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Let $K/F$ be a field extension. I am interested in the situation where there exists a field extension $L/F$ such that the ring $L \otimes_FK$ is not a field.

If there exists $z\in K \setminus F$ such that $z$ is algebraic over $F$, then I think I can show that $K \otimes_F K$ is not a field. So my question is: If no such element exists in $K$, does it follow that for any field extension $L$ of $F$, $L \otimes_F K$ is a field?

M.Sina
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neilme
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  • Related: http://math.stackexchange.com/questions/1979476/the-tensor-product-of-non-algebraic-extensions-is-not-a-field – user26857 Oct 22 '16 at 18:42

2 Answers2

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If $K/F$ is a non-trivial extension, then $K\otimes_FK$ is not a field. We have a natural $K$-algebra map $K\otimes_FK\rightarrow K$ with $x\otimes y\mapsto xy$. Take $x\in K\setminus F$. Then $x\otimes 1\neq 1\otimes x$ because there is a $K$-basis for $K\otimes_FK$ containing $1\otimes x$ and $1\otimes 1$. But $x\otimes 1-1\otimes x$ is in the kernel of the map in question, so $K\otimes_FK$ has a non-trivial, proper ideal, so it isn't a field.

Keenan Kidwell
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  • Of course; how silly of me. Actually what I was really interested in was conditions for $L \otimes_F K$ to always be an integral domain, but for some reason I thought I could state it purely on the level of fields. Thank you. I guess I'll ask separately for what I really wanted. – neilme Feb 01 '13 at 16:15
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    When $K/F$ is finite $L\otimes_FK$ is a domain if and only if it is a field, so in this case the answer is still no. This addresses the case of an extension containing a non-trivial algebraic extension. – Keenan Kidwell Feb 01 '13 at 16:24
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    For $K/F$ purely transcendental, $K=F(S)$, $S$ a set of variables, $L\otimes_FK$ is a localization of the polynomial ring $L[S]$, so it is a domain, but it can never be a field if $L$ is not algebraic over $F$. – Keenan Kidwell Feb 01 '13 at 16:30
  • Great! So that leaves open the case where your extension looks like $F \subseteq G \subseteq K$, $G$ purely transcendental over $F$, and $K$ algebraic over $G$ (e.g. $F=\mathbb{Q}$, $G = F(\pi)$, $K$ the algebraic closure of $G$). – neilme Feb 01 '13 at 16:41
  • @neilme The case left open is in fact the general case, since any field extension can be written as a tower of two extensions, one of them is purely transcendental and the other one is algebraic. –  Feb 01 '13 at 17:05
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As a graduate student I remember being disappointed that it was hard to find much information concerning tensor products of fields. Later, as with many things, I realized that it depends a good bit on knowing where to look: it turns out that the more standard topic (found in most "serious" treatments of field theory) of linear disjointness is closely related.

You can find some material on when $L \otimes_F K$ is a field (and also when it is a domain) in $\S 12$ of my field theory notes.

Community
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Pete L. Clark
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