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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.

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Simone
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    You can reduce to the case where $L$ is generated by a single element by looking at the obvious subalgebra of $L \otimes_k L$. – Zhen Lin Jan 20 '14 at 13:43
  • Yes, I can consider $\alpha\in L$ and ask if it's true that $L\otimes_k k[\alpha]$ reduced implies $\alpha$ separable, but I think the problem remains. – Simone Jan 20 '14 at 14:07
  • @Simone Could you explain how the "only if" part of this works after you conclude that the extension is simple? – baltazar Dec 03 '14 at 22:26
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    @baltazar Sure! By PET,you know that $L\cong k[\alpha]\cong k[X]/(f(X))$, where $f(X)$ is the minimum polynomial of $\alpha$, the Primitive Element. So you get $L\otimes _{k} L\cong L\otimes _{k} k[X]/(f(X))\cong L[X]/(f(X))$, and this is obviously reduced since $f(X)$ doesn't have any multiple root. I hope I've been clear! – Simone Dec 06 '14 at 22:04

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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.

user26857
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    I think you can't say this, because $L$ is just an extension of $k$, not its algebraic closure, so you don't know if $f$ completely splits in $L$. It can happen that, in $L$, $f(X)=(X-x)h(X)$, where $h(X)\in L[X]$ still has multiple roots. – Simone Jan 20 '14 at 14:04
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    @Simone: It cannot happen that $f(X)$ has multiple roots unless $x$ is also a multiple root. For if $f(X)$ has a multiple root, then $\gcd(f(X),f'(X))$ is non-trivial. But that gcd $\in k[X]$, and is a factor of $f(X)$. That contradicts the assumption that $f(X)$ is irreducible as a minimal polynomial. – Jyrki Lahtonen Jan 20 '14 at 20:00
  • Minor nitpick: In the first line of the edit, $x$ not separable over $k$ should imply $f'(x) = 0$ (not $f'(X) = 0$). Excellent answer though, +1 – zcn Jan 22 '14 at 16:22
  • @user115654: You're wrong, one of the equivalent conditions for inseparability is that the derivative is identically zero. You can find this result in Isaacs' "Algebra" in Chapter 19. – Simone Jan 24 '14 at 10:58
  • @Simone: What I meant to say (but expressed poorly) was that in this case, the weaker claim $f'(x) = 0$ is enough to show that $x$ is a multiple root of $f$. I am well aware of the fact you state (for the minimal polynomial, otherwise the derivative will not be identically zero). – zcn Jan 24 '14 at 18:42
  • Amazing: I thought that the equivalence evoked by Simone was false, but this answer proves I was wrong . Bravo and +1. By the way, I would suggest you used your real name or at least a more memorable handle than user115654. – Georges Elencwajg Jan 25 '14 at 20:56