ s$ be the minimal primes of $R^\wedge$. Then we see that

$Q(R^\wedge ) = R^\wedge _{\mathfrak p_1} \times \ldots \times R^\wedge _{\mathfrak p_ s}$

with each $R^\wedge _{\mathfrak p_ i}$ a field as $R^\wedge$ is reduced (see Lemma 10.25.4). Hence the integral closure $S$ of $R^\wedge$ in $Q(R^\wedge )$ is equal to $S = S_1 \times \ldots \times S_ s$ with $S_ i$ the integral closure of $R^\wedge /\mathfrak p_ i$ in its fraction field. In particular $S$ is finite over $R^\wedge$. Denote $R'$ the integral closure of $R$ in $Q(R)$. As $R \to R^\wedge$ is flat we see that $R' \otimes _ R R^\wedge \subset Q(R) \otimes _ R R^\wedge \subset Q(R^\wedge )$. Moreover $R' \otimes _ R R^\wedge$ is integral over $R^\wedge$ (Lemma 10.36.13). Hence $R' \otimes _ R R^\wedge \subset S$ is a $R^\wedge$-submodule. As $R^\wedge$ is Noetherian it is a finite $R^\wedge$-module. Thus we may find $f_1, \ldots , f_ n \in R'$ such that $R' \otimes _ R R^\wedge$ is generated by the elements $f_ i \otimes 1$ as a $R^\wedge$-module. By faithful flatness we see that $R'$ is generated by $f_1, \ldots , f_ n$ as an $R$-module. This proves (4).

Part (5) is a special case of part (4). $\square$

Lemma 10.162.11. Let $R$ be a Noetherian local ring. Let $\mathfrak p \subset R$ be a prime. Assume

1. $R_{\mathfrak p}$ is a discrete valuation ring, and

2. $\mathfrak p$ is analytically unramified.

Then for any associated prime $\mathfrak q$ of $R^\wedge /\mathfrak pR^\wedge$ the local ring $(R^\wedge )_{\mathfrak q}$ is a discrete valuation ring.

Proof. Assumption (2) says that $R^\wedge /\mathfrak pR^\wedge$ is a reduced ring. Hence an associated prime $\mathfrak q \subset R^\wedge$ of $R^\wedge /\mathfrak pR^\wedge$ is the same thing as a minimal prime over $\mathfrak pR^\wedge$. In particular we see that the maximal ideal of $(R^\wedge )_{\mathfrak q}$ is $\mathfrak p(R^\wedge )_{\mathfrak q}$. Choose $x \in R$ such that $xR_{\mathfrak p} = \mathfrak pR_{\mathfrak p}$. By the above we see that $x \in (R^\wedge )_{\mathfrak q}$ generates the maximal ideal. As $R \to R^\wedge$ is faithfully flat we see that $x$ is a nonzerodivisor in $(R^\wedge )_{\mathfrak q}$. Hence we win. $\square$

Lemma 10.162.12. Let $(R, \mathfrak m)$ be a Noetherian local domain. Let $x \in \mathfrak m$. Assume

1. $x \not= 0$,

2. $R/xR$ has no embedded primes, and

3. for each associated prime $\mathfrak p \subset R$ of $R/xR$ we have

1. the local ring $R_{\mathfrak p}$ is regular, and

2. $\mathfrak p$ is analytically unramified.

Then $R$ is analytically unramified.

Proof. Let $\mathfrak p_1, \ldots , \mathfrak p_ t$ be the associated primes of the $R$-module $R/xR$. Since $R/xR$ has no embedded primes we see that each $\mathfrak p_ i$ has height $1$, and is a minimal prime over $(x)$. For each $i$, let $\mathfrak q_{i1}, \ldots , \mathfrak q_{is_ i}$ be the associated primes of the $R^\wedge$-module $R^\wedge /\mathfrak p_ iR^\wedge$. By Lemma 10.162.11 we see that $(R^\wedge )_{\mathfrak q_{ij}}$ is regular. By Lemma 10.65.3 we see that

$\text{Ass}_{R^\wedge }(R^\wedge /xR^\wedge ) = \bigcup \nolimits _{\mathfrak p \in \text{Ass}_ R(R/xR)} \text{Ass}_{R^\wedge }(R^\wedge /\mathfrak pR^\wedge ) = \{ \mathfrak q_{ij}\} .$

