I asked this question a couple of weeks ago.

The question arose while looking for a criterion of separability for extensions of fraction fields $K(A)\to K(B)$ induced by a faithfully flat morphism $A\to B$ between algebras over a field of positive characteristic which are domains with $A$ Noetherian, and such that the induced morphism $A/m\to B/mB$ is bijective for every maximal ideal $m$ of $A$.

G.Leuschke gave me this reference, whose results (concretely, the Theorem 1.8) allows one to conclude that the separability is guaranteed when $B$ is also Noetherian by the argument that follows:

1. The next theorem is an exercise from Bourbaki's Algebra II:

Theorem 1 (Bourbaki Alg II, V.15. Ex.11): Let $K$ be a field of characteristic $p>0$ and let $C$ be a $K$-algebra. Then $C$ is separable if and only if for every family of elements $\{k_{i} \}\subset K$ linearly free over $K^{p}$ and every family $\{c_{i}\}\subset C$ (with $c_{i}=0$ except for a finite number of subindices) the equality $$\sum_{i} k_{i}c_{i}^{p}=0$$ implies that $c_{i}=0$ (for every $i$).

2. Now, it is not hard to see that this exercise gives the following:

Theorem 2: Let $A\hookrightarrow B$ be a flat extension of algebras over a field of characteristic $p>0$ which are domains, and denote by $A^{1/p}$ (resp, by $B^{1/p}$) the algebra $A$ (resp. $B$) seen as an $A$ (resp $B$) Module via the Frobenius map. Then the field extension $K(A)\to K(B)$ is separable if and only every finite set $a_{1},\dots, a_{n}\in A$ of free elements in $A^{1/p}$ is free in $B^{1/p}$. This happens if and only if the canonical map
$$B \otimes_{A} \left\langle a_{1},\dots ,a_{n} \right\rangle_{B^{1/p}}\to \left\langle a_{1},\dots ,a_{n} \right\rangle_{B^{1/p}}$$ is injective.

3. Here comes Frankild et al's paper: when $A, B$ are Noetherian, $A\to B$ is faithfully flat and the induced map $A/m\to B/mB$ is bijective for every maximal ideal $m$ of $A$, an easy application of the local-global principle together with Theorem 1.8 in the paper guarantees the separability of $K(A)\to K(B)$.

I thought the criterion was extendible to the case in which $B$ is not Noetherian, but there were mistakes in my argument. Does anybody have an idea on how to prove (or refute) the corresponding affirmation in such a case? (Frankild et al's paper can give you some hints, but I don't write them down in order to avoid bias).

P.S: The argument can be extended easily, I think, to the case in which the localization of $B$ at every maximal ideal $\eta$ is $\eta$-adically separated.

I asked this question a couple of weeks ago.

The question arose while looking for a criterion of separability for extensions of fraction fields $K(A)\to K(B)$ induced by a faithfully flat morphism $A\to B$ between algebras over a field of positive characteristic which are domains with $A$ Noetherian, and such that the induced morphism $A/m\to B/mB$ is bijective for every maximal ideal $m$ of $A$.

G.Leuschke gave me this reference, whose results (concretely, the Theorem 1.8) allows one to conclude that the separability is guaranteed when $B$ is also Noetherian by the argument that follows:

1. The next theorem is an exercise from Bourbaki's Algebra II:

Theorem 1 (Bourbaki Alg II, V.15. Ex.11): Let $K$ be a field of characteristic $p>0$ and let $C$ be a $K$-algebra. Then $C$ is separable if and only if for every family of elements $\{k_{i} \}\subset K$ linearly free over $K^{p}$ and every family $\{c_{i}\}\subset C$ (with $c_{i}=0$ except for a finite number of subindices) the equality $$\sum_{i} k_{i}c_{i}^{p}=0$$ implies that $c_{i}=0$ (for every $i$).

