The Stacks project

Proof. This follows from the fact that

as algebras over $\kappa (\mathfrak q)$. $\square$

Proof. Let $\mathfrak p \subset R$ be a prime. Then the fibre over $\mathfrak p$ is the ring $\Lambda \otimes _ R \kappa (\mathfrak p)$ whose spectrum maps bijectively onto the subset of $\mathop{\mathrm{Spec}}(\Lambda )$ consisting of primes $\mathfrak q$ lying over $\mathfrak p$, see Algebra, Remark 10.17.8. For such a prime $\mathfrak q$ choose a maximal ideal $\mathfrak q \subset \mathfrak m'$ and set $\mathfrak m = R \cap \mathfrak m'$. Then $\mathfrak p \subset \mathfrak m$ and we have

as $\kappa (\mathfrak q)$-algebras. Thus (1), (2), and (3) are equivalent because by (B) we can check property $P$ on local rings. $\square$

Proof. It is clear that (1) $\Rightarrow $ (2) $\Rightarrow $ (3). Assume $P$ holds for $\kappa (\mathfrak q) \to R_\mathfrak p^\wedge \otimes _ R \kappa (\mathfrak q)$. By Algebra, Lemma 10.124.3 we see that

for some $R_\mathfrak p^\wedge $-algebra $B$. Hence $R'_{\mathfrak p'} \to (R'_{\mathfrak p'})^\wedge $ is a factor of a base change of the map $R_\mathfrak p \to R_\mathfrak p^\wedge $. It follows that $(R'_{\mathfrak p'})^\wedge \otimes _{R'} \kappa (\mathfrak q')$ is a factor of

Thus the result follows from the assumptions on $P$. $\square$

Proof. Assume the formal fibres of $R_\mathfrak m$ have $P$ for all maximal ideals $\mathfrak m$ of $R$. Let $\mathfrak p$ be a prime of $R$ and choose a maximal ideal $\mathfrak p \subset \mathfrak m$. Since $R_\mathfrak m \to R_\mathfrak m^\wedge $ is faithfully flat we can choose a prime $\mathfrak p'$ if $R_\mathfrak m^\wedge $ lying over $\mathfrak pR_\mathfrak m$. Consider the commutative diagram

By assumption the fibres of the ring map $R_\mathfrak m \to R_\mathfrak m^\wedge $ have $P$. By Proposition 15.50.6 $(R_\mathfrak m^\wedge )_{\mathfrak p'} \to (R_\mathfrak m^\wedge )_{\mathfrak p'}^\wedge $ is regular. The localization $R_\mathfrak m^\wedge \to (R_\mathfrak m^\wedge )_{\mathfrak p'}$ is regular. Hence $R_\mathfrak m^\wedge \to (R_\mathfrak m^\wedge )_{\mathfrak p'}^\wedge $ is regular by Lemma 15.41.4. Hence the fibres of $R_\mathfrak m \to (R_\mathfrak m^\wedge )_{\mathfrak p'}^\wedge $ have $P$ by (C). Since $R_\mathfrak m \to (R_\mathfrak m^\wedge )_{\mathfrak p'}^\wedge $ factors through the localization $R_\mathfrak p$, also the fibres of $R_\mathfrak p \to (R_\mathfrak m^\wedge )_{\mathfrak p'}^\wedge $ have $P$. Thus we may apply (D) to see that the fibres of $R_\mathfrak p \to R_\mathfrak p^\wedge $ have $P$. $\square$

Proof. Since being a $P$-ring is a property of the local rings it is clear that a localization of a $P$-ring is a $P$-ring. Conversely, if every localization at a prime is a $P$-ring, then the ring is a $P$-ring. Thus it suffices to show that $S_\mathfrak q$ is a $P$-ring for every finite type $R$-algebra $S$ and every prime $\mathfrak q$ of $S$. Writing $S$ as a quotient of $R[x_1, \ldots , x_ n]$ we see from Lemma 15.51.3 that it suffices to prove that $R[x_1, \ldots , x_ n]$ is a $P$-ring. By induction on $n$ it suffices to prove that $R[x]$ is a $P$-ring. Let $\mathfrak q \subset R[x]$ be a maximal ideal. By Lemma 15.51.4 it suffices to show that the fibres of

have $P$. If $\mathfrak q$ lies over $\mathfrak p \subset R$, then we may replace $R$ by $R_\mathfrak p$. Hence we may assume that $R$ is a Noetherian local $P$-ring with maximal ideal $\mathfrak m$ and that $\mathfrak q \subset R[x]$ lies over $\mathfrak m$. Note that there is a unique prime $\mathfrak q' \subset R^\wedge [x]$ lying over $\mathfrak q$. Consider the diagram

Since $R$ is a $P$-ring the fibres of $R[x] \to R^\wedge [x]$ have $P$ because they are base changes of the fibres of $R \to R^\wedge $ by a finitely generated field extension so (A) applies. Hence the fibres of the lower horizontal arrow have $P$ for example by Lemma 15.51.2. The right vertical arrow is regular because $R^\wedge $ is a G-ring (Propositions 15.50.6 and 15.50.10). It follows that the fibres of the composition $R[x]_\mathfrak q \to (R^\wedge [x]_{\mathfrak q'})^\wedge $ have $P$ by (C). Hence the fibres of the left vertical arrow have $P$ by (D) and the proof is complete. $\square$

