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Tag 032E

10.156. Nagata rings

Here is the definition.

Definition 10.156.1. 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 10.156.2. 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 10.30.6. 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 10.24.4 and 10.24.1. 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 $\subset K_i$ is finite over $\kappa(\mathfrak p_i)$, see Fields, Lemma 9.26.10. 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 10.156.3. 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 10.35.16). We conclude that $R$ is universally Japanese. $\square$

Lemma 10.156.4. 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 10.155.3. $\square$

Lemma 10.156.5. 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 10.155.5 as desired. $\square$

Lemma 10.156.6. A localization of a Nagata ring is a Nagata ring.

Proof. Clear from Lemma 10.155.3. $\square$

Lemma 10.156.7. 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 10.155.4. 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 10.155.4. This proves (2). $\square$

Lemma 10.156.8. A Noetherian complete local ring is a Nagata ring.

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 10.154.2. Hence it suffices to show that a Noetherian complete local domain $R$ is N-2. By Lemmas 10.155.5 and 10.154.11 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 10.155.17. 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 10.155.16 once more with $x = p \in \Lambda$ we reduce yet again to the case of a field. Thus we win. $\square$

Definition 10.156.9. Let $(R, \mathfrak m)$ be a Noetherian local ring. We say $R$ is analytically unramified if its completion $R^\wedge = \mathop{\rm lim}\nolimits_n R/\mathfrak m^n$ 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 10.96.3). The completion $R^\wedge$ is Noetherian (Lemma 10.96.5) and complete (Lemma 10.96.4). Hence the completion $R^\wedge$ is a Nagata ring (Lemma 10.156.8). Moreover, we have seen in Section 10.154 that $R^\wedge$ is a quotient of a regular local ring (Theorem 10.154.8), and hence universally catenary (Remark 10.154.9).

Lemma 10.156.10. 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 10.156.9. 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 10.62.6). 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 10.96.2). 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 10.96.2) 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 10.24.4). 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 10.35.13). 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 10.156.11. 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 10.156.12. 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 10.156.11 we see that $(R^\wedge)_{\mathfrak q_{ij}}$ is regular. By Lemma 10.64.3 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 10.105.2) 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 10.62.19. 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 10.50.4). $\square$

Lemma 10.156.13. 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 10.156.5 we see that $S$ is Nagata. By Lemma 10.111.4 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 10.96.2 $R^\wedge \to S^\wedge$ is an injective map, and by the Chinese Remainder Lemma 10.14.3 combined with Lemma 10.95.9 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 10.156.12. 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 10.151.6. 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 10.118.7). Thus (3)(a) holds. Finally (3)(b) holds by induction hypothesis, since $R/\mathfrak p$ is Nagata (by Lemma 10.156.5 or directly from the definition). Thus we conclude $R$ is analytically unramified. $\square$

Lemma 10.156.14. 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 10.156.5. Hence the local ring $S_{\mathfrak m'}$ is Nagata (Lemma 10.156.6). Thus it is analytically unramified by Lemma 10.156.13. 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 10.156.10 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 10.156.15 (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 10.155.7). 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 $I = \mathop{\rm Ker}(R[X] \to S)$ 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. $\square$

Proposition 10.156.16. 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 10.156.8. 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 10.155.11 kicks in. $\square$

Example 10.156.17. A discrete valuation ring is Nagata if and only if it is N-2 (this follows immediately from the definition). The discrete valuation ring $A$ of Example 10.118.5 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 10.50.4), a contradiction.

Lemma 10.156.18. 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 10.156.15) and hence analytically unramified by Lemma 10.156.13. $\square$

    The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 43032–43710 (see updates for more information).

    \section{Nagata rings}
    \label{section-nagata}
    
    \noindent
    Here is the definition.
    
    \begin{definition}
    \label{definition-nagata}
    Let $R$ be a ring.
    \begin{enumerate}
    \item We say $R$ is {\it 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).
    \item We say that $R$ is a {\it Nagata ring} if $R$ is Noetherian and
    for every prime ideal $\mathfrak p$ the ring $R/\mathfrak p$ is N-2.
    \end{enumerate}
    \end{definition}
    
    \noindent
    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.
    
