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Recall that a prime $\mathfrak{p}$ is called nonsingular (regular) if the localization at that prime is a regular local ring. If all primes of a ring $R$ are nonsingular, $R$ is called regular. Let $S\subseteq R$ be a ring extension, and let $\mathfrak{q}$ be a prime ideal of $S$. Then $R$ is called regular above $\mathfrak{q}$ if any prime contracting to $\mathfrak{q}$ is nonsingular.

Recall that a ring $R$ is said to have regular fibre over $\mathfrak{q}$ if the scheme-theoretic fibre $R\otimes_S \frac{S_\mathfrak{q}}{\mathfrak{q}S_\mathfrak{q}}=R\otimes_S \kappa(\mathfrak{q})$ is regular.

Are these conditions equivalent? There is an obvious implication that regularity above $\mathfrak{q}$ implies regularity of the fibre, but the converse is not obvious to me. If it's not true in general, is it true when $R$ is a subring of a number field and $S=\mathbf{Z}$ (with $\mathfrak{q}$ a nonzero prime)?

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up vote 6 down vote accepted

First note that it is not true that regularity above $\mathfrak q$ implies regularity of the fibre.

For example, consider the map $\mathbb k[s] \to \mathbb k[t]$ given by $s \mapsto t^2$. Each prime in $k[t]$ is regular, and so in particular each prime of $k[t]$ above the prime $(s)$ is reqular. (In fact there is just one of them, namely $(t)$.) On the other hand, the fibre over $(s)$ is the ring $k[t]/(t^2)$, which is a non-regular local ring.

If you want an arithmetic example instead, consider the inclusion $\mathbb Z \subset \mathbb Z[i]$, and take the prime $\mathfrak q = (2)$ downstairs, which has a unique prime $(1+i)$ lying over it. Again, every prime in the PID $\mathbb Z[i]$ is regular, but the fibre $\mathbb Z[i]/2 = \mathbb Z[i]/(1+i)^2$ is a non-regular local ring.

(The general phenomenon is that a map $X \to Y$ between smooth spaces can have non-smooth fibres: these occur at the critical points of the map.)

As for your question, your asking if the fact that a map $X\to Y$ has a smooth fibres over some point implies that the target is smooth at that point.

This is also false in general.

Consider for example the identity map $k[t^2,t^3] \to k[t^2,t^3]$. The fibre over $(t^2,t^3)$ is just $k$, which is a regular local ring. But $K[t^2,t^3]$ is not regular at $(t^2,t^3).$ (Of course this example is cheap, but its existence foreshadows the existence of many other counterexamples, for example for any etale map $k[t^2,t^3] \to R$, of which there are many non-trivial examples, as well as my trivial example.)

On the other hand, when the base is Spec $\mathbb Z$, which is very nicely behaved (regular, Noetherian, excellent, perfect residue fields, ... ), if the map $X \to $ Spec $\mathbb Z$ is flat and of finite type (e.g. arising from an inclusion $\mathbb Z \subset R$ of the form you envisage) then having regular (and hence smooth) fibre at one point implies being smooth in a neighbourhood of that point, and a smooth map over a regular base has a regular total space --- thus $X$ will be regular in a neighbourhood of the regular fibre. In particular, the points of the regular fibre will themselves be regular in $X$.

In particular, in your special case $\mathbb Z \subset R$, the answer is "yes".

Edit: I should note that in your situation, where $R$ is a subring of a number field, this "yes" is easily proved directly: one combines the fact that $R/\mathfrak q$ is regular with the fact that it is a priori finite (in cardinality) to see that it is a product of finite field extensions of $\mathbb Z/\mathfrak q$, and hence that the completion of $R$ at $\mathfrak q$ is a product of DVRs, and hence that $R$ is a DVR --- and thus regular --- after localization at each prime above $\mathfrak q$. The point of the more highbrow explanation above is to indicate how one thinks about such questions geometrically --- which is normally the easiest way to see what should be true and what should be false for these kinds of questions.

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Great, thanks! – Harry Gindi Oct 16 '10 at 2:57
Matt, beautiful answer, as usual! – Hailong Dao Oct 16 '10 at 3:02
Harry, the general principle underlying Matt's answer is that if $f:X \rightarrow S$ is an flat map of finite type between noetherian schemes then for many interesting "homological" properties $P$ of local noetherian rings the validity of $P$ for $O_{X,x}$ is equivalent to the same for $O_{S,s}$ and $O_{X_s,x}$ together, where $s = f(x)$. This is just a principle rather than a "meta-theorem" (whatever that may mean), but it (or a mild variant) is true for many $P$. Look at the section "flatness and fibers" in Matsumura's "Commutative Ring Theory" for examples (including regularity). – BCnrd Oct 16 '10 at 3:17
Thanks, I'll check it out. – Harry Gindi Oct 16 '10 at 3:23

How about we try the simple case $R=S$ and the identity map. Let $\mathfrak q$ be a singular prime. The fibre over $\mathfrak q$ is $\kappa (\mathfrak q)$ which is regular (it is a field) but $R_{\mathfrak q}$ is not regular.

Edit: let me add a few more details since my original terse answer hardly deserves the upvotes! Let $P$ be the set of primes in $\text{Spec}(R)$ which contracts to $\mathfrak q $ (the set-theoretic fibre). Your first condition says that:

(1) $R_{\mathfrak p}$ is regular for all $\mathfrak p \in P$.

while the second says:

(2) $R_{\mathfrak p}/\mathfrak qR_{\mathfrak p}$ is regular for all $\mathfrak p \in P$.

So you can see where the problems come from: in general, for a local ring $R$, regularity of $R$ has nothing to do with regularity of $R/I$ for some ideal $I$, unless if $I$ is very special. That is how I think about the counter-examples, hopefully it helps.

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Dear Long, We gave the same counterexample! What does it say about mathematicians' training that the first map we consider is always the identity map? Best wishes, Matt – Emerton Oct 16 '10 at 2:54
(-;, we are simple folks. – Hailong Dao Oct 16 '10 at 3:01

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