Tensor product of inseparable field extensions 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? 
 A: 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. 
