Conditions for tensor product of fields to be a field

Question

Suppose $k$ is a field of characteristic 0. $L/k$ is any field extension, $K/k$ is a finite extension (not assumed to be Galois). If we know that $L\otimes_k K$ is not a field, can we conclude that there exists a finite subextension $L_0/k$ inside $L/k$ such that $L_0\otimes_k K$ is not a field as well?

Answer 1

(By the primitive element theorem) $K=k(\alpha) \cong k[x]/(f)$ with $f$ monic irreducible

That $L\otimes_k K\cong L[x]/(f)$ is not a field means that $f$ is not irreducible over $L$.

ie. $f= gh$ with $g,h\in L[x]_{monic}$.

The coefficients of $g,h$ are algebraic over $k$ so they lie in a finite extension $E/k,E\subset L$.

and $E\otimes_k K \cong E[x]/(f)$ is not a field.

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The primitive element theorem is not really needed because if $K=k(\alpha,\beta)$ then $L\otimes_k K= (L\otimes_k k(\alpha))\otimes_{k(\alpha)} K$

Answer 2

I think maybe the following argument works. Suppose for every finite subextension $L_0/k$ we have $L_0\otimes_k K$ is a field, we want to show that $L\otimes_k K$ is a field as well. First, WLOG ${\rm tr.deg}(L/k)$ is finite. Take $X$ to be a proper normal integral scheme over $k$ with rational function field $L$. We know ${\rm H}^0(X,\mathcal{O}_X)=L_0$ is a finite extension of $k$. Then ${\rm H}^0(X_K, \mathcal{O}_{X_K})=L_0\otimes_k K$ is a field by our assumption which implies that $X_K$ is connected. But $X$ is normal implies that $X_K$ is also normal. Then $X_K$ is both normal and connected hence integral. Now, ${\rm Spec}(L\otimes_k K)$ is nothing but the rational function field of $X_K$.