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Currently I am studying varieties over $\mathbb{C}$ and I know some scheme theory.

My professor mentioned the other day that given a morphism of varieties over an alg. closed field $k$: $f: X \rightarrow Y$, one should count fibres "with multiplicity". I have been trying to make sense of this. It is especially important for me if $Y$ is a curve and the fibres are divisors of $X$.

So I looked in Hartshorne, dusted off my scheme knowledge, and rediscovered the scheme theoretic fibre $X_y := X \times_Y \text{Spec}(k(y))$ for some $y \in Y$. I worked out an example: $f: \mathbb{A}^2 \rightarrow \mathbb{A}^2$, $(a,b) \mapsto (a^2,b)$. A calculation shows that the fibre over a point $(p,q)$ is the spectrum of $k[x,y]/(x^2-p,y-q) \cong k[x]/(x^2-p)$, which is reduced and consists of two different points if $p \neq 0$, and a nonreduced one point scheme if $p=0$.

Really similarly $g: \mathbb{A}^1 \rightarrow \mathbb{A}^1$, $a \mapsto a^2$, then the fibre is the spectrum of $k[x]/(x^2-p)$. I mention these two examples since in one case, the points and fibres are divisors, in the other they are not.

This is all no problem at all, however i am trying to translate this back into varieties, trying to make sense of what "counting with multiplicities" should mean.

Now the questions.

  • How does one translate this into the language of varieties? It seems obvious from the above that the fibre of $(0,q)$ should be $2(0,q)$, since this "doubling" is hidden in the structure sheaf. This would be especially important if the points were divisors, since then 2 times a point makes a lot of sense. But how to make this mathematically precise? As in, what is the actual mathematical procedure? I guess if we can factor the ideal by which we are modding out into prime ideals, we can count the prime factors (i.e. varieties) with multiplicity, BUT: is this factoring always possible? And is this the right method?

  • I did a course on Riemann surfaces, there multiplicity was also defined: a holomorphic map is locally at a point always of the form $z \mapsto z^n$, then $n$ is the multiplicity of the map at that point. I'm quite sure the definitions agree, of course assuming $k=\mathbb{C}$ and the varieties to be smooth. Can anyone give an argument why?

An ideal answer could include a reference to the two examples, point out differences if there are any (coming from the fact that fibres are/are not divisors) and a general method for writing down a fibre in the language of varieties. If this is only possible in the case of fibres being divisors, i'd still be really happy.

Thanks a lot!

Edit: Thanks a lot for your answers, Andrew and Froggie! There is a small question left. I should give you a motivation for my question: somewhere in Beauville there is a map $f: S \rightarrow C$ where $C$ is a curve, then the inverse image of a point turns out to be the divisor $nE$, where $E$ is a curve on $S$ and $n>1$. So this is why i expected that there at least would be a notion of "multiplicity" in the language of varieties if the fibres were divisors. But as i realize now, there is also the notion of $f^*: \text{Div}(C) \rightarrow \text{Div}(S)$, which is most probably what Beauville used (i havent checked this). So new question: does this $n$ always agree with the multiplicity that you defined Froggie? Is there any other way to relate the scheme theoretic multiplicity and the multiplicity of divisors?

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By the way, i know this is a long post, but i thought including my thoughts on it so far would help potential answerers to pinpoint my exact problem.. –  Joachim Aug 14 '12 at 19:08

2 Answers 2

up vote 7 down vote accepted

I'm not really an algebraic geometer, so in my answer I'll stick to the simple situation where $f\colon X\to Y$ is a finite surjective morphism between smooth irreducible varieties over $k$. Both of your examples fall into this category.

