I feel completely in the dark, like I am totally missing what is going on behind the scenes in this section. I apologize in advance.

Prop. 6.9: Let $X \to Y$ be a finite morphism of non-singular curves, then for any divisor $D$ on $Y$ we have $\deg f^*D=\deg f\deg D$.

I don't understand some of the statements in the second paragraph:

Why are points $P_i$ of $X$ such that $f(P_i)=Q$ in 1-1 correspondence with maximal ideals $m_i$ of $A'$? Somehow we have killed the other maximal ideals by localizing?

And why is $\dim_k \mathcal{O}_{P_i}/t\mathcal{O}_{P_i}=v_{P_i}(t)$?

I'm confused about some other things but I figure if I ask too many details nobody will want to answer so I'll stop here ;-)

EDIT: I think I actually figured out my first question, using that $B\hookrightarrow A$, that $f(I)=I\cap A$ and the ideal correspondence. Second question stands (and a bunch of other stuff I'm still trying to figure out).

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    $\begingroup$ I just realized how much I want a "hartshorne" tag. $\endgroup$ Oct 17, 2014 at 1:46
  • $\begingroup$ I second that. $ $ $\endgroup$
    – Seth
    Oct 17, 2014 at 1:53
  • $\begingroup$ I added the tag and hope that they grant us the wish. $\endgroup$
    – quantum
    Aug 12, 2020 at 10:40
  • $\begingroup$ @quantum I have removed the tag, following the previous discussion on meta. See here: Adding (hartshorne) Tag? and other posts linked there. $\endgroup$ Aug 12, 2020 at 12:18

1 Answer 1


If we have an injective morphism of rings $\phi:B\to A$, and we localize at a prime ideal $\mathcal{P}$ of $B$, what we're doing to $A$ is forcing the elements of $\phi(B\setminus\mathcal{P})$ to be invertible.

What does this mean for the ideals of $A$? If some ideal $I$ happens to satisfy $I \cap \phi(B\setminus\mathcal{P}) \neq \emptyset$, that means that one of its elements gets inverted, so $I$ is "localized away".

In general, the ideals of an arbitrary localization $S^{-1}A$ correspond to the ideals $I$ of $A$ with $I\cap S = \emptyset$.

So the maximal ideals we get in the localization of $A$ at the maximal ideal $\mathfrak{m} \subset B$, which Hartshorne calls $A'$, can be identified with the maximal ideals $\mathfrak{m}_i$ of $A$ with $\mathfrak{m}_i \cap (B\setminus \mathfrak{m}) = \emptyset $, i.e. $\mathfrak{m}_i \cap B \subset \mathfrak{m}$.

But $\mathfrak{m}_i \cap B$ is exactly $f (\mathfrak{m}_i)$. Since closed points map to closed points, this means that the maximal ideals of $A_{\mathfrak{m}}$ are exactly the points of $f^{-1}(\{\mathfrak{m}\})$.

As for the dimension thing, the point is that the local ring $\mathcal{O}_{P_i}$ is a DVR. We want to properly count the multiplicities of the various preimages, and the way to do that is to take a local parameter at $Q$, i.e. a generator of the maximal ideal of the local ring, and see what its multiplicity is at the various preimages. (this is actually nothing but the definition of the pullback of a divisor)

We end up with this element $t$ in the DVR $\mathcal{O}_{P_i}$, with nonzero ideals $(1) \supset (\pi) \supset (\pi^2) \supset \cdots$, and $\nu_{P_i} (t)$ is exactly the largest $n$ such that $t \in (\pi^n)$.

But we can write $t = u\pi^n$, where $u$ is a unit, so in fact $(t) = (\pi^n)$. So one way to find $n$ is just take the quotient by $t$, and see how long the chain of ideals is.

The under-the-table trick here is that each quotient $(\pi^j) / (\pi^{j+1})$, for $j\geq 0$, is a $1$-dimensional vector space over the base field $k$ (think about this; things can go wrong here if $k$ is not algebraically closed). So we can treat the chain of ideals as a filtration of a vector space, and just read off $n$ as the dimension.

See also the discussion on this question regarding this filtration, and the relationship between length and vector space dimension.

Okay, I think that I just answered your questions—or tried to; you never can tell when you're done understanding Hartshorne. Let me know if anything's unclear. (sounds like you've already got a handle on the first part, though)

  • $\begingroup$ Thank you for that very generously detailed explanation. The main thing I was missing was the "under-the-table trick" (In fact I had forgotten $k$ was assumed algebraically closed here). I'm trying to figure out why successive quotients are isomorphic to $k$. Since $k$ is algebraically closed I guess the residue field of the DVR is $k$, since the DVR is a f.g. $k$-algebra (is this last part true)? And then I guess I should show successive quotients are $1$-dimensional vector spaces over the residue field? $\endgroup$
    – Seth
    Oct 16, 2014 at 22:23
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    $\begingroup$ @Seth You might have missed my edit, but I linked to another thread where Georges Elencwajg and I discussed this filtration to death (though the comments were mostly deleted). The basic point is that, if $(R,\mathfrak{m})$ is local, then $\mathfrak{m}^j / \mathfrak{m}^{j+1}$ is a vector space over the field $R/\mathfrak{m}$. $R/\mathfrak{m}$ is a finite extension of $k$, so if $k$ is algebraically closed, they coincide. When $\mathfrak{m}$ is principal, each vector space has one generator, so is either trivial or one-dimensional. $\endgroup$ Oct 16, 2014 at 22:29
  • $\begingroup$ @Seth I probably shouldn't have said "isomorphic", because only $A/\mathfrak{m}$ is naturally a field. The others are modules over this field. $\endgroup$ Oct 16, 2014 at 22:31
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    $\begingroup$ @Seth I don't think a DVR over $k$ can ever be finitely generated, unless it's a field (often excluded from the definitions). Are you just trying to conclude that $\mathcal{O}_P / \mathfrak{m}_P\mathcal{O}_P$ is a finite extension of $k$? That follows from the fact that it's isomorphic to $\mathcal{O}/\mathfrak{m}_P$, and $\mathcal{O}$ is finitely generated. $\endgroup$ Oct 16, 2014 at 22:46
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    $\begingroup$ @Seth Great. This was a good chance for me to dust off the old toolbox also. :) About the DVR thing: I haven't put a lot of thought into the details, but you should be able to quickly tell whether something is a finitely-generated $k$-algebra, at least after killing nilpotents, by seeing whether its spectrum is a variety. The spectrum of a local $k$-algebra has a unique closed point, but the only variety with just one closed point is $\operatorname{Spec}$ of a field. So local $k$-algebras are pretty much never finitely generated, field extensions being the possible exception. $\endgroup$ Oct 17, 2014 at 1:05

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