Suppose $K$ is a field and $L, L'$ are finite extensions of $K$. It is known that if $L/K$ (or $L'/K$) is separable, then $L \otimes_K L'$ is a product of finitely many fields. Is there a counterexample if $L/K$ and $L'/K$ are both inseparable?

That is, for what $L,L'$ is $L \otimes_K L'$ not a product of finitely many fields?

Edit: I think I may have figured out a solution. Let $L=L'= \mathbb{F}_p(t), K = \mathbb{F}_p(t^p)$. Then $L \otimes_K L'$ has non-zero nilpotent elements - for example, $t\otimes 1 - 1\otimes t$. However, a product of finitely many fields does not have nilpotent elements. Does this work?

  • $\begingroup$ Yes, this works. I was composing my answer while you were editing your question by adding your example. Since you are new here, I will advise you to think carefully before posting a question, so as not to make users lose their time answering questions to which you know the answer. Checking whether a proof is correct is an entirely different matter since it only needs to be answered by yes or no. However, since you are new here, this will not be held against you and I wish you welcome to our site in the name of all users. $\endgroup$ – Georges Elencwajg Mar 29 '13 at 8:46
  • $\begingroup$ Yes. I apologize for that. After posting my edit, I was surprised to see your answer and realized that you must have been composing your answer in the meantime. I sent several hours trying to figure this issue out and for some reason saw the solution quite quickly after posting. Maybe I should have thought more about it..Thank you for your help! $\endgroup$ – bob Mar 29 '13 at 8:58
  • $\begingroup$ Dear bob: yes it happens that one may think a long time about a problem and then see the light while explaining it to someone. After your explanation I can only welcome you here even more warmly! $\endgroup$ – Georges Elencwajg Mar 29 '13 at 9:21

Take for $K$ an imperfect field of characteristic $p\gt 0 $ : this means that some $a\in K$ has no $p$-th root in $K$. An example would be $a=x$ in $K=\mathbb F_p(x)$ .

A $p$-th root $\alpha\in K^\text {alg} , \alpha^p=a$ exists however in an algebraic closure $K^\text {alg} $ of $K$ .
The minimal polynomial of $\alpha$ over $K$ is $T^p-a\in K[T]$ (beware that its irreducibility is not a trivial result), so that $$K[\alpha]=K[T]/(T^p-a).$$

You can then take for your example $L=L'=K[\alpha]$, a finite extension of degree $p$ of $K$, and compute $$L\otimes_KL'=L\otimes_K K[T]/(T^p-a)=L[T]/(T^p-a)=L[T]/(T-\alpha)^p.$$ The last expression shows tha $L\otimes_KL'$ has some non-zero nilpotent elements (for example the class of $T-\alpha$) which prevent it from being isomorphic to a product of fields
(since such a product has zero as its only nilpotent element).

Summing up, $L\otimes_KL'$ is an example of a tensor product of finite field extensions of a field $K$ which is not isomorphic to any product of fields.


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