    \begin{lemma}
    \label{lemma-nagata-in-reduced-finite-type-finite-integral-closure}
    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 $R$.
    \end{lemma}
    
    \begin{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 \ref{lemma-Noetherian-irreducible-components}.
    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 \ref{lemma-total-ring-fractions-no-embedded-points}
    and \ref{lemma-minimal-prime-reduced-ring}.
    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 $\subset K_i$
    is finite over $\kappa(\mathfrak p_i)$, see
    Fields, Lemma \ref{fields-lemma-algebraic-closure-in-finitely-generated}.
    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.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-check-universally-japanese}
    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.
    \end{lemma}
    
    \begin{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 \ref{lemma-finite-transitive})
    and $S''$ is the integral closure of $S$ in $L$
    (Lemma \ref{lemma-integral-closure-transitive}).
    We conclude that $R$ is universally Japanese.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-universally-japanese}
    If $R$ is universally Japanese then any algebra essentially of finite type
    over $R$ is universally Japanese.
    \end{lemma}
    
    \begin{proof}
    The case of an algebra of finite type over $R$ is immediate from
    the definition. The general case follows on applying
    Lemma \ref{lemma-localize-N}.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-quasi-finite-over-nagata}
    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.
    \end{lemma}
    
    \begin{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 \ref{lemma-quasi-finite-over-Noetherian-japanese}
    as desired.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-nagata-localize}
    A localization of a Nagata ring is a Nagata ring.
    \end{lemma}
    
    \begin{proof}
    Clear from Lemma \ref{lemma-localize-N}.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-nagata-local}
    Let $R$ be a ring. Let $f_1, \ldots, f_n \in R$ generate the
    unit ideal.
    \begin{enumerate}
    \item  If each $R_{f_i}$ is universally Japanese then so is $R$.
    \item  If each $R_{f_i}$ is Nagata then so is $R$.
    \end{enumerate}
    \end{lemma}
    
    \begin{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 \ref{lemma-Japanese-local}. This proves (1).
    
    \medskip\noindent
    If each $R_{f_i}$ is Nagata, then each $R_{f_i}$ is Noetherian and
    hence $R$ is Noetherian, see Lemma \ref{lemma-cover}. 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 \ref{lemma-Japanese-local}. This proves (2).
    \end{proof}
    
    \begin{lemma}
    \label{lemma-Noetherian-complete-local-Nagata}
    A Noetherian complete local ring is a Nagata ring.
    \end{lemma}
    
    \begin{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 \ref{lemma-quotient-complete-local}.
    Hence it suffices to show that a Noetherian complete local
    domain $R$ is N-2. By
    Lemmas \ref{lemma-quasi-finite-over-Noetherian-japanese}
    and \ref{lemma-complete-local-Noetherian-domain-finite-over-regular}
    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.
    
    \medskip\noindent
    In the case $k[[X_1, \ldots, X_d]]$ we reduce to the statement that a
    field is N-2 by Lemma \ref{lemma-power-series-over-N-2}. 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 \ref{lemma-tate-japanese}
    once more with $x = p \in \Lambda$ we reduce yet again to the case
    of a field. Thus we win.
    \end{proof}
    
    \begin{definition}
    \label{definition-analytically-unramified}
    Let $(R, \mathfrak m)$ be a Noetherian local ring.
    We say $R$ is {\it analytically unramified} if its completion
    $R^\wedge = \lim_n R/\mathfrak m^n$ is reduced.
    A prime ideal $\mathfrak p \subset R$ is said to be
    {\it analytically unramified} if $R/\mathfrak p$ is analytically
    unramified.
    \end{definition}
    
    \noindent
    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 \ref{lemma-completion-faithfully-flat}).
    The completion $R^\wedge$ is Noetherian
    (Lemma \ref{lemma-completion-Noetherian})
    and complete (Lemma \ref{lemma-completion-complete}).
    Hence the completion $R^\wedge$ is a Nagata ring
    (Lemma \ref{lemma-Noetherian-complete-local-Nagata}).
    Moreover, we have seen in Section \ref{section-cohen-structure-theorem}
    that $R^\wedge$ is
    a quotient of a regular local ring
    (Theorem \ref{theorem-cohen-structure-theorem}), and hence
    universally catenary
    (Remark \ref{remark-Noetherian-complete-local-ring-universally-catenary}).
    
    \begin{lemma}
    \label{lemma-analytically-unramified-easy}
    Let $(R, \mathfrak m)$ be a Noetherian local ring.
    \begin{enumerate}
    \item If $R$ is analytically unramified, then $R$ is reduced.
    \item If $R$ is analytically unramified, then each minimal prime of
    $R$ is analytically unramified.
    \item 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.
    \item If $R$ is analytically unramified, then the integral closure
    of $R$ in its total ring of fractions $Q(R)$ is finite over $R$.
    \item If $R$ is a domain and analytically unramified, then $R$ is N-1.
    \end{enumerate}
    \end{lemma}
    
    \begin{proof}
    In this proof we will use the remarks immediately following
    Definition \ref{definition-analytically-unramified}.
    As $R \to R^\wedge$ is a faithfully flat local ring homomorphism
    it is injective and (1) follows.
    
