Lemma 10.44.1. 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 10.43.6. 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!

Comment #5562 by DatPham on

I don't understand why $x_1$ is algebraic over $k(x_2,\ldots,x_r,\alpha)$. I know that $x_1$ is a root of the polynomial $P(X,x_2,\ldots,x_n,\alpha)\in k(x_2,\ldots,x_r,\alpha)[X]$. But how do we know this polynomial is nonzero? If we write $P(x_1,\ldots,x_n,T)=X_1^dg(x_2,\ldots,x_n,T)+h(x_1,x_2,\ldots,x_n,T)$ for $d>0$, $g(x_2,\ldots,x_n,T)\ne 0$ and $h(x_1,x_2,\ldots,x_n,T)$ of degree $, then it might happen that $g(x_2,\ldots,x_n,\alpha)=0$.

Comment #5821 by DatPham on

@#5745: Dear Professor Johan, I did read comment #764 many times, but I still cannot understand why $g(x_2.\ldots,x_n,\alpha)\ne 0$ when $g(x_2,\ldots,x_n,T)\ne 0$ because ${x_2,\ldots,x_n,\alpha}$ may be algebraically dependent over k (of course if $x_1$ is algebraic over $k(x_2,\ldots,x_n,\alpha)$ then ${x_2,\ldots,x_n,\alpha}$ has to be algebraically independent but this seems to be circular ...) .

Comment #5823 by on

To help me understand the confusion, please read the proof from the beginning and point out the first sentence in the proof where you do not understand the assertion. Because each time I read your comment, I think you are pointing to something in the second paragraph of the proof where we already know that $P \in k[x_1, \ldots, x_r, T]$ is monic in $T$, $\partial P / \partial x_1$ is nonzero and $P$ is irreducible in $T$ over $k(x_1, \ldots, x_r)$. Hence $P$ is irreducible in $x_1$ over $k(x_2, \ldots, x_r, T)$ by Gauss lemma. Hence $x_2, \ldots, x_r, \alpha$ are algebraically independent. So the polynomial $P(X, x_2, \ldots, x_r, \alpha)$ cannot be zero as a polynomial in $X$ because this would mean we get an algebraic relation between $x_2, \ldots, x_r, \alpha$ by looking at the coefficients of powers of $X$. (I am just repeating the proof here, so this probably doesn't help.)

Comment #5824 by DatPham on

@#5823: Dear Professor Johan, thank you for being patient with me. In your comment above, I start getting confused from the sentence ''Hence $x_2,\ldots,x_r,\alpha$ are algebraically independent...'' I don't understand how we can deduce this from the fact that $P$ is irreducible in $x_1$ over $k(x_2,\ldots,x_r,T)$.

Comment #5825 by on

Otherwise the transcendence degree of the field generated by $x_1, \ldots, x_r, \alpha$ over $k$ would be less than $r$!

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