Let $y \in R^\wedge$ with $y^2 = 0$. As $(R^\wedge )_{\mathfrak q_{ij}}$ is regular, and hence a domain (Lemma 10.106.2) we see that $y$ maps to zero in $(R^\wedge )_{\mathfrak q_{ij}}$. Hence $y$ maps to zero in $R^\wedge /xR^\wedge$ by Lemma 10.63.19. Hence $y = xy'$. Since $x$ is a nonzerodivisor (as $R \to R^\wedge$ is flat) we see that $(y')^2 = 0$. Hence we conclude that $y \in \bigcap x^ nR^\wedge = (0)$ (Lemma 10.51.4). $\square$

Lemma 10.162.13. Let $(R, \mathfrak m)$ be a local ring. If $R$ is Noetherian, a domain, and Nagata, then $R$ is analytically unramified.

Proof. By induction on $\dim (R)$. The case $\dim (R) = 0$ is trivial. Hence we assume $\dim (R) = d$ and that the lemma holds for all Noetherian Nagata domains of dimension $< d$.

Let $R \subset S$ be the integral closure of $R$ in the field of fractions of $R$. By assumption $S$ is a finite $R$-module. By Lemma 10.162.5 we see that $S$ is Nagata. By Lemma 10.112.4 we see $\dim (R) = \dim (S)$. Let $\mathfrak m_1, \ldots , \mathfrak m_ t$ be the maximal ideals of $S$. Each of these lies over the maximal ideal $\mathfrak m$ of $R$. Moreover

$(\mathfrak m_1 \cap \ldots \cap \mathfrak m_ t)^ n \subset \mathfrak mS$

for sufficiently large $n$ as $S/\mathfrak mS$ is Artinian. By Lemma 10.97.2 $R^\wedge \to S^\wedge$ is an injective map, and by the Chinese Remainder Lemma 10.15.4 combined with Lemma 10.96.9 we have $S^\wedge = \prod S^\wedge _ i$ where $S^\wedge _ i$ is the completion of $S$ with respect to the maximal ideal $\mathfrak m_ i$. Hence it suffices to show that $S_{\mathfrak m_ i}$ is analytically unramified. In other words, we have reduced to the case where $R$ is a Noetherian normal Nagata domain.

Assume $R$ is a Noetherian, normal, local Nagata domain. Pick a nonzero $x \in \mathfrak m$ in the maximal ideal. We are going to apply Lemma 10.162.12. We have to check properties (1), (2), (3)(a) and (3)(b). Property (1) is clear. We have that $R/xR$ has no embedded primes by Lemma 10.157.6. Thus property (2) holds. The same lemma also tells us each associated prime $\mathfrak p$ of $R/xR$ has height $1$. Hence $R_{\mathfrak p}$ is a $1$-dimensional normal domain hence regular (Lemma 10.119.7). Thus (3)(a) holds. Finally (3)(b) holds by induction hypothesis, since $R/\mathfrak p$ is Nagata (by Lemma 10.162.5 or directly from the definition). Thus we conclude $R$ is analytically unramified. $\square$

Lemma 10.162.14. Let $(R, \mathfrak m)$ be a Noetherian local ring. The following are equivalent

1. $R$ is Nagata,

2. for $R \to S$ finite with $S$ a domain and $\mathfrak m' \subset S$ maximal the local ring $S_{\mathfrak m'}$ is analytically unramified,

3. for $(R, \mathfrak m) \to (S, \mathfrak m')$ finite local homomorphism with $S$ a domain, then $S$ is analytically unramified.

Proof. Assume $R$ is Nagata and let $R \to S$ and $\mathfrak m' \subset S$ be as in (2). Then $S$ is Nagata by Lemma 10.162.5. Hence the local ring $S_{\mathfrak m'}$ is Nagata (Lemma 10.162.6). Thus it is analytically unramified by Lemma 10.162.13. It is clear that (2) implies (3).

