Lemma 10.43.1. Let $k$ be a field of characteristic $p > 0$. Let $k \subset K$ be a field extension. The following are equivalent:

1. $K$ is separable over $k$,

2. the ring $K \otimes _ k k^{1/p}$ is reduced, and

3. $K$ is geometrically reduced over $k$.

Proof. The implication (1) $\Rightarrow$ (3) follows from Lemma 10.42.6. The implication (3) $\Rightarrow$ (2) is immediate.

Assume (2). Let $k \subset L \subset K$ be a subextension such that $L$ is a finitely generated field extension of $k$. We have to show that we can find a separating transcendence basis of $L$. The assumption implies that $L \otimes _ k k^{1/p}$ is reduced. Let $x_1, \ldots , x_ r$ be a transcendence basis of $L$ over $k$ such that the degree of inseparability of the finite extension $k(x_1, \ldots , x_ r) \subset L$ is minimal. If $L$ is separable over $k(x_1, \ldots , x_ r)$ then we win. Assume this is not the case to get a contradiction. Then there exists an element $\alpha \in L$ which is not separable over $k(x_1, \ldots , x_ r)$. Let $P(T) \in k(x_1, \ldots , x_ r)[T]$ be the minimal polynomial of $\alpha$ over $k(x_1, \ldots , x_ r)$. After replacing $\alpha$ by $f \alpha$ for some nonzero $f \in k[x_1, \ldots , x_ r]$ we may and do assume that $P$ lies in $k[x_1, \ldots , x_ r, T]$. Because $\alpha$ is not separable $P$ is a polynomial in $T^ p$, see Fields, Lemma 9.12.1. Let $dp$ be the degree of $P$ as a polynomial in $T$. Since $P$ is the minimal polynomial of $\alpha$ the monomials

$x_1^{e_1} \ldots x_ r^{e_ r} \alpha ^ e$

for $e < dp$ are linearly independent over $k$ in $L$. We claim that the element $\partial P/\partial x_ i \in k[x_1, \ldots , x_ r, T]$ is not zero for at least one $i$. Namely, if this was not the case, then $P$ is actually a polynomial in $x_1^ p, \ldots , x_ r^ p, T^ p$. In that case we can consider $P^{1/p} \in k^{1/p}[x_1, \ldots , x_ r, T]$. This would map to $P^{1/p}(x_1, \ldots , x_ r, \alpha )$ which is a nilpotent element of $k^{1/p} \otimes _ k L$ and hence zero. On the other hand, $P^{1/p}(x_1, \ldots , x_ r, \alpha )$ is a $k^{1/p}$-linear combination the monomials listed above, hence nonzero in $k^{1/p} \otimes _ k L$. This is a contradiction which proves our claim.

Thus, after renumbering, we may assume that $\partial P/\partial x_1$ is not zero. As $P$ is an irreducible polynomial in $T$ over $k(x_1, \ldots , x_ r)$ it is irreducible as a polynomial in $x_1, \ldots , x_ r, T$, hence by Gauss's lemma it is irreducible as a polynomial in $x_1$ over $k(x_2, \ldots , x_ r, T)$. Since the transcendence degree of $L$ is $r$ we see that $x_2, \ldots , x_ r, \alpha$ are algebraically independent. Hence $P(X, x_2, \ldots , x_ r, \alpha ) \in k(x_2, \ldots , x_ r, \alpha )[X]$ is irreducible. It follows that $x_1$ is separably algebraic over $k(x_2, \ldots , x_ r, \alpha )$. This means that the degree of inseparability of the finite extension $k(x_2, \ldots , x_ r, \alpha ) \subset L$ is less than the degree of inseparability of the finite extension $k(x_1, \ldots , x_ r) \subset L$, which is a contradiction. $\square$

Comment #387 by Filip Chindea on

Is it really obvious that $G^{1/p}$ maps to a nonzero element of $k^{1/p} \otimes L$? As far as I know there is a vanishing criterion for elements in a tensor product of modules, but nothing for fields (algebras). I've been thinking at this for some time, but I may be missing something. If the answer is trivial you can delete this altogether after my apologies, anyway thank you for your time.

Comment #388 by on

Yes, this needs a small argument. Say $G$ has degree $pd$ in $T$. Then the elements $X_1^{e_1}, \ldots, X_r^{e-r} T^e$ for $e < pd$ of $k[X_1, \ldots, X_r, T]$ map to $k$-linearly independent elements of $L$. By construction of the tensor product, the same elements map to $k^{1/p}$-linearly independent elements of $k^{1/p} \otimes_k L$. In particular, $G^{1/p}$ does not map to zero as the $T$-degree of $G^{1/p}$ is $d < pd$. OK?