2. Now, it is not hard to see that this exercise gives the following:

Theorem 2: Let $A\hookrightarrow B$ be a flat extension of algebras over a field of characteristic $p>0$ which are domains, and denote by $A^{1/p}$ (resp, by $B^{1/p}$) the algebra $A$ (resp. $B$) seen as an $A$ (resp $B$) Module via the Frobenius map. Then the field extension $K(A)\to K(B)$ is separable if and only every finite set $a_{1},\dots, a_{n}\in A$ of free elements in $A^{1/p}$ is free in $B^{1/p}$. This happens if and only if the canonical map
$$B \otimes_{A} \left\langle a_{1},\dots ,a_{n} \right\rangle_{B^{1/p}}\to \left\langle a_{1},\dots ,a_{n} \right\rangle_{B^{1/p}}$$ is injective.

3. Here comes Frankild et al's paper: when $A, B$ are Noetherian, $A\to B$ is faithfully flat and the induced map $A/m\to B/mB$ is bijective for every maximal ideal $m$ of $A$, an easy application of the local-global principle together with Theorem 1.8 in the paper guarantees the separability of $K(A)\to K(B)$.

I thought the criterion was extendible to the case in which $B$ is not Noetherian, but there were mistakes in my argument. Does anybody have an idea on how to prove (or refute) the corresponding affirmation in such a case? (Frankild et al's paper can give you some hints, but I don't write them down in order to avoid bias).

P.S: The argument can be extended easily, I think, to the case in which the localization of $B$ at every maximal ideal $\eta$ is $\eta$-adically separated.

Hi,

I made this question a couple of weeks ago.

The question arose while looking for a criterion of separability for extensions of fraction fields $K(A)\to K(B)$ induced by a faithfully flat morphism $A\to B$ between algebras over a field of positive characteristic which are domains with $A$ Noetherian, and such that the induced morphism $A/m\to B/mB$ is bijective for every maximal ideal $m$ of $A$.

G.Leuschke gave me this reference, whose results (concretely, the Theorem 1.8) allows one to conclude that the separability is guaranteed when $B$ is also Noetherian by the argument that follows:

1. The next theorem is an exercise from Bourbaki's Algebra II:

Theorem 1 (Bourbaki Alg II, V.15. Ex.11): Let $K$ be a field of characteristic $p>0$ and let $C$ be a $K$-algebra. Then $C$ is separable if and only if for every family of elements $\{k_{i} \}\subset K$ linearly free over $K^{p}$ and every family $\{c_{i}\}\subset C$ (with $c_{i}=0$ except for a finite number of subindices) the equality $$\sum_{i} k_{i}c_{i}^{p}=0$$ implies that $c_{i}=0$ (for every $i$).

2. Now, it is not hard to see that this exercise gives the following:

Theorem 2: Let $A\hookrightarrow B$ be a flat extension of algebras over a field of characteristic $p>0$ which are domains, and denote by $A^{1/p}$ (resp, by $B^{1/p}$) the algebra $A$ (resp. $B$) seen as an $A$ (resp $B$) Module via the Frobenius map. Then the field extension $K(A)\to K(B)$ is separable if and only every finite set $a_{1},\dots, a_{n}\in A$ of free elements in $A^{1/p}$ is free in $B^{1/p}$. This happens if and only if the canonical map
$$B \otimes_{A} \left\langle a_{1},\dots ,a_{n} \right\rangle_{B^{1/p}}\to \left\langle a_{1},\dots ,a_{n} \right\rangle_{B^{1/p}}$$ is injective.

3. Here comes Frankild et al's paper: when $A, B$ are Noetherian, $A\to B$ is faithfully flat and the induced map $A/m\to B/mB$ is bijective for every maximal ideal $m$ of $A$, an easy application of the local-global principle together with Theorem 1.8 in the paper guarantees the separability of $K(A)\to K(B)$.

I thought the criterion was extendible to the case in which $B$ is not Noetherian, but there were mistakes in my argument. Does anybody have an idea on how to prove (or refute) the corresponding affirmation in such a case? (Frankild et al's paper can give you some hints, but I don't write them down in order to avoid bias).

P.S: The argument can be extended easily, I think, to the case in which the localization of $B$ at every maximal ideal $\eta$ is $\eta$-adically separated.