    \medskip\noindent
    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 \ref{proposition-minimal-primes-associated-primes}).
    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 \ref{lemma-completion-flat}).
    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.
    
    \medskip\noindent
    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 \ref{lemma-completion-flat})
    we see that
    $R^\wedge \subset (R/\mathfrak q_1)^\wedge \times \ldots \times
    (R/\mathfrak q_t)^\wedge$. Hence (3) is clear.
    
    \medskip\noindent
    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 \ref{lemma-total-ring-fractions-no-embedded-points}).
    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 \ref{lemma-base-change-integral}).
    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).
    
    \medskip\noindent
    Part (5) is a special case of part (4).
    \end{proof}
    
    \begin{lemma}
    \label{lemma-codimension-1-analytically-unramified}
    Let $R$ be a Noetherian local ring.
    Let $\mathfrak p \subset R$ be a prime.
    Assume
    \begin{enumerate}
    \item $R_{\mathfrak p}$ is a discrete valuation ring, and
    \item $\mathfrak p$ is analytically unramified.
    \end{enumerate}
    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.
    \end{lemma}
    
    \begin{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.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-criterion-analytically-unramified}
    Let $(R, \mathfrak m)$ be a Noetherian local domain.
    Let $x \in \mathfrak m$. Assume
    \begin{enumerate}
    \item $x \not = 0$,
    \item $R/xR$ has no embedded primes, and
    \item for each associated prime $\mathfrak p \subset R$
    of $R/xR$ we have
    \begin{enumerate}
    \item the local ring $R_{\mathfrak p}$ is regular, and
    \item $\mathfrak p$ is analytically unramified.
    \end{enumerate}
    \end{enumerate}
    Then $R$ is analytically unramified.
    \end{lemma}
    
    \begin{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 \ref{lemma-codimension-1-analytically-unramified}
    we see that $(R^\wedge)_{\mathfrak q_{ij}}$ is regular.
    By Lemma \ref{lemma-bourbaki} 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 \ref{lemma-regular-domain})
    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 \ref{lemma-zero-at-ass-zero}.
    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 \ref{lemma-intersect-powers-ideal-module-zero}).
    \end{proof}
    
    \begin{lemma}
    \label{lemma-local-nagata-domain-analytically-unramified}
    Let $(R, \mathfrak m)$ be a local ring.
    If $R$ is Noetherian, a domain, and Nagata, then $R$ is
    analytically unramified.
    \end{lemma}
    
    \begin{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 $< d$.
    
    \medskip\noindent
    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 \ref{lemma-quasi-finite-over-nagata} we see that
    $S$ is Nagata. By Lemma \ref{lemma-integral-sub-dim-equal} 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 \ref{lemma-completion-flat} $R^\wedge \to S^\wedge$
    is an injective map, and by the Chinese Remainder
    Lemma \ref{lemma-chinese-remainder} combined with
    Lemma \ref{lemma-change-ideal-completion} 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.
    
    \medskip\noindent
    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 \ref{lemma-criterion-analytically-unramified}.
    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 \ref{lemma-normal-domain-intersection-localizations-height-1}.
    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 \ref{lemma-characterize-dvr}). Thus (3)(a) holds.
    Finally (3)(b) holds by induction hypothesis, since
    $R/\mathfrak p$ is Nagata (by Lemma \ref{lemma-quasi-finite-over-nagata}
    or directly from the definition).
    Thus we conclude $R$ is analytically unramified.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-local-nagata-and-analytically-unramified}
    Let $(R, \mathfrak m)$ be a Noetherian local ring. The following
    are equivalent
    \begin{enumerate}
    \item $R$ is Nagata,
    \item for $R \to S$ finite with $S$ a domain and $\mathfrak m' \subset S$
    maximal the local ring $S_{\mathfrak m'}$ is analytically unramified,
    \item for $(R, \mathfrak m) \to (S, \mathfrak m')$ finite
    local homomorphism with $S$ a domain, then $S$ is analytically
    unramified.
    \end{enumerate}
    \end{lemma}
    
    \begin{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 \ref{lemma-quasi-finite-over-nagata}.
    Hence the local ring $S_{\mathfrak m'}$ is Nagata
    (Lemma \ref{lemma-nagata-localize}). Thus it is analytically
    unramified by Lemma \ref{lemma-local-nagata-domain-analytically-unramified}.
    It is clear that (2) implies (3).
    
