Proposition 17.4.5 In Vakil's Foundations of Algebraic geometry states:

Suppose that $\pi: C \rightarrow C'$ is a finite morphism, where $C$ is a (pure dimension $1$) curve with no embedded points, and $C'$ a regular curve. Then $\pi_*\mathcal{O}_C$ is locally free of finite rank.

The proof boils down to showing that $C'$ is a regular integral affine curve $\operatorname{Spec}(A')$ and $C$ = $\operatorname{Spec}(A)$, then for any maximal ideal $\mathfrak{m}$ of $A'$, $A_{\mathfrak{m}}$ is a free $A_{\mathfrak{m}}$ module. $A'_{\mathfrak{m}}$ is a DVR, so we may take a uniformiser $t$ and it suffices to prove the case where $t \in A'$. Then since $A_{\mathfrak{m}}$ is a finitely generated $A'_{\mathfrak{m}}$ module, it is a direct sum of a free module and modules of the form $A'_{\mathfrak{m}}/(t^n)$ for some $n$. But if there were any components of the latter form, then the image of $t$ in $A$ would be a zero divisor.

The part that I don't understand is what follows, he says that, since any zero divisor is contained in some associated prime, there is some associated point of $C$ in $\pi^{-1}([\mathfrak{m}])$, which contradicts the fact that $C$ has no embedded points.

Firstly, I don't understand why the fact that we know the image of $t$ is contained in an associated prime of $C$ means that we know that there is an associated point in the fibre. Is it the case that any prime containing the image of $t$ maps to $[\mathfrak{m}]$? Certainly any such prime maps to a prime containing $t$, but is $\mathfrak{m}$ the only one?.

Secondly, why couldn't a generic point of $C$ map to $[\mathfrak{m}]$? I know that continuity would imply that everything in that component would then also map to $[\mathfrak{m}]$, but I'm not sure why this couldn't happen. I thought it might have something to do with the fact that finite morphisms have finite fibres, so this component must be finite, but there also exist curves with finitely many points so I don't think this leads anywhere.

  • $\begingroup$ @Hoot For Vakil, a curve in general is just an equidimensional, one dimensional scheme. As far as I can tell there were no restrictions for this section. Also, with regards to your second comment, there was a typo, I'll fix it. $\endgroup$ – Tom Oldfield Jul 20 '16 at 3:02
  • $\begingroup$ For Q1: This follows from the fact that $A'$ has dimension $1$. The worry is that such a prime could pull back to something not maximal, but the only such primes are $0$. $\endgroup$ – Hoot Jul 20 '16 at 3:31
  • $\begingroup$ @Hoot Ah, I think there was another typo. It should read "to $[\mathfrak{m}]$" rather than "a $[\mathfrak{m}]$", the point is I don't understand why this point would be in the fibre over $\mathfrak{m}$. $\endgroup$ – Tom Oldfield Jul 20 '16 at 3:39
  • $\begingroup$ I think Q2 follows from facts about dimension in integral extensions. For the first: I'm a little worried that he needs to say more. I think I have a counterexample otherwise — maybe after this talk. $\endgroup$ – Hoot Jul 20 '16 at 17:21
  • $\begingroup$ @Hoot You're right about Q2, Dtseng's answer reminded me that integral morphisms have finite fibres (because a $k$ scheme with integral structure morphism is $0$ dimensional), so it does in fact hold even for finite curves, which is nice. I'm still unsure about Q1, please let me know if you have a counterexample or some additional hypotheses where it works. I think there should be some sort of locally Noetherian hypothesis, anyway, since Vakil talks about associated points which he only defines for locally Noetherian schemes. $\endgroup$ – Tom Oldfield Jul 20 '16 at 21:31

I've now understood what was confusing me, and will post this answer for future readers who may have the same problem:

A regular scheme is, by definition, locally Noetherian! The confusion arose from the fact that we only define regular points for locally Noetherian schemes, and so $C'$ is in fact locally Noetherian, which I didn't realise at the time.

Given this, I can answer my questions as follows:

Firstly, at the same time as assuming $t \in A'$, we may assume that $\mathfrak{m} = (t)$, since $\mathfrak{m}$ has finitely many generators, and considering each as an element of $A'_{\mathfrak{m}}$ we see that each is of the form $t \cdot f/g$ for some $f/g \in A'_{\mathfrak{m}}$ and so by considering the distinguished affine open piece corresponding to the product of all such $g$ (of which there are finitely many) we have that $(t) = \mathfrak{m}$. It is now clear that any prime of $A$ containing the image of $t$ pulls back to a prime ideal containing $(t)$, and so equaling $\mathfrak{m}$ by maximality.

Secondly, any non-empty fibre of an integral morphism has dimension $0$ (exercise $11.1.E$ in Vakil) so, under finite (even integral) morphisms, only closed points can map to closed points (if a non-closed point mapped to a closed point, so would everything in it's closure by continuity and then the fibre would have dimension strictly positive).


For question 1, if we write it out in terms of rings, we have $t\in A$ that maps to a zero divisor in $B$. That zero divisor must be contained in some associated prime. Take the inverse image of that associated prime in $A$. It contains $t$, but the ideal generated by $t$ is maximal (if we restrict to an open subset disjoint from the zeros of $t$ away from [m]). Therefore our associated prime pulls back to exactly our maximal ideal [m].

For the second question, finite morphisms have finite fibers. In particular, this means the fibers are zero-dimensional. Finite fibers means more than the fibers are finite as sets. In any case, I think finite type over a field and positive dimension means you will have infinitely many points.

  • $\begingroup$ Ah, I now understand the second question, finite morphisms have zero-dimensional fibres (since they're integral) which gives us what we want. I'm still not sure about the first question though, firstly I'm not sure why the $(t)$ is maximal in the localisation you describe, and secondly I'm not sure why the pullback of the ideal we started with should like within this open set? $\endgroup$ – Tom Oldfield Jul 20 '16 at 13:26
  • $\begingroup$ @Dsteng Thanks for the answer, but this by itself wasn't enough for me to solve the problem, since I don't really understand your first part. It did definitely help me work out a solution though, which I posted as another answer, so thank you! $\endgroup$ – Tom Oldfield Jul 20 '16 at 22:50

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