I asked this question a couple of weeks ago.

The question arose while looking for a criterion of separability for extensions of fraction fields $K(A)\to K(B)$ induced by a faithfully flat morphism $A\to B$ between algebras over a field of positive characteristic which are domains with $A$ Noetherian, and such that the induced morphism $A/m\to B/mB$ is bijective for every maximal ideal $m$ of $A$.

G.Leuschke gave me this reference, whose results (concretely, the Theorem 1.8) allows one to conclude that the separability is guaranteed when $B$ is also Noetherian by the argument that follows:

1. The next theorem is an exercise from Bourbaki's Algebra II:

Theorem 1 (Bourbaki Alg II, V.15. Ex.11): Let $K$ be a field of characteristic $p>0$ and let $C$ be a $K$-algebra. Then $C$ is separable if and only if for every family of elements $\{k_{i} \}\subset K$ linearly free over $K^{p}$ and every family $\{c_{i}\}\subset C$ (with $c_{i}=0$ except for a finite number of subindices) the equality $$\sum_{i} k_{i}c_{i}^{p}=0$$ implies that $c_{i}=0$ (for every $i$).

2. Now, it is not hard to see that this exercise gives the following:

Theorem 2: Let $A\hookrightarrow B$ be a flat extension of algebras over a field of characteristic $p>0$ which are domains, and denote by $A^{1/p}$ (resp, by $B^{1/p}$) the algebra $A$ (resp. $B$) seen as an $A$ (resp $B$) Module via the Frobenius map. Then the field extension $K(A)\to K(B)$ is separable if and only every finite set $a_{1},\dots, a_{n}\in A$ of free elements in $A^{1/p}$ is free in $B^{1/p}$. This happens if and only if the canonical map
$$B \otimes_{A} \left\langle a_{1},\dots ,a_{n} \right\rangle_{B^{1/p}}\to \left\langle a_{1},\dots ,a_{n} \right\rangle_{B^{1/p}}$$ is injective.

3. Here comes Frankild et al's paper: when $A, B$ are Noetherian, $A\to B$ is faithfully flat and the induced map $A/m\to B/mB$ is bijective for every maximal ideal $m$ of $A$, an easy application of the local-global principle together with Theorem 1.8 in the paper guarantees the separability of $K(A)\to K(B)$.

I thought the criterion was extendible to the case in which $B$ is not Noetherian, but there were mistakes in my argument. Does anybody have an idea on how to prove (or refute) the corresponding affirmation in such a case? (Frankild et al's paper can give you some hints, but I don't write them down in order to avoid bias).

P.S: The argument can be extended easily, I think, to the case in which the localization of $B$ at every maximal ideal $\eta$ is $\eta$-adically separated.

Hi,

I made this question a couple of weeks ago.

The question arose while looking for a criterion of separability for extensions of fraction fields $K(A)\to K(B)$ induced by a faithfully flat morphism $A\to B$ between algebras over a field of positive characteristic which are domains with $A$ Noetherian, and such that the induced morphism $A/m\to B/mB$ is bijective for every maximal $m$ of $A$.

G.Leuschke gave me this reference, whose results (concretely, the Theorem 1.8) allows one to conclude that the separability is guaranteed when $B$ is also Noetherian by the argument that follows:

1. The next theorem is an exercise from Bourbaki's Algebra II:

Theorem 1 (Bourbaki Alg II, V.15. Ex.11): Let $K$ be a field of characteristic $p>0$ and let $C$ be a $K$-algebra. Then $C$ is separable if and only if for every family of elements $\{k_{i} \}\subset K$ linearly free over $K^{p}$ and every family $\{c_{i}\}\subset C$ (with $c_{i}=0$ except for a finite number of subindices) the equality $$\sum_{i} k_{i}c_{i}^{p}=0$$ implies that $c_{i}=0$ (for every $i$).

