# showing locally that a diagram commutes

When showing that a (natural family of) diagram of $R$-algebras for all rings $R$ commutes, why does it suffice to show that it commutes for all $R$ local with algebraically closed residue field?

My idea: Reduce to local rings (fpqc descent), then to strictly henselian rings (strict henselisation is faithfully flat).

Is this correct?

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Strict henselianization demands that the field be separably closed, which I think is not what you want. –  Akhil Mathew Jan 15 '11 at 14:20

You don't need anything as fancy or as hard as the general fpqc descent theorem (or even the affine case, which is easier), but only basic facts about faithfully flat maps. So first of all it is fairly well-known and easy to prove (i.e., from the definition) that two maps of $R$-modules are equal iff the base-changes to the localizations $R_{\mathfrak{p}}$ for $\mathfrak{p}$ prime are equal, for every prime $\mathfrak{p}$. This is why you can reduce to the case of a local ring.

The more interesting part of your question is why we can reduce to the case where the residue field is algebraically closed. Here I quote a lemma from EGA $0_{III}$:

Lemma: Let $(R, \mathfrak{m})$ be a noetherian local ring with residue field $k$, and let $K$ be an extension of $k$. Then there exists a flat, local noetherian $R$-algebra $R'$ such that $\mathfrak{m} R'$ is the maximal ideal of $R'$ and such that the residue field is $K$.

If $R$ is not assumed noetherian, then $R'$ will not be assumed noetherian, but the argument still works. Actually, it is not hard if you forget about noetherianness, so let me sketch it. We write $K$ as an inductive (by an ordinal whose cardinality is generally large) colimit of towers of extensions generated by one element. By transfinite induction (as a colimit of flat things is flat), we reduce to the case where $K$ is generated by one element over $k$, say some $\alpha$. If $\alpha$ is transcendental, then we can take $R' = (R[t])_{\mathfrak{m} R[t]}$. If $\alpha$ is algebraic, satisfying some monic polynomial $\overline{P} \in k[X]$ that lifts to $P \in R[X]$, then let $R' = R[X]/P$. One can check that this works (the maximal ideals of $R'$ are in correspondence with those of $R[X]/P \otimes_R k$ by Nakayama, and this last thing is clearly a field.) In either of these cases, it is easy to check that $R'$ is flat (even free in the second case).

OK. So if $R'$ is flat over $R$ (assumed local as above, in the situation of the last paragraph), then $R'$ is faithfully flat because we are working with local rings. So if $M \rightrightarrows N$ are two maps of $R$-modules, they are equal if and only if $M \otimes_{R} R' \rightrightarrows N \otimes_R R'$. It follows that if we want to prove something about commutativity of diagrams of $R$-modules, we can reduce to proving something about $R'$-modules.

Now, as I explained above, any local ring admits a flat local homomorphism into a local ring whose residue field is algebraically closed. So if you want to prove some kind of commutativity for local rings, you can reduce to the case where the residue field is algebraically closed. And, as I explained at the very beginning, if you want to prove something about commuting diagrams of modules over a ring, you can reduce to the case of local rings.

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Two questions: 1) Would it be correct to refer to fpqc descent? 2) How do you prove directly "that two maps of R-modules are equal iff the base-changes to the localizations Rp for p prime are equal"? I know that a module is 0 iff its localisations are 0. –  user5262 Jan 15 '11 at 15:09
@user5262: Dear user5262, 1) fpqc descent is much stronger than the elementary facts I referred to above (that if $R'$ is a f.f. $R$-algebra, then for any $R$-module $M$, the map $M \to M \otimes_R R'$ is injective -- this is in Bourbaki, Commutative algebra, ch. 1 if memory serves). I don't see why fpqc descent is necessary. 2) It suffices to show that if $m \in M$ (for $M$ an $R$-module), then $m=0$ if and only if $m$ goes to zero in every localization (for then it would follow that the difference of the two maps went to zero in every localization)... –  Akhil Mathew Jan 15 '11 at 15:43
...and this in turn can be seen by considering the annihilator $Ann(m) \subset R$. If $m \neq 0$, then this is a proper ideal, so is contained in a maximal ideal $\mathfrak{m} \subset R$. Then it follows that $m$ does not go to zero in $M_{\mathfrak{m}}$. Alternatively, you could argue that if $f,g: M \rightrightarrows M'$ were the two maps, then the kernel of $f-g$ vanishes at every localization (because localization is an exact functor), so the kernel is zero by the fact you referred to in your comment. –  Akhil Mathew Jan 15 '11 at 15:45
Thank you very much! –  user5262 Jan 15 '11 at 16:22
Great answer, Akhil! –  Mariano Suárez-Alvarez Jan 15 '11 at 17:11