Timeline for Is a "separable" algebra over a field finite-dimensional?
Current License: CC BY-SA 4.0
23 events
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| Feb 10, 2022 at 1:34 | history | edited | Benjamin Steinberg | CC BY-SA 4.0 |
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| Feb 10, 2022 at 1:07 | history | edited | Benjamin Steinberg | CC BY-SA 4.0 |
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| Feb 10, 2022 at 0:58 | history | edited | Benjamin Steinberg | CC BY-SA 4.0 |
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| Feb 10, 2022 at 0:54 | comment | added | Benjamin Steinberg | @H.E., I think I have now fixed the proof. If $F/k$ is infnitely generated, I can use Vamos. Otherwise, it is finitely generated and infinite dimensional and so much have positive trancendence degree. Then $F\otimes_k F$ is not artinian by a result of Sharp and I can use that. | |
| Feb 10, 2022 at 0:52 | history | edited | Benjamin Steinberg | CC BY-SA 4.0 |
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| Feb 10, 2022 at 0:48 | comment | added | Benjamin Steinberg | I misread it as finite extension. But if you have a nonalgebraic extension, then $F\otimes_k F$ is not artinian. So let me fix. | |
| Feb 10, 2022 at 0:22 | comment | added | H. E. | Thank you for your answer! But I have a question on the part when you use Vamos's paper. You claim that if $F/k$ is an infinite extension, then $F \otimes_k F$ is not noetherian. I check the paper and found that $F \otimes_k F$ is noetherian iff $F$ is a finitely generated extension, not a finite extension. Actually, I think that for $F = k(x)$, which is an infinite extension of $k$, $F \otimes_k F$ is noetherian. | |
| Feb 9, 2022 at 17:46 | comment | added | Benjamin Steinberg | @UriyaFirst, yes that fixed the D algebraic over k issue but your comment removed my other obstacle which was more serious. | |
| Feb 9, 2022 at 17:43 | comment | added | Uriya First | The proof is now correct as far as I can tell. Going to the algebraic closure at first indeed bypasses the complications arising if $D$ is algebraic over $k$. | |
| Feb 9, 2022 at 16:52 | history | edited | Benjamin Steinberg | CC BY-SA 4.0 |
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| Feb 9, 2022 at 16:38 | comment | added | Benjamin Steinberg | For the commutative case there is an easier proof. If $A$ is commutative and satisfies 2., then $A$ is a direct product of fields and so $A\otimes_k A$ will be semisimple since tensor products commute with direct product. Thus $A$ is a projective $A\otimes_k A$-module and so $A$ satisfies 1 and hence 3. | |
| Feb 9, 2022 at 16:07 | history | edited | Benjamin Steinberg | CC BY-SA 4.0 |
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| Feb 9, 2022 at 16:02 | comment | added | Benjamin Steinberg | @UriyaFirst, I found what I think is a correct proof now using your idea. | |
| Feb 9, 2022 at 16:01 | history | edited | Benjamin Steinberg | CC BY-SA 4.0 |
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| Feb 9, 2022 at 15:41 | history | edited | Benjamin Steinberg | CC BY-SA 4.0 |
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| Feb 9, 2022 at 15:38 | comment | added | Benjamin Steinberg | @Uriya First, The previous version was slightly wrong. What I really get are subfields of unbounded dimension. I guess they could still all be finite. It is an open question I believe whether a division algebra all of whose elements are algebraic over its center is necessarily finite dimensional over the center. | |
| Feb 9, 2022 at 15:29 | comment | added | Uriya First | In a previous version of your answer you said that if $D$ is infinite-dimensional over $Z(D)$, then there is a subfield $F$ of $D$ of infinite dimension over $k$. The ring $D$ is a free right $F$-module. Consequently, $D\otimes_kF$ is a free right $F\otimes_kF$-module. The latter ring is not noetherian, so there is a chain of ideals $I_1\subsetneq I_2 \subsetneq I_3\subsetneq\dots$ in $F\otimes_kF$. Since $D\otimes_kF$ is free over $F\otimes_kF$, we get a strictly increasing chain of left ideals $(D\otimes_k F)I_1\subsetneq(D\otimes_k F)I_2\subsetneq\dots$, so $D\otimes_k F$ is not noetherian. | |
| Feb 9, 2022 at 15:22 | history | edited | Benjamin Steinberg | CC BY-SA 4.0 |
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| Feb 9, 2022 at 15:15 | history | edited | Benjamin Steinberg | CC BY-SA 4.0 |
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| Feb 9, 2022 at 15:13 | history | edited | Benjamin Steinberg | CC BY-SA 4.0 |
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| Feb 9, 2022 at 15:08 | history | edited | Benjamin Steinberg | CC BY-SA 4.0 |
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| Feb 9, 2022 at 15:02 | history | edited | Benjamin Steinberg | CC BY-SA 4.0 |
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| Feb 9, 2022 at 14:27 | history | answered | Benjamin Steinberg | CC BY-SA 4.0 |