    \medskip\noindent
    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 \ref{lemma-analytically-unramified-easy}
    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.
    \end{proof}
    
    \noindent
    The following proposition says in particular that an algebra of finite
    type over a Nagata ring is a Nagata ring.
    
    \begin{proposition}[Nagata]
    \label{proposition-nagata-universally-japanese}
    Let $R$ be a ring. The following are equivalent:
    \begin{enumerate}
    \item $R$ is a Nagata ring,
    \item any finite type $R$-algebra is Nagata, and
    \item $R$ is universally Japanese and Noetherian.
    \end{enumerate}
    \end{proposition}
    
    \begin{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.
    
    \medskip\noindent
    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
    $S \cong R[x]/I$.
    
    \medskip\noindent
    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.
    
    \medskip\noindent
    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 \ref{lemma-finite-extension-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 then $S$ is N-2.
    
    \medskip\noindent
    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 \ref{lemma-polynomial-ring-N-2} 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.
    
    \medskip\noindent
    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 \ref{lemma-quasi-finite-over-nagata}).
    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.
    
    \medskip\noindent
    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 \ref{lemma-characterize-N-1} it suffices to show that
    $S_{\mathfrak m}$ is N-1 for every maximal ideal $\mathfrak m$ of $S$.
    
    \medskip\noindent
    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 \ref{theorem-nullstellensatz}) 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 \ref{lemma-quasi-finite-over-nagata}).
    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 \ref{lemma-characterize-N-1}, which in turn
    implies that $S_{\mathfrak m}$ is N-1 by
    Lemma \ref{lemma-finite-extension-N-2}.
    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 $x \in \mathfrak m$.
    
    \medskip\noindent
    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.
    
    \medskip\noindent
    We will show that Lemma \ref{lemma-criterion-analytically-unramified}
    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 \ref{lemma-analytically-unramified-easy}.
    
    \medskip\noindent
    We have to check properties (1), (2), (3)(a) and (3)(b).
    Property (1) is trivial.
    Let $I = \Ker(R[X] \to S)$ 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 \ref{lemma-make-integral-trivial})
    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 \ref{lemma-normal-domain-intersection-localizations-height-1}
    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 \ref{lemma-normal-domain-intersection-localizations-height-1}).
    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 \ref{lemma-quasi-finite-over-nagata}
    and \ref{lemma-nagata-localize})
    and hence analytically unramified
    (Lemma \ref{lemma-local-nagata-domain-analytically-unramified}).
    This shows (3)(b) holds and we are done.
    \end{proof}
    
    \begin{proposition}
    \label{proposition-ubiquity-nagata}
    The following types of rings are Nagata and in particular universally Japanese:
    \begin{enumerate}
    \item fields,
    \item Noetherian complete local rings,
    \item $\mathbf{Z}$,
    \item Dedekind domains with fraction field of characteristic zero,
    \item finite type ring extensions of any of the above.
    \end{enumerate}
    \end{proposition}
    
    \begin{proof}
    The Noetherian complete local ring case is
    Lemma \ref{lemma-Noetherian-complete-local-Nagata}.
    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 \ref{lemma-domain-char-zero-N-1-2} kicks in.
    \end{proof}
    
    \begin{example}
    \label{example-nonjapanese-dvr}
    A discrete valuation ring is Nagata if and only if it is N-2
    (this follows immediately from the definition). The discrete valuation
    ring $A$ of Example \ref{example-bad-dvr-char-p} 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 \ref{lemma-intersect-powers-ideal-module-zero}), a contradiction.
    \end{example}
    
    \begin{lemma}
    \label{lemma-nagata-pth-roots}
    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 $A$.
    \end{lemma}
    
    \begin{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 \ref{proposition-nagata-universally-japanese})
    and hence analytically unramified by
    Lemma \ref{lemma-local-nagata-domain-analytically-unramified}.
    \end{proof}

    Comments (2)

    Comment #1520 by kollar on June 19, 2015 a 8:12 pm UTC

    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."

    Comment #1522 by Johan (site) on June 20, 2015 a 8:41 pm UTC

    OK, thanks. This is fixed here.

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