# Does regularity of a prime ideal in the fibre imply regularity of the prime?

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)?

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.

• Matt, beautiful answer, as usual! 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). Oct 16 '10 at 3:17

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.

• 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 Oct 16 '10 at 2:54
• (-;, we are simple folks. Oct 16 '10 at 3:01