why separable algebra is a generalisation of separable field extension?
They are defined differently.
Suppose we have a field $K$.
We say a finite field extension $L$ is separable over $K$ iff the number of embeddings $Lhookrightarrow bar K$ into the algebraic closure $bar K$ which is invariant on $K$ is $[L:K]$.
We say a finitely generated $K$-algebra $A$ is separable iff for any field extension $E/K$, $Aotimes_KE$ is reduced.
I have read that separable algebra is a generalisation of separable field extension. I think there is a proposition that
For a finite field extension $L/K$, if the field extension is
separable, then for any other field extension $E/K$, the $E$-algebra
$Lotimes_KE$ is reduced.
Can someone give some suggestions, I have no idea how to apply separability of field extensions. Or even a simpler case about $K[alpha]$ for $alpha$ separable over $K$.
commutative-algebra field-theory
asked Jan 4 at 19:09
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add a comment |
They are defined differently.
Suppose we have a field $K$.
We say a finite field extension $L$ is separable over $K$ iff the number of embeddings $Lhookrightarrow bar K$ into the algebraic closure $bar K$ which is invariant on $K$ is $[L:K]$.
We say a finitely generated $K$-algebra $A$ is separable iff for any field extension $E/K$, $Aotimes_KE$ is reduced.
I have read that separable algebra is a generalisation of separable field extension. I think there is a proposition that
For a finite field extension $L/K$, if the field extension is
separable, then for any other field extension $E/K$, the $E$-algebra
$Lotimes_KE$ is reduced.
Can someone give some suggestions, I have no idea how to apply separability of field extensions. Or even a simpler case about $K[alpha]$ for $alpha$ separable over $K$.
commutative-algebra field-theory
asked Jan 4 at 19:09
Display NameDisplay Name
427212
See wikipedia and its references. A separable algebra is a generalization to associative algebras of the notion of a separable field extension.
– Dietrich Burde
Jan 4 at 20:15
1
Gosh are you using some weird definitions. Do you know how to define separable algebras in terms of separability idempotents? And how to define separable field extensions in terms of minimal polynomials? In that case, you can look up the proof of Theorem 23 in my A problem on bilinear maps (to be precise, I only prove Proposition 24 there; but the primitive element theorem says that any separable field extension $L$ of $K$ can be written as $Kleft[xright]$ for some $x in L$, and thus Proposition 24 yields Theorem 23).
– darij grinberg
Jan 4 at 20:19
add a comment |
They are defined differently.
Suppose we have a field $K$.
We say a finite field extension $L$ is separable over $K$ iff the number of embeddings $Lhookrightarrow bar K$ into the algebraic closure $bar K$ which is invariant on $K$ is $[L:K]$.
We say a finitely generated $K$-algebra $A$ is separable iff for any field extension $E/K$, $Aotimes_KE$ is reduced.
I have read that separable algebra is a generalisation of separable field extension. I think there is a proposition that
For a finite field extension $L/K$, if the field extension is
separable, then for any other field extension $E/K$, the $E$-algebra
$Lotimes_KE$ is reduced.
Can someone give some suggestions, I have no idea how to apply separability of field extensions. Or even a simpler case about $K[alpha]$ for $alpha$ separable over $K$.
commutative-algebra field-theory
asked Jan 4 at 19:09
Display NameDisplay Name
427212
They are defined differently.
Suppose we have a field $K$.
We say a finite field extension $L$ is separable over $K$ iff the number of embeddings $Lhookrightarrow bar K$ into the algebraic closure $bar K$ which is invariant on $K$ is $[L:K]$.
We say a finitely generated $K$-algebra $A$ is separable iff for any field extension $E/K$, $Aotimes_KE$ is reduced.
I have read that separable algebra is a generalisation of separable field extension. I think there is a proposition that
For a finite field extension $L/K$, if the field extension is
separable, then for any other field extension $E/K$, the $E$-algebra
$Lotimes_KE$ is reduced.
