Is it true that a finite degree field extension $L/k$ is separable if and only if $L\otimes_{k}L$ is a reduced $L$-algebra?
Surely the "only if" part is true because if the extension is separable, we have the Primitive Element Theorem and everything follows. But I'm asking if it's true, and how to prove the "if" part. Thanks.
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.