Timeline for When is the tensor product of two fields a field?
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| Jun 22, 2022 at 8:13 | history | edited | CommunityBot |
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| Jul 22, 2020 at 23:05 | comment | added | benblumsmith | Dear @GeorgesElencwajg, what is the general principle you are applying in the proof of the sufficient condition in concluding that if one of $K,L$ is primary then the quotient of $K\otimes_k L$ by its nilradical is a domain? Can you point me to a reference? I can't find anything on this in any of my usual references or by searching. (The wikipedia entry on primary extensions points to a book of Fried and Jarden, but I couldn't find anything on interaction between primary extensions and tensor products in that book...) | |
| Apr 25, 2020 at 10:45 | history | edited | David Roberts♦ | CC BY-SA 4.0 |
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| Feb 12, 2020 at 4:42 | answer | added | R. van Dobben de Bruyn | timeline score: 37 | |
| Feb 9, 2020 at 23:44 | history | edited | YCor |
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| Jan 7, 2020 at 10:11 | comment | added | Georges Elencwajg | Dear @Fawzy: By $K \cdot L$ I mean the $k$-algebra generated by $K$ and $L$, not the $k$-field extension generated by those fields. I think this is the prevailibng convention: Bastida for example adopts it and writes $K\vee L$ for the field extension, so that $K\vee L=\operatorname {Frac} (K \cdot L)$ (the field of fractions of the domain $K \cdot L$). | |
| Jan 7, 2020 at 4:41 | comment | added | FNH | Small question: why is the canonical morphism $K \otimes_k L \to KL$ surjective? For example, if $x\in K, y\in L$, then, what is the preimage of $(x+y)^{-1}$? As far as I understand, $K.L$ is the smallest field containing both $K$ and $L$. The problem is in expressing an inverse of the sum of two elements as a sum itself. | |
| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Apr 24, 2016 at 9:28 | history | edited | Georges Elencwajg | CC BY-SA 3.0 |
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| Feb 14, 2016 at 20:47 | history | edited | Ali Taghavi |
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| Jul 16, 2014 at 14:25 | history | edited | filmor | CC BY-SA 3.0 |
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| Jan 29, 2014 at 7:39 | history | edited | Andrés E. Caicedo |
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| Feb 15, 2013 at 7:06 | history | edited | Georges Elencwajg | CC BY-SA 3.0 |
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| Feb 15, 2013 at 6:38 | history | edited | Georges Elencwajg | CC BY-SA 3.0 |
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| Feb 15, 2013 at 6:28 | history | edited | Georges Elencwajg | CC BY-SA 3.0 |
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| Dec 14, 2011 at 23:35 | answer | added | Will Sawin | timeline score: 3 | |
| Nov 30, 2011 at 6:50 | comment | added | Will Sawin | This suggests that the Galois closure of $K$ sharing a subfield with $L$ is enough to cause problems. | |
| Nov 30, 2011 at 6:46 | comment | added | Will Sawin | Can we pinpoint exactly which elements of $R$ prevent this from being a field? For instance it's enough to adjoin the $x$ and $y$ coordinates of the root in the complex plane. | |
| Nov 29, 2011 at 18:24 | comment | added | Georges Elencwajg | Dear @François: 1) yes, if one of $K,L$ is algebraic over $k$ the question whether they are linearly disjoint after embedding into a composite field does not depend on the composite field chosen. 2) Your beautiful example is quite illuminating and confirms that the problem we are investigating is rather subtle. Thank you very much for this interesting contribution. | |
| Nov 29, 2011 at 17:15 | comment | added | François Brunault | @Will : One can take a degree 4 irreducible polynomial $f$ with Galois group $S_4$ and no real roots, then $K=\mathbf{Q}[X]/f$ and $L=\mathbf{R}$ will work ($K$ has no non-trivial subfield and $K \otimes_{\mathbf{Q}} L$ is not a field). | |
| Nov 29, 2011 at 14:53 | comment | added | Will Sawin | The obvious conjecture is "whenever the first theorem does not hold", that is, when there is no subfield $K'\in K$, strictly larger than $k$ and corresponding strictly larger subfield of $L$ that are isomorphic. But presumably there's a counterexample to this. What is it? | |
| Nov 29, 2011 at 14:44 | comment | added | François Brunault | Assume $K/k$ and $L/k$ are finite and suppose we see $K$ and $L$ inside $\overline{k}$. Then isn't it true that the condition $[KL:k]=[K:k][L:k]$ is equivalent to $K \otimes L$ being a field (and thus, this condition is independent of the given embeddings) ? | |
| Nov 29, 2011 at 14:21 | comment | added | Georges Elencwajg | Dear @Tommaso, I forgot to answer your question, sorry. No, I don't suppose $K$ finite over $k$ in the Proposition. If it isn't I just use dimensions over $K$, and consider $K\otimes_k K$ as a $K$-vector space by letting scalars from $K$ act on the left of the tensor product. Multiplication is then $K$-linear and the argument remains valid. | |
| Nov 29, 2011 at 14:09 | history | edited | Georges Elencwajg | CC BY-SA 3.0 |
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| Nov 29, 2011 at 11:37 | answer | added | Hagen | timeline score: 23 | |
| Nov 29, 2011 at 10:13 | comment | added | Jose Capco | Didn't Abhyankhar once worked on this (he once worked on compositum of algebraically closed field when he was a student)? | |
| Nov 29, 2011 at 7:56 | comment | added | Georges Elencwajg | Dear Ho, prompted by your comment, I have just completed my reference to Pete's notes. Thanks a lot. | |
| Nov 29, 2011 at 7:53 | history | edited | Georges Elencwajg | CC BY-SA 3.0 |
Completed the reference to Pete's notes.