2. Now, it is not hard to see that this exercise gives the following:

Theorem 2: Let $A\hookrightarrow B$ be a flat extension of algebras over a field of characteristic $p>0$ which are domains, and denote by $A^{1/p}$ (resp, by $B^{1/p}$) the algebra $A$ (resp. $B$) seen as an $A$ (resp $B$) Module via the Frobenius map. Then the field extension $K(A)\to K(B)$ is separable if and only every finite set $a_{1},\dots, a_{n}\in A$ of free elements in $A^{1/p}$ is free in $B^{1/p}$. This happens if and only if the canonical map
$$B \otimes_{A} \left\langle a_{1},\dots ,a_{n} \right\rangle_{B^{1/p}}\to \left\langle a_{1},\dots ,a_{n} \right\rangle_{B^{1/p}}$$ is injective.

3. Here comes Frankild et al's paper: when $A, B$ are Noetherian, $A\to B$ is faithfully flat and the induced map $A/m\to B/mB$ is bijective for every maximal ideal $m$ of $A$, an easy application of the local-global principle together with Theorem 1.8 in the paper guarantees the separability of $K(A)\to K(B)$.

I thought the criterion was extendible to the case in which $B$ is not Noetherian, but there were mistakes in my argument. Does anybody have an idea on how to prove (or refute) the corresponding affirmation in such a case? (Frankild et al's paper can give you some hints, but I don't write them down in order to avoid bias).

P.S: The argument can be extended easily, I think, to the case in which the localization of $B$ at every maximal ideal $\eta$ is $\eta$-adically separated.

Hi,

I made this question a couple of weeks ago.

The question arose while looking for a criterion of separability for extensions of fraction fields $K(A)\to K(B)$ induced by a faithfully flat morphism $A\to B$ between algebras over a field of positive characteristic which are domains with $A$ Noetherian, and such that the induced morphism $A/m\to B/mB$ is bijective for every maximal ideal $m$ of $A$.

G.Leuschke gave me this reference, whose results (concretely, the Theorem 1.8) allows one to conclude that the separability is guaranteed when $B$ is also Noetherian by the argument that follows:

1. The next theorem is an exercise from Bourbaki's Algebra II:

Theorem 1 (Bourbaki Alg II, V.15. Ex.11): Let $K$ be a field of characteristic $p>0$ and let $C$ be a $K$-algebra. Then $C$ is separable if and only if for every family of elements $\{k_{i} \}\subset K$ linearly free over $K^{p}$ and every family $\{c_{i}\}\subset C$ (with $c_{i}=0$ except for a finite number of subindices) the equality $$\sum_{i} k_{i}c_{i}^{p}=0$$ implies that $c_{i}=0$ (for every $i$).

2. Now, it is not hard to see that this exercise gives the following:

Theorem 2: Let $A\hookrightarrow B$ be a flat extension of algebras over a field of characteristic $p>0$ which are domains, and denote by $A^{1/p}$ (resp, by $B^{1/p}$) the algebra $A$ (resp. $B$) seen as an $A$ (resp $B$) Module via the Frobenius map. Then the field extension $K(A)\to K(B)$ is separable if and only every finite set $a_{1},\dots, a_{n}\in A$ of free elements in $A^{1/p}$ is free in $B^{1/p}$. This happens if and only if the canonical map
$$B \otimes_{A} \left\langle a_{1},\dots ,a_{n} \right\rangle_{B^{1/p}}\to \left\langle a_{1},\dots ,a_{n} \right\rangle_{B^{1/p}}$$ is injective.

3. Here comes Frankild et al's paper: when $A, B$ are Noetherian, $A\to B$ is faithfully flat and the induced map $A/m\to B/mB$ is bijective for every maximal ideal $m$ of $A$, an easy application of the local-global principle together with Theorem 1.8 in the paper guarantees the separability of $K(A)\to K(B)$.

I thought the criterion was extendible to the case in which $B$ is not Noetherian, but there were mistakes in my argument. Does anybody have an idea on how to prove (or refute) the corresponding affirmation in such a case? (Frankild et al's paper can give you some hints, but I don't write them down in order to avoid bias).

P.S: The argument can be extended easily, I think, to the case in which the localization of $B$ at every maximal ideal $\eta$ is $\eta$-adically separated.