Assume (3) holds. Let $\mathfrak p \subset R$ be a prime ideal and let $L/\kappa (\mathfrak p)$ be a finite extension of fields. To prove (1) we have to show that the integral closure of $R/\mathfrak p$ is finite over $R/\mathfrak p$. Choose $x_1, \ldots , x_ n \in L$ which generate $L$ over $\kappa (\mathfrak p)$. For each $i$ let $P_ i(T) = T^{d_ i} + a_{i, 1} T^{d_ i - 1} + \ldots + a_{i, d_ i}$ be the minimal polynomial for $x_ i$ over $\kappa (\mathfrak p)$. After replacing $x_ i$ by $f_ i x_ i$ for a suitable $f_ i \in R$, $f_ i \not\in \mathfrak p$ we may assume $a_{i, j} \in R/\mathfrak p$. In fact, after further multiplying by elements of $\mathfrak m$, we may assume $a_{i, j} \in \mathfrak m/\mathfrak p \subset R/\mathfrak p$ for all $i, j$. Having done this let $S = R/\mathfrak p[x_1, \ldots , x_ n] \subset L$. Then $S$ is finite over $R$, a domain, and $S/\mathfrak m S$ is a quotient of $R/\mathfrak m[T_1, \ldots , T_ n]/(T_1^{d_1}, \ldots , T_ n^{d_ n})$. Hence $S$ is local. By (3) $S$ is analytically unramified and by Lemma 10.162.10 we find that its integral closure $S'$ in $L$ is finite over $S$. Since $S'$ is also the integral closure of $R/\mathfrak p$ in $L$ we win. $\square$

The following proposition says in particular that an algebra of finite type over a Nagata ring is a Nagata ring.

Proposition 10.162.15 (Nagata). Let $R$ be a ring. The following are equivalent:

1. $R$ is a Nagata ring,

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

3. $R$ 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 10.161.7). 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 10.161.13 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 $M/L$ 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 10.162.5). 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 10.161.15 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 $\kappa (\mathfrak m)/\kappa (\mathfrak n)$ is finite (Theorem 10.34.1) 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 $K''/K$ 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 10.162.5). 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 10.161.15, which in turn implies that $S_{\mathfrak m}$ is N-1 by Lemma 10.161.7. 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 10.162.12 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 10.162.10.

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 10.123.1) 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 10.157.6 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 10.157.6). 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 10.162.5 and 10.162.6) and hence analytically unramified (Lemma 10.162.13). This shows (3)(b) holds and we are done. $\square$

Proposition 10.162.16. The following types of rings are Nagata and in particular universally Japanese:

1. fields,

2. Noetherian complete local rings,

3. $\mathbf{Z}$,

4. Dedekind domains with fraction field of characteristic zero,

5. finite type ring extensions of any of the above.

Proof. The Noetherian complete local ring case is Lemma 10.162.8. In the other cases you just check if $R/\mathfrak p$ is N-2 for every prime ideal $\mathfrak p$ of the ring. This is clear whenever $R/\mathfrak p$ is a field, i.e., $\mathfrak p$ is maximal. Hence for the Dedekind ring case we only need to check it when $\mathfrak p = (0)$. But since we assume the fraction field has characteristic zero Lemma 10.161.11 kicks in. $\square$

Example 10.162.17. A discrete valuation ring is Nagata if and only if it is N-2 (because the quotient by the maximal ideal is a field and hence N-2). The discrete valuation ring $A$ of Example 10.119.5 is not Nagata, i.e., it is not N-2. Namely, the finite extension $A \subset R = A[f]$ is not N-1. To see this say $f = \sum a_ i x^ i$. For every $n \geq 1$ set $g_ n = \sum _{i < n} a_ i x^ i \in A$. Then $h_ n = (f - g_ n)/x^ n$ is an element of the fraction field of $R$ and $h_ n^ p \in k^ p[[x]] \subset A$. Hence the integral closure $R'$ of $R$ contains $h_1, h_2, h_3, \ldots$. Now, if $R'$ were finite over $R$ and hence $A$, then $f = x^ n h_ n + g_ n$ would be contained in the submodule $A + x^ nR'$ for all $n$. By Artin-Rees this would imply $f \in A$ (Lemma 10.51.4), a contradiction.

Lemma 10.162.18. Let $(A, \mathfrak m)$ be a Noetherian local domain which is Nagata and has fraction field of characteristic $p$. If $a \in A$ has a $p$th root in $A^\wedge$, then $a$ has a $p$th root in $A$.

Proof. Consider the ring extension $A \subset B = A[x]/(x^ p - a)$. If $a$ does not have a $p$th root in $A$, then $B$ is a domain whose completion isn't reduced. This contradicts our earlier results, as $B$ is a Nagata ring (Proposition 10.162.15) and hence analytically unramified by Lemma 10.162.13. $\square$

Comment #1520 by kollar on

Small point, but I think that "B is a domain whose completion isn't a domain." should be changed to "B is a domain whose completion isn't reduced."

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