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| Nov 29, 2011 at 2:12 | comment | added | user709 | Dear Georges, I was actually suggesting to use Prop 107 of Pete's notes, if linear disjointness is considered well-understood. | |
| Nov 28, 2011 at 23:34 | comment | added | Georges Elencwajg | Dear Ho, yes linear disjointness is a well-understood condition for subextensions of a big extension. See my Edit. | |
| Nov 28, 2011 at 23:29 | history | edited | Georges Elencwajg | CC BY-SA 3.0 |
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| Nov 28, 2011 at 21:11 | comment | added | user709 | Is linear disjointness considered a well-understood condition? If so, it suffices to check that images of $K$ and $L$ in $E$ are linearly disjoint for any extension $K,L \to E$. For this we can check dimensions as Ralph said, at least for finite extensions. | |
| Nov 28, 2011 at 18:26 | comment | added | Georges Elencwajg | Dear @Ralph, concerning a): yes you can $k$-embed $K$ and $L$ into $\bar k$ . The problem is that this is non canonical and the $KL$ you obtain depends on the embeddings. For example, take $K=\mathbb Q, K=\mathbb Q(\sqrt [3] 2), L=\mathbb Q(\omega\sqrt [3] 2)$ where $\omega =e^{2i\pi/3}$. You can embed $K,L$ naturally into $\bar {\mathbb Q}\subset \mathbb C$ in which case you obtain $KL=\mathbb Q(\omega,\sqrt [3] 2)$. But you can also embed $L$ onto $\mathbb Q(\sqrt [3] 2)$, in which case you obtain $KL=\mathbb Q(\sqrt [3] 2)$. (Anyway $K\otimes_{\mathbb Q} L$ is not a field by my Corollary) | |
| Nov 28, 2011 at 17:47 | comment | added | Ralph | @Georges: a) If $K|k$, $L|k$ are finite, they are algebraic and thus, they can be considered as a subfield of the algebraic closure $\bar{k}$ of $k$. Formally, you may choose subfields $K', L' \le \bar{k}$ that are isomorphic to $K$ resp. $L$ and use that $K \otimes_k L \cong K' \otimes_k L'$ as rings. c) Right (I was thinking of the chain of finite fields that constitutes the algebraic closure of the prime field, without realizing that there dividing degrees are choosen). | |
| Nov 28, 2011 at 17:28 | comment | added | Tommaso Centeleghe | Your proof of the proposition uses $K/k$ finite, right? I assume this condition can be removed, right? | |
| Nov 28, 2011 at 16:55 | history | edited | Georges Elencwajg | CC BY-SA 3.0 |
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| Nov 28, 2011 at 16:52 | comment | added | Georges Elencwajg | Dear Ralph, a) your $KL$ only makes sense if $K,L$ are subfields of a big extension of $k$. In this case your condition on dimensions is equivalent to linear disjointness. b) Your remark that $K\otimes _k L$ is not a field if $K\leq L$ also follows from my Corollary. c) However your last sentence "In particular..." is not correct: for example the tensor product of two finite extensions of a finite field is a field as soon as the two extensions have relatively prime dimensions. (The simplest case is $\mathbb F_4 \otimes_{\mathbb F_2} \mathbb F_8=\mathbb F_{64}$.) | |
| Nov 28, 2011 at 15:53 | comment | added | Ralph | (continuation) For example, if $k \lneq K \le L$ then $K \otimes_k L$ is not a field. In particular, the tensor product of finite fields that aren't the prime field, is never a field. | |
| Nov 28, 2011 at 14:50 | comment | added | Ralph | If $K|k, L|k$ are finite dimensional extensions, then (as in your Proposition) multiplication $K \otimes_k L \to KL$ is a surjective ring homomorphism. Thus $K \otimes_k L$ is a field (in effect is $KL$) iff $\dim_k(KL) = (\dim_kK) \cdot (\dim_kL)$. This generalizes the case when the $k$-dimensions of $K,L$ are relaively prime. | |
| Nov 28, 2011 at 13:58 | history | edited | Georges Elencwajg | CC BY-SA 3.0 |
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| Nov 28, 2011 at 13:58 | comment | added | Georges Elencwajg | Right. Thanks a lot, @Vladimir: I have fixed that typo . | |
| Nov 28, 2011 at 13:51 | comment | added | Vladimir Dotsenko | In Grothendieck's formula, the last K should be L, presumably? | |
| Nov 28, 2011 at 13:44 | history | asked | Georges Elencwajg | CC BY-SA 3.0 |