Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

While having lunch today with my advisor I tried to come up with a proof of the following fact:

EGA 0-IV (17.3.3): Let $\phi : (A,\mathfrak{m}) \to (B,\mathfrak{n})$ be a flat local homomorphism of local Noetherian rings (necessarily faithfully flat). Then $B$ regular implies $A$ is regular.

Let me say I am aware of the usual proof of the fact above using Serre's characterization of regular local rings as those with finite global dimension. However my own idea to prove this is much simpler; I want to induct on the dimension of $B$ and use the following facts:

  1. Flatness is stable under base change.
  2. Flat morphisms send non-zero divisors to non-zero divisors.
  3. Let $(B,\mathfrak{n})$ be a Noetherian local ring and $x$ an element of $ \mathfrak{n}\setminus \mathfrak{n}^2$. Then $B$ is regular iff $B/(x)$ is regular and $x$ is a non-zero divisor.

With this the proof should supposedly be easy. When $\dim B = 0$; $A$ has to be a domain otherwise a zero divisor $x \in A$ would be sent to a zero divisor in $B$ ($\phi$ being faithfully flat is necessarily injective) contradicting the fact that $B$ is a field. The dimension of $B$ being zero implies $A$ is zero dimensional (fiber dimension formula for flat morphisms) and thus is a field.

Now suppose $\dim B > 0$. We choose an $x \in \mathfrak{m}\setminus \mathfrak{m}^2$ that is a non-zero divisor so that $\phi(x) \notin \mathfrak{n}^2$. Then $A/(x) \to B/\phi(x)$ is flat with $B/\phi(x)$ regular and by induction we find $A/(x)$ is regular. Hence $A$ is regular by (3) above.

The only place I can see where the proof could fail is the statement in bold. This leads me to the following questions.

  1. Is there a Noetherian local ring of positive depth with every element in $\mathfrak{m}\setminus \mathfrak{m}^2$ a zero divisor?

  2. Is there an example of a $\phi : (A,\mathfrak{m}) \to (B,\mathfrak{n})$ with conditions as above such that $x \in \mathfrak{m}\setminus \mathfrak{m}^2$ but $\phi(x) \in \mathfrak{n}^2$?

share|improve this question

1 Answer 1

up vote 16 down vote accepted

(1) No. Use prime avoidance: if the maximal ideal is not associated, it is not contained in the union of the associated primes together with $\mathfrak{m}^2$.

(2) Yes, consider $k[[x^2]] \subset k[[x]]$.

share|improve this answer
    
Dear Graham, thanks for your answer. Thus it seems like my proof above fails. Do you know of a proof of the fact above that avoids homological methods (such as Serre's characterization of regular local rings as those local rings with finite global definition)? Thanks. –  Ben Lim Jun 6 at 14:49
7  
No, I don't, and I don't think I would expect one. Such an argument would, it seems to me, be close in spirit to proving that regularity localizes, a notoriously intractable problem before the advent of homological methods. If any similar argument existed, I would expect to find it in Zariski-Samuel, but of course they don't mention flatness at all. –  Graham Leuschke Jun 6 at 15:37

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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