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Let $A \subseteq B$ be commutative noetherian rings. I have found the following claim: "Separability implies smoothness" with the following explanation: "The natural thing is to prove that a separable algebra is a regular homomorphism (and so smooth if it is of finite type). In general ...".

Remark: I am only interested in finitely generated algebras, so did not quote the rest of the explanation.

However, it seems to me, at least with my definitions, that this claim is false: If $A \subseteq B$ is separable, then (assuming the claim is true) $A \subseteq B$ is smooth, then by a result of Grothendieck, $A \subseteq B$ is flat, but this seems too much (especially in view of Examples on pages 95-97 and Corollary 9; I do not see why separability should imply flatness of $B$ over $A$. Notice that actually, this is what the explanation suggests: it says that separability implies $A \to B$ is a regular homomorphism, and by definition, a regular homomorphism is flat).

I will really appreciate if one can either post a sketch of proof/tell me where I can find a proof, or a counterexample.

Actually, I have already posted this question here, but got no comments thus far.

EDIT: I work with the definition of separability of this book, and with the definition of smoothness of this paper.

Let $A \subseteq B$ be commutative noetherian rings. I have found the following claim: "Separability implies smoothness" with the following explanation: "The natural thing is to prove that a separable algebra is a regular homomorphism (and so smooth if it is of finite type). In general ...".

Remark: I am only interested in finitely generated algebras, so did not quote the rest of the explanation.

However, it seems to me, at least with my definitions, that this claim is false: If $A \subseteq B$ is separable, then (assuming the claim is true) $A \subseteq B$ is smooth, then by a result of Grothendieck, $A \subseteq B$ is flat, but this seems too much (especially in view of Examples on pages 95-97 and Corollary 9; I do not see why separability should imply flatness of $B$ over $A$. Notice that actually, this is what the explanation suggests: it says that separability implies $A \to B$ is a regular homomorphism, and by definition, a regular homomorphism is flat).

I will really appreciate if one can either post a sketch of proof/tell me where I can find a proof, or a counterexample.

Actually, I have already posted this question here, but got no comments thus far.

EDIT: I work with the definition of separability of this book, and with the definition of smoothness of this paper.

Let $A \subseteq B$ be commutative noetherian rings. I have found the following claim: "Separability implies smoothness" with the following explanation: "The natural thing is to prove that a separable algebra is a regular homomorphism (and so smooth if it is of finite type). In general ...".

Remark: I am only interested in finitely generated algebras, so did not quote the rest of the explanation.

However, it seems to me, at least with my definitions, that this claim is false: If $A \subseteq B$ is separable, then (assuming the claim is true) $A \subseteq B$ is smooth, then by a result of Grothendieck, $A \subseteq B$ is flat, but this seems too much (especially in view of Examples on pages 95-97 and Corollary 9; I do not see why separability should imply flatness of $B$ over $A$. Notice that actually, this is what the explanation suggests: it says that separability implies $A \to B$ is a regular homomorphism, and by definition, a regular homomorphism is flat).

I will really appreciate if one can either post a sketch of proof/tell me where I can find a proof, or a counterexample.

Actually, I have already posted this question here, but got no comments thus far.

Let $A \subseteq B$ be commutative noetherian rings. I have found the following claim: "Separability implies smoothness" with the following explanation: "The natural thing is to prove that a separable algebra is a regular homomorphism (and so smooth if it is of finite type). In general ...".

Remark: I am only interested in finitely generated algebras, so did not quote the rest of the explanation.

However, it seems to me, at least with my definitions, that this claim is false: If $A \subseteq B$ is separable, then (assuming the claim is true) $A \subseteq B$ is smooth, then by a result of Grothendieck, $A \subseteq B$ is flat, but this seems too much (especially in view of Examples on pages 95-97 and Corollary 9; I do not see why separability should imply flatness of $B$ over $A$. Notice that actually, this is what the explanation suggests: it says that separability implies $A \to B$ is a regular homomorphism, and by definition, a regular homomorphism is flat).

I will really appreciate if one can either post a sketch of proof/tell me where I can find a proof, or a counterexample.

Actually, I have already posted this question here, but got no comments thus far.

EDIT: I work with the definition of separability of this book, and with the definition of smoothness of this paper.

Let $A \subseteq B$ be commutative noetherian rings. I have found the following claim: "Separability implies smoothness".

However, it seems to me that, at least with my definitions, this claim is false: If $A \subseteq B$ is separable, then (assuming the claim is true) $A \subseteq B$ is smooth, then by a result of Grothendieck, $A \subseteq B$ is flat, but this seems too much (especially in view of Examples on pages 95-97 and Corollary 9).

I will really appreciate if one can either post a sketch of proof/tell me where I can find a proof, or a counterexample.

Actually, I have already posted this question here, but got no comments thus far.

Let $A \subseteq B$ be commutative noetherian rings. I have found the following claim: "Separability implies smoothness" with the following explanation: "The natural thing is to prove that a separable algebra is a regular homomorphism (and so smooth if it is of finite type). In general ...".

Remark: I am only interested in finitely generated algebras, so did not quote the rest of the explanation.

However, it seems to me, at least with my definitions, that this claim is false: If $A \subseteq B$ is separable, then (assuming the claim is true) $A \subseteq B$ is smooth, then by a result of Grothendieck, $A \subseteq B$ is flat, but this seems too much (especially in view of Examples on pages 95-97 and Corollary 9; I do not see why separability should imply flatness of $B$ over $A$. Notice that actually, this is what the explanation suggests: it says that separability implies $A \to B$ is a regular homomorphism, and by definition, a regular homomorphism is flat).

I will really appreciate if one can either post a sketch of proof/tell me where I can find a proof, or a counterexample.

Actually, I have already posted this question here, but got no comments thus far.