Proof. The ring map $A \to A^\wedge $ is flat by Algebra, Lemma 10.97.2. The ring $A^\wedge $ is Noetherian by Algebra, Lemma 10.97.6. Thus it suffices to check the third condition of Lemma 15.51.2. Let $\mathfrak m' \subset A^\wedge $ be a maximal ideal lying over $\mathfrak m \subset A$. By Algebra, Lemma 10.96.6 we have $IA^\wedge \subset \mathfrak m'$. Since $A^\wedge /IA^\wedge = A/I$ we see that $I \subset \mathfrak m$, $\mathfrak m/I = \mathfrak m'/IA^\wedge $, and $A/\mathfrak m = A^\wedge /\mathfrak m'$. Since $A^\wedge /\mathfrak m'$ is a field, we conclude that $\mathfrak m$ is a maximal ideal as well. Then $A_\mathfrak m \to A^\wedge _{\mathfrak m'}$ is a flat local ring homomorphism of Noetherian local rings which identifies residue fields and such that $\mathfrak m A^\wedge _{\mathfrak m'} = \mathfrak m'A^\wedge _{\mathfrak m'}$. Thus it induces an isomorphism on complete local rings, see Lemma 15.43.9. Let $(A_\mathfrak m)^\wedge $ be the completion of $A_\mathfrak m$ with respect to its maximal ideal. The ring map

is faithfully flat (Algebra, Lemma 10.97.3). Thus we can apply (D) to the ring maps

to conclude because the fibres of $A_\mathfrak m \to (A_\mathfrak m)^\wedge $ have $P$ as $A$ is a $P$-ring. $\square$

Proof. Let $\mathfrak m^ h \subset A^ h$ be a maximal ideal. We have to show that the fibres of $A^ h_{\mathfrak m^ h} \to (A^ h_{\mathfrak m^ h})^\wedge $ have $P$, see Lemma 15.51.4. Let $\mathfrak m$ be the inverse image of $\mathfrak m^ h$ in $A$. Note that $I^ h \subset \mathfrak m^ h$ and hence $I \subset \mathfrak m$ as $(A^ h, I^ h)$ is a henselian pair. Recall that $A^ h$ is Noetherian, $I^ h = IA^ h$, and that $A \to A^ h$ induces an isomorphism on $I$-adic completions, see Lemma 15.12.4. Then the local homomorphism of Noetherian local rings

induces an isomorphism on completions at maximal ideals by Lemma 15.43.9 (details omitted). Let $\mathfrak q^ h$ be a prime of $A^ h_{\mathfrak m^ h}$ lying over $\mathfrak q \subset A_\mathfrak m$. Set $\mathfrak q_1 = \mathfrak q^ h$ and let $\mathfrak q_2, \ldots , \mathfrak q_ t$ be the other primes of $A^ h$ lying over $\mathfrak q$, so that $A^ h \otimes _ A \kappa (\mathfrak q) = \prod \nolimits _{i = 1, \ldots , t} \kappa (\mathfrak q_ i)$, see Lemma 15.45.12. Using that $(A^ h)_{\mathfrak m^ h}^\wedge = (A_\mathfrak m)^\wedge $ as discussed above we see

Hence, looking at local rings and using (B), we see that

has $P$ as $\kappa (\mathfrak q) \to (A_\mathfrak m)^\wedge \otimes _{A_\mathfrak m} \kappa (\mathfrak q)$ does by assumption on $A$. Since $\kappa (\mathfrak q^ h)/\kappa (\mathfrak q)$ is separable algebraic, by (E) we find that $\kappa (\mathfrak q^ h) \to (A^ h_{\mathfrak m^ h})^\wedge \otimes _{A^ h_{\mathfrak m^ h}} \kappa (\mathfrak q^ h)$ has $P$ as desired. $\square$

Proof. We have seen this for the henselization in Lemma 15.51.7. To prove it for the strict henselization, it suffices to show that the formal fibres of $R^{sh}$ have $P$, see Lemma 15.51.4. Let $\mathfrak r \subset R^{sh}$ be a prime and set $\mathfrak p = R \cap \mathfrak r$. Set $\mathfrak r_1 = \mathfrak r$ and let $\mathfrak r_2, \ldots , \mathfrak r_ s$ be the other primes of $R^{sh}$ lying over $\mathfrak p$, so that $R^{sh} \otimes _ R \kappa (\mathfrak p) = \prod \nolimits _{i = 1, \ldots , s} \kappa (\mathfrak r_ i)$, see Lemma 15.45.13. Then we see that

