Separability and tensor product of fields

Solution 1:

Let $x\in L$ and $f\in k[X]$ the minimal polynomial of $x$ over $k$. Then $k(x)\subset L$ and therefore $L\otimes_kL$ contains $k(x)\otimes_kL$. Since $k(x)\simeq k[X]/(f)$ we get $k(x)\otimes_kL\simeq L[X]/(f)$. If $x$ is not separable over $k$, then the polynomial $f$ has multiple roots in $L$, so the ring $L[X]/(f)$ is not reduced, a contradiction.

Edit. Since $x$ is not separable over $k$ we have $f'(X)=0$. If $f(X)=(X-x)g(X)$ in $L[X]$, then $f'(X)=g(X)+(X-x)g'(X)$, so $X-x\mid g(X)$. This means that $x$ is a multiple root of $f$ (in $L$), so $f(X)=(X-x)^th(X)$ with $t\ge 2$ and $h\in L[X]$ with $h(x)\ne 0$. (Eventually $h=1$.) Then $\gcd((X-x)^t,h(X))=1$ in $L[X]$, so by Chinese Remainder Theorem $L[X]/(f)\simeq L[X]/(X-x)^t\times L[X]/(h)$, and therefore the ring $L[X]/(f)$ is not reduced.