If $x\in X$ is a (not necessarily closed) point and $y = f(x)$, then the multiplicity you are probably looking for is the integer I'll denote by $m_f(x)$, which is $$m_f(x):= \dim_{\kappa(y)}\mathcal{O}_{X,x}/\mathfrak{m}_y\mathcal{O}_{X,x} = \dim_{\kappa(y)}\mathcal{O}_{X,x}\otimes_{\mathcal{O}_{Y,y}}\kappa(y),$$ where here you use $f$ to make $\mathcal{O}_{X,x}$ into a $\mathcal{O}_{Y,y}$-module.

Another way of computing this integer is the following. The space $X_y$ you defined previously is a scheme, and its generic points correspond to the preimages of $y$. If $x$ is a generic point of $X_y$, then $\mathcal{O}_{X_y,x}$ is an artinian ring, and one can take its length, which I will denote $v_f(x):= length\,\mathcal{O}_{X_y,x}$. Then one can derive $m_f(x) = v_f(x)\times [\kappa(x):\kappa(y)]$. In both of your examples, $x$ and $y$ were closed points, so $\kappa(x) = \kappa(y) = k$, and the factor $[\kappa(x):\kappa(y)]$ is $1$. Thus $m_f(x) = v_f(x)$.

Let's apply these definitions to your examples.

(1) $f(a,b) = (a^2,b)$. Let $x = (p,q)$ and $y = (p^2,q) = f(x)$. Then $\mathcal{O}_{X,x} = k[u,v]_{(u-p,v-q)}$ and $\mathfrak{m}_y\mathcal{O}_{X,x} = (u^2-p^2,v-q)\subset k[u,v]_{(u-p,v-q)}$. If $p\neq 0$, then $$\mathcal{O}_{X,x}/\mathfrak{m}_y\mathcal{O}_{X,x}\cong k[u,v]_{(u-p,v-q)}/(u^2-p^2,v-q) = k,$$ which has dimension 1. If $p = 0$, then $$\mathcal{O}_{X,x}/\mathfrak{m}_y\mathcal{O}_{X,x}\cong k[u,v]_{(u,v-q)}/(u^2,v-q) = k[u]/(u^2),$$ which has dimension 2. Thus $m_f(x) = 1$ if $p\neq 0$ and $m_f(x) = 2$ if $p = 0$.

To compute this multiplicity the other way, we start from your observation that $X_y$ has two components so long as $p\neq 0$ and $1$ component (a double point) if $p = 0$. In the first case, $\mathcal{O}_{X_y,x}\cong k[t]/(t\pm p)$ for a preimage $x$ of $y$, which has length $1$, so $v_f(x) = m_f(x) = 1$. If $p = 0$, however, then $\mathcal{O}_{X_y,x}\cong k[t]/(t^2)$, which has length $2$. Thus $v_f(x) = m_f(x) = 2$.

Your second example is worked out similarly. The fact that $x$ and $y$ are or are not divisors doesn't make any real difference.

One can show that in the specific situation I'm considering (finite morphism between smooth varieties), every (not necessarily closed) point $y\in Y$ has the same number of preimages when counted with the multiplicity $m_f$, and that integer is the degree $d$ of the map $f$. A sketch of the proof of this fact is the following: Finite morphisms between smooth irreducible varieties over any algebraically closed field are flat, and hence $f_*\mathcal{O}_X$ is a locally free $\mathcal{O}_Y$-module of rank $d$. The fiber of $f_*\mathcal{O}_X$ at $y\in Y$ is exactly $\bigoplus_{f(x) = y} \mathcal{O}_{X,x}/\mathfrak{m}_y\mathcal{O}_{X,x}$, so that $$d = \sum_{f(x) = y} \dim\mathcal{O}_{X,x}/\mathfrak{m}_y\mathcal{O}_{X,x} = \sum_{f(x) = y} m_f(x).$$

Edit: Multiplicities when pulling back divisors. Let $f\colon X\to Y$ be a (surjective) morphism between smooth irreducible varieties. Let $D$ be a prime divisor on $Y$, and let $f^{-1}(D)$ have irreducible decomposition $D_1\cup\cdots\cup D_n$. