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

My problem is the following:

I have a finite surjective morphism $f: X\rightarrow Y$ of noetherian schemes and know that $Y$ is a regular scheme. (Indeed, in my situation, the two schemes are topologically the same and the arrow is topologically the identity.)

I don't know if $f$ is étale or smooth. But I know that $f$ has a section.

I want to conclude somehow that $X$ is also a regular scheme. What further minimal conditions (on $f$ or $X$) would imply this?

I would be glad about just some hints of what one could do here.

share|improve this question
    
Is f the Frobenius, in the special case you are interested in? –  Mahdi Majidi-Zolbanin Feb 4 '12 at 17:24
    
Is the morphism $f$ birational? If so, I recommend you look up "Zariski's Main Theorem". –  Jason Starr Feb 4 '12 at 17:27
1  
The only other thing I know of is Theorem 23.7 in Matsumura's Commutative Ring Theory where they prove the result for flat with regular closed fiber. –  Karl Schwede Feb 4 '12 at 18:12
3  
In characteristic 2, the following inclusion is flat $$ k[x^2] \subseteq k[x^2, x^3] $$ although and although $k[x^2]$ is regular, $k[x^2,x^3]$ is not. There is no section, although there is a map from $X = \text{Spec} k[x^2]$ to $Y = \text{Spec} k[x^2,x^3]$ such that the composition with $Y \to X$ is the Frobenius. –  Karl Schwede Feb 4 '12 at 18:22
add comment

2 Answers

up vote 3 down vote accepted

I may be misunderstanding the question, but it seems rather straightforward to me.

a) As stated, without assuming, that $X$ is, say, reduced, it is certainly false: Let $X=\mathrm{Spec}k[\varepsilon]=k[x]/(x^2)$, $Y=\mathrm{Spec} k$ and $f:X\to Y$ the structure map of $X$ as a $Y$-scheme. This is obviously a homeomorphism topologically and it has a section that maps $Y$ isomorphically to $X_{\mathrm{red}}$.

b) It seems to me that pretty much this is the only thing that can go wrong: If $f$ has a section, then by definition of a section, $Y$ is isomorphic to the image of the section.

As $Y$ is regular, it is necessarily reduced, so if $f$ is one-to-one, then $f$ and the section induce an isomorphism between $Y$ and $X_{\mathrm{red}}$. So, if $X$ is reduced, the statement is true, if $X$ is not reduced, then the question is whether there exists a morphism $X\to X_{\mathrm{red}}$ that's an isomorphism on $X_{\mathrm{red}}$. The above example shows that this can happen, so some assumption is required to rule out this possibility.

If $f$ is not one-to-one, something similar would still happen. The section would still induce an isomorphism between $Y$ and the image of the section which would have to be an irreducible component of $X_{\mathrm{red}}$. On the other hand, in this case you would probably want to assume that $X$ and $Y$ are irreducible, since otherwise you can just add an additional component that screws things up. However, then (assuming that $X$ is irreduciblke) this implies that $Y\simeq X_{\mathrm{red}}$ and you end up in the previous case again.

share|improve this answer
add comment

Here is a possible affirmative answer. At any point $P\in X$, write $f_P^{\#}$ for the induced local homomorphism of stalks $\mathcal{O}_{Y,f(P)}\rightarrow\mathcal{O}_{X,P}$. Assume:

  1. $X$ is Cohen-Macaulay;
  2. At any point $P\in X$, we have: $\dim \mathcal{O}_{X,P}=\dim {\mathcal{O}}_{{Y,f(P)}}+\dim (\mathcal{O}_{X,P}/f_P^{\#}(\mathfrak{m}_{f(P)}){\mathcal{O}}_{{X,P}})$.
  3. At any point $P\in X$, the ring $\mathcal{O}_{X,P}/f_P^{\#}(\mathfrak{m}_{f(P)}){\mathcal{O}}_{{X,P}}$ is regular.

Then 1) and 2), and the assumption that $Y$ is regular imply that for every $P\in X$, $f_P^{\#}$ is flat (see, e.g., Matsumura, Commutative ring theory, Theorem 23.1, p. 179). This flatness plus condition 3) plus the assumption that $Y$ is regular imply that $X$ is also regular (see, e.g., Matsumura, Commutative ring theory, Theorem 23.7, p. 182).

I don't know if conditions 1)-3) are satisfied in the case you are considering, but maybe you can say more about your special case.

share|improve this answer
add comment

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.