Comment #389 by Filip Chindea on

Thanks; you can delete this. It was just my ridiculous expectation that the irreducibility of $G$ in $T$ should turn up while calculating in the tensor product.

Comment #390 by on

OK, I went ahead and added the extra argument. The change is here.

If there is a misunderstanding about what the argument is supposed to be, then probably the argument deserves to be updated and/or extended.

Comment #740 by Keenan Kidwell on

Is it clear that the polynomial $G(T)=P(T,x_2,\ldots,x_r,\alpha)\in k[x_2,\ldots,x_r,\alpha][T]$ is non-zero, and that the fact that $\partial P/\partial x_1\neq 0$ implies that $\partial G/\partial T\neq 0$ (for some reason this kind of thing always really confuses me, so this might be a very dumb question)? I thought perhaps it followed from the linear independence of the monomials in the displayed equation, but I can't seem to make that work.

Comment #751 by on

It took me a while to understand your question, but you do have a valid point. I've tried to address your concern by adding a couple of lines explaing that $P$ is irreducible as a polynomial in $x_1$ over $k(x_2, \ldots, x_r, \alpha)$ which is a purely transcendental extension of $k$. Here is the commit which also includes fixing the other typos you found.

It seems that there is a curse on this proof as it keeps having to be modified! There is a still a last step in the proof that perhaps should be clarified a bit more, namely, why the inseparable degree is lessened...

Comment #756 by Keenan Kidwell on

I looked at the commit, but I'm still confused. How are we concluding that $x_2,\ldots,x_r,\alpha$ is algebraically independent before knowing that $x_1$ is algebraic over $K=k(x_2,\ldots,x_r,\alpha)$? Once we know that $x_1$ is algebraic over $K$, then because $L$ is algebraic over $k(x_1,\ldots,x_r)\subseteq K(x_1)$, $L$ is algebraic over $K(x_1)$, and thus $r=\mathrm{trdeg}_k(L)=\mathrm{trdeg}_k(K(x_1))$, which forces $x_2,\ldots,x_r,\alpha$ to be algebraically independent. The argument as it is now uses this algebraic independence to prove that $x_1$ is algebraic over $K$ (by proving that the relevant polynomial relation with coefficients in $K$ is non-trivial). Also, in the commit, there is a typo: the second $n$ is missing from "independent."

Comment #759 by on

There is a nontrivial algebraic relation between $x_1, \ldots, x_r, \alpha$ which involves $x_1$, namely $P$. The fact that $P$ involves $x_1$ is proved before we point out that $x_2, \ldots, x_r, \alpha$ are algebraically independent. (When you read the commit you have to remove the red lines.) OK?

Comment #760 by on

OK, I guess what I said wasn't good enough because for example, if $P = x_1 Q$ for some $Q$, then it wouldn't work. But we also know that $P$ is irreducible in the polynomial ring on $r + 1$ variables and then it is enough.

Actually the way I think about the situation of the proof is, as soon as we have found $P$ in $k[x_1, \ldots, x_r, T]$ a minimal polynomial for $\alpha$ over $k(x_1, \ldots, x_r)$, then I think of the irreducible hypersurface and I replace $k(x_1, \ldots, x_r, \alpha)$ by the function field of $V$. Then after we show that $\partial P(y_1, \ldots, y_{r + 1}) / \partial y_1$ is nonzero, I think of that as saying that the projection $V \to \mathbf{A}^r$ gotten by forgetting the first variable, is generically \'etale, i.e., that the function field extension $k(V) \supset k(\mathbf{A}^r)$ is (finite) separable. So certainly, as soon as you agree that at least one variable occurs in a term of $P$ with an exponent not divisible by $p$, then I am completely sure that the proof is correct.

The problem is that we also need to keep the proof readable, understandable, etc. A good way to do this would probably be to have a discussion of the relationship between multivariable polynomials and (nonalgebraic) field extensions and then to refer to that. I encourage you to write your own and submit it! Thanks.

Comment #761 by on

The displayed equation should be

Comment #763 by on

Another thing we need is a statement and proof of Gauss's lemma, maybe somewhere in the chapter on fields?

Comment #764 by Keenan Kidwell on

Yes! I understand now. The whole point is that the derivative in $x_1$ is non-zero, so $x_1$ shows up, and we can write $P(x_1,\ldots,x_r,T)=x_1^dg(x_2,\ldots,x_r,T)+h(x_1,\ldots,x_r,T)$ with $h$ of degree less than $d$ in $x_1$, $d>0$. Written this way, it's then obvious that $P(X,x_1,\ldots,x_r,\alpha)\neq 0$ (as a polynomial in $k(x_2,\ldots,x_r,\alpha)[X]$). I don't know why this kind of thing confuses me so much. Thank you for explaining it!

Comment #765 by on

OK, yes, that is a good way to see it. I really appreciate reporting back here. Thanks!

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