Can someone give some suggestions, I have no idea how to apply separability of field extensions. Or even a simpler case about $K[alpha]$ for $alpha$ separable over $K$.
commutative-algebra field-theory
commutative-algebra field-theory
asked Jan 4 at 19:09
Display NameDisplay Name
427212
asked Jan 4 at 19:09
Display NameDisplay Name
427212
asked Jan 4 at 19:09
Display NameDisplay Name
427212
asked Jan 4 at 19:09
Display NameDisplay Name
427212
asked Jan 4 at 19:09
Display NameDisplay Name
427212
427212
See wikipedia and its references. A separable algebra is a generalization to associative algebras of the notion of a separable field extension.
– Dietrich Burde
Jan 4 at 20:15
1
Gosh are you using some weird definitions. Do you know how to define separable algebras in terms of separability idempotents? And how to define separable field extensions in terms of minimal polynomials? In that case, you can look up the proof of Theorem 23 in my A problem on bilinear maps (to be precise, I only prove Proposition 24 there; but the primitive element theorem says that any separable field extension $L$ of $K$ can be written as $Kleft[xright]$ for some $x in L$, and thus Proposition 24 yields Theorem 23).
– darij grinberg
Jan 4 at 20:19
add a comment |
See wikipedia and its references. A separable algebra is a generalization to associative algebras of the notion of a separable field extension.
– Dietrich Burde
Jan 4 at 20:15
1
Gosh are you using some weird definitions. Do you know how to define separable algebras in terms of separability idempotents? And how to define separable field extensions in terms of minimal polynomials? In that case, you can look up the proof of Theorem 23 in my A problem on bilinear maps (to be precise, I only prove Proposition 24 there; but the primitive element theorem says that any separable field extension $L$ of $K$ can be written as $Kleft[xright]$ for some $x in L$, and thus Proposition 24 yields Theorem 23).
– darij grinberg
Jan 4 at 20:19
See wikipedia and its references. A separable algebra is a generalization to associative algebras of the notion of a separable field extension.
– Dietrich Burde
Jan 4 at 20:15
See wikipedia and its references. A separable algebra is a generalization to associative algebras of the notion of a separable field extension.
– Dietrich Burde
Jan 4 at 20:15
1
1
Gosh are you using some weird definitions. Do you know how to define separable algebras in terms of separability idempotents? And how to define separable field extensions in terms of minimal polynomials? In that case, you can look up the proof of Theorem 23 in my A problem on bilinear maps (to be precise, I only prove Proposition 24 there; but the primitive element theorem says that any separable field extension $L$ of $K$ can be written as $Kleft[xright]$ for some $x in L$, and thus Proposition 24 yields Theorem 23).
– darij grinberg
Jan 4 at 20:19
Gosh are you using some weird definitions. Do you know how to define separable algebras in terms of separability idempotents? And how to define separable field extensions in terms of minimal polynomials? In that case, you can look up the proof of Theorem 23 in my A problem on bilinear maps (to be precise, I only prove Proposition 24 there; but the primitive element theorem says that any separable field extension $L$ of $K$ can be written as $Kleft[xright]$ for some $x in L$, and thus Proposition 24 yields Theorem 23).
– darij grinberg
Jan 4 at 20:19
add a comment |
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See wikipedia and its references. A separable algebra is a generalization to associative algebras of the notion of a separable field extension.
– Dietrich Burde
Jan 4 at 20:15
1
Gosh are you using some weird definitions. Do you know how to define separable algebras in terms of separability idempotents? And how to define separable field extensions in terms of minimal polynomials? In that case, you can look up the proof of Theorem 23 in my A problem on bilinear maps (to be precise, I only prove Proposition 24 there; but the primitive element theorem says that any separable field extension $L$ of $K$ can be written as $Kleft[xright]$ for some $x in L$, and thus Proposition 24 yields Theorem 23).
– darij grinberg
Jan 4 at 20:19