Note that $R^\wedge \to (R^{sh})^\wedge $ is formally smooth in the $\mathfrak m_{(R^{sh})^\wedge }$-adic topology, see Lemma 15.45.3. Hence $R^\wedge \to (R^{sh})^\wedge $ is regular by Proposition 15.49.2. We conclude that property $P$ holds for $\kappa (\mathfrak p) \to (R^{sh})^\wedge \otimes _ R \kappa (\mathfrak p)$ by (C) and our assumption on $R$. Using property (B), using the decomposition above, and looking at local rings we conclude that property $P$ holds for $\kappa (\mathfrak p) \to (R^{sh})^\wedge \otimes _{R^{sh}} \kappa (\mathfrak r)$. Since $\kappa (\mathfrak r)/\kappa (\mathfrak p)$ is separable algebraic, it follows from (E) that $P$ holds for $\kappa (\mathfrak r) \to (R^{sh})^\wedge \otimes _{R^{sh}} \kappa (\mathfrak r)$. $\square$

Proof. Part (A) follows from the definition of geometrically reduced algebras (Algebra, Definition 10.43.1). Part (B) follows too: a ring is reduced if and only if all local rings are reduced. Part (C). This follows from Lemma 15.42.1. Part (D). This follows from Algebra, Lemma 10.164.2. Part (E). This follows from Algebra, Lemma 10.43.9. $\square$

Proof. Part (A) follows from the definition of geometrically normal algebras (Algebra, Definition 10.165.2). Part (B) follows too: a ring is normal if and only if all of its local rings are normal. Part (C). This follows from Lemma 15.42.2. Part (D). This follows from Algebra, Lemma 10.164.3. Part (E). This follows from Algebra, Lemma 10.165.6. $\square$

Proof. Let $k \to R$ be a ring map where $k$ is a field and $R$ a Noetherian ring. Let $k'/k$ be a finitely generated field extension. Then the fibres of the ring map $R \to R \otimes _ k k'$ are Cohen-Macaulay by Algebra, Lemma 10.167.1. Hence we may apply Algebra, Lemma 10.163.4 to the ring map $R \to R \otimes _ k k'$ to see that if $R$ has $(S_ n)$ so does $R \otimes _ k k'$. This proves (A). Part (B) follows too: a Noetherian rings has $(S_ n)$ if and only if all of its local rings have $(S_ n)$. Part (C). This follows from Algebra, Lemma 10.163.4 as the fibres of a regular homomorphism are regular and in particular Cohen-Macaulay. Part (D). This follows from Algebra, Lemma 10.164.5. Part (E). This is immediate as the condition does not refer to the ground field. $\square$

Proof. Follows immediately from Lemma 15.51.11 and the fact that a Noetherian ring is Cohen-Macaulay if and only if it satisfies conditions $(S_ n)$ for all $n$. $\square$

Proof. Let $k \to R$ be a ring map where $k$ is a field and $R$ a Noetherian ring. Assume $P(k \to R)$ is true. Let $K/k$ be a finitely generated field extension. By Algebra, Lemma 10.45.3 we can find a diagram

where $k'/k$, $K'/K$ are finite purely inseparable field extensions such that $K'/k'$ is separable. By Algebra, Lemma 10.158.10 there exists a smooth $k'$-algebra $B$ such that $K'$ is the fraction field of $B$. Now we can argue as follows: Step 1: $R \otimes _ k k'$ satisfies $(S_ n)$ because we assumed $P$ for $k \to R$. Step 2: $R \otimes _ k k' \to R \otimes _ k k' \otimes _{k'} B$ is a smooth ring map (Algebra, Lemma 10.137.4) and we conclude $R \otimes _ k k' \otimes _{k'} B$ satisfies $(S_ n)$ by Algebra, Lemma 10.163.5 (and using Algebra, Lemma 10.140.3 to see that the hypotheses are satisfied). Step 3. $R \otimes _ k k' \otimes _{k'} K' = R \otimes _ k K'$ satisfies $(R_ n)$ as it is a localization of a ring having $(R_ n)$. Step 4. Finally $R \otimes _ k K$ satisfies $(R_ n)$ by descent of $(R_ n)$ along the faithfully flat ring map $K \otimes _ k A \to K' \otimes _ k A$ (Algebra, Lemma 10.164.6). This proves (A). Part (B) follows too: a Noetherian ring has $(R_ n)$ if and only if all of its local rings have $(R_ n)$. Part (C). This follows from Algebra, Lemma 10.163.5 as the fibres of a regular homomorphism are regular (small detail omitted). Part (D). This follows from Algebra, Lemma 10.164.6 (small detail omitted).

Part (E). Let $l/k$ be a separable algebraic extension of fields and let $l \to R$ be a ring map with $R$ Noetherian. Assume that $k \to R$ has $P$. We have to show that $l \to R$ has $P$. Let $l'/l$ be a finite extension. First observe that there exists a finite subextension $l/m/k$ and a finite extension $m'/m$ such that $l' = l \otimes _ m m'$. Then $R \otimes _ l l' = R \otimes _ m m'$. Hence it suffices to prove that $m \to R$ has property $P$, i.e., we may assume that $l/k$ is finite. If $l/k$ is finite, then $l'/k$ is finite and we see that

is a localization (by Algebra, Lemma 10.43.8) of the Noetherian ring $l' \otimes _ k R$ which has property $(R_ n)$ by assumption $P$ for $k \to R$. This proves that $l' \otimes _ l R$ has property $(R_ n)$ as desired. $\square$