Timeline for When is the tensor product of two fields a field?
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| Feb 13, 2020 at 18:37 | comment | added | R. van Dobben de Bruyn | Whoops, you're absolutely right! Sorry. | |
| Feb 13, 2020 at 17:10 | comment | added | Will Sawin | @R.vanDobbendeBruyn If you look at my statement, my assumption is that the Galois closures do not contain a common subfield, so I don't think that's a problem. | |
| Feb 13, 2020 at 14:48 | comment | added | R. van Dobben de Bruyn | The reduction to the Galois case doesn't seem to work, because they may have a subfield in common (or even agree!) even when $K$ and $L$ do not. See Example 1 in my answer. | |
| May 30, 2012 at 17:56 | history | edited | Will Sawin | CC BY-SA 3.0 |
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| May 30, 2012 at 17:55 | comment | added | Will Sawin | Yes. I do that a lot. | |
| May 30, 2012 at 17:19 | comment | added | Martin Brandenburg | @Will: You actually mean $K \otimes L$ instead of $K \times L$, right? | |
| Mar 20, 2012 at 7:30 | comment | added | name | By the way, in the purely inseparable case there is a unique embedding in any algebraic closure and so we can meaningfully talk about $K \cap L$. Another equivalent criterion would be $K \cap L = k$. | |
| Mar 20, 2012 at 3:56 | comment | added | name | I guess we could also rephrase as: there is no pair of elements $(a, b) \in (K - k) \times (L - k)$ with the same minimal polynomial. | |
| Mar 20, 2012 at 3:54 | comment | added | name | For the purely inseparable case, how about some criterion like: there does not exist $a \in (K - k)$ and $b \in (L - k)$ such that $a^p = b^p$? This is also implied by the statement: the fields $L$ and $K$ do not contain any nontrivial extensions of $k$ which are isomorphic. | |
| Dec 15, 2011 at 2:56 | comment | added | B R | (The results are on pages 66 and 67) | |
| Dec 15, 2011 at 2:55 | comment | added | B R | Apologies, I had forgotten why I didn't make my comment. I should have said "follows from linear-disjointness". See Pete's linked notes, results 107, 108, and 111. Though I agree that this covers most of the interesting stuff (and I'm not exactly clear what is left over). | |
| Dec 15, 2011 at 2:08 | history | edited | Will Sawin | CC BY-SA 3.0 |
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| Dec 15, 2011 at 1:28 | comment | added | Will Sawin | Not true. My argument proves that things are fields if they satisfy certain conditions. | |
| Dec 15, 2011 at 0:11 | comment | added | B R | (Will, you have actually stated the contrapositive of the Corollary in the original post (I nearly made the same mistake myself, awhile back)). | |
| Dec 14, 2011 at 23:35 | history | answered | Will Sawin | CC BY-SA 3.0 |