I asked this question on math.stackexchange but maybe it fits here better. If not, I apologize in advance and will remove the question.

Let $K$ be a global field and $\upsilon$ a prime of $K$. Then in a paper, which I read, it is used that $K_{\upsilon}|K$ is separable, where $K_{\upsilon}|K$ is the completion. This is obviously true if $charK=0$.

Why is $K_{\upsilon}|K$ separable if $charK=p>0$?

Things I tried but could not get to work (maybe I missed something and they would work):

  • Use that $\hat A \cap Frac(A)^{alg} \subset A^h$ for an excellent ring $A$ (this is mentioned here henselization and completion)
  • Use Exercise 1 of section II.4 of Serre's Local Fields: If $A$ is a DVR for which every finite purely inseparable extension $Frac(A)$ has integral closure finite over $A$ then the completion of $Frac(A)$ is separable over $Frac(A)$
  • Try to use that $K_{\upsilon}=\mathbb F_{p^n}((T))$

The statement seems to me like it should be in a book somewhere but I just can't find it.

Thanks in advance.

  • 1
    $\begingroup$ I do not have with me Serre's Local Fields. However, I would just like to point out that for the only positive characteristic global fields, i.e., for finite extensions $K$ of $\mathbb{F}_p(T)$, for every finite extension $L/K$, the integral closure of the DVR $\mathbb{F}_p[T]_{\langle T\rangle}$ in $L$ is finite over the DVR. This is a corollary of the Noether Normalization Theorem. You can find an explicit statement in Eisenbud's Commutative Algebra soon after the proof of the Noether Normalization Theorem. $\endgroup$ – Jason Starr Dec 6 '17 at 18:11
  • 1
    $\begingroup$ Also, closely related to this is the fact that an excellent Noetherian local ring that is reduced is analytically reduced, i.e., the completion is also reduced. $\endgroup$ – Jason Starr Dec 6 '17 at 18:13

By definition an extension of fields $K'/K$ is separable when $K' \otimes_K F$ is reduced for all field extensions $F/K$, and by limit considerations it is the same to say that all finitely generated subextensions are separable. But for finitely generated extensions, separability can be tested via scalar extension against all algebraic extensions (see Theorem 26.2 in Matsumura's Commutative Ring Theory), so again by limit considerations it is the same to test using just finite extensions $F/K$. Taking direct limits again in $K'/K$, we see that $K'/K$ is separable if and only if $K' \otimes_K F$ is reduced for all finite extensions $F/K$. (See EGA IV$_2$, 4.3.5 and especially 4.6.1 for further discussion on this issue.)

So now consider a normal connected scheme $X$ of finite type over a field $k$ and a non-generic point $x$ of $X$. Let $K$ be the function field of $X$, and $R$ the local ring $O_{X,x}$. The local noetherian completion $\widehat{R}$ is a normal domain by excellence, though for $X$ of dimension 1 (so $R$ is a discrete valuation ring) it is elementary. Let's see that the fraction field $K_x$ of $\widehat{R}$ is separable over $K$ (so applying this to Dedekind $X$ with finite $k$ would settle the question posed).

Letting $F/K$ be a finite extension, we want to show that $K_x \otimes_K F$ is reduced. Let $Y \to X$ be the normalization of $X$ in $F$; if $X = {\rm{Spec}}(A)$ were affine then $Y$ would be ${\rm{Spec}}(B)$ for the integral closure $B$ of $A$ in $F$. Note that $Y \to X$ is finite; indeed, by denominator-chasing we can write $F$ as the fraction field of some $A$-finite subalgebra $B_0$, so $B$ is the integral closure of $B_0$ in its own fraction field, and the $B_0$-finiteness of $B$ is then a special case of the assertion that the normalization of a domain finitely generated over a field is always module finite: see section 33 of Matsumura's Commutative Ring Theory for a proof.

Now letting ${\rm{Spec}}(A)$ be an open affine neighborhood of $x$ in $X$, and ${\rm{Spec}}(B)$ its preimage in $Y$, we have $F = K \otimes_A B$, so $$K_x \otimes_K F = K_x \otimes_A B = K_x \otimes_R R'$$ where $R = O_{X,x}$ and $R'$ is the localization of $B$ at the prime ideal of $A$ corresponding to $x$. Since localization preserves reducedness, it suffices to show that $\widehat{R} \otimes_R R'$ is reduced.

By $R$-finiteness of $R'$, the maximal ideals of $R'$ are precisely the primes of $R'$ over the maximal ideal of the local $R$. Thus, $\widehat{R} \otimes_R R'$ is the product of the completions of the semi-local $R'$ at each of its maximal ideals (use Theorem 8.15 in Matsumura's Commutative Ring Theory applied to $R'$, whose $\mathfrak{m}_R$-adic topology is the same as its topology for its Jacobson radical). Hence, we just need to show that those completions of $R'$ are reduced. But these are completions of local rings on $Y$, so they're even normal domains (by the same argument used for $X$: excellence in general, or easy considerations with Dedekind domains in the 1-dimensional case).

  • $\begingroup$ It took me a little time to work through this as I am not so familiar with algebraic geometry but I think I understand it now. Thank you very much for writing this. $\endgroup$ – Frida Dec 9 '17 at 15:31

Since there is already an answer, I want to give my slightly different answer in a special case: Say $K = k(t)$ where $k$ is a finite field and $K_v = k((t))$. We only need to check that $$ K_v \otimes_K K^{1/p} $$ is reduced, see Tag 030W. Since $K^{1/p} = k(t^{1/p})$ this is true because the tensor product is visibly equal to $k((t^{1/p}))$. I claim this argument works in general too, but since we have a fully written out proof above, I won't insist. Enjoy!

  • 2
    $\begingroup$ I am not sure what you mean by "in general", but for arbitrary $k$ of char. $p>0$ and $K:=k(t)$ we have $K^{1/p}=k^{1/p}(t^{1/p})$. The natural map $k(\!(t)\!)\otimes_K K^{1/p}\to k^{1/p}(\!(t^{1/p})\!)$ is bijective iff $[k:k^p]<\infty$. However, it is always injective, so the conclusion still holds. $\endgroup$ – Laurent Moret-Bailly Dec 7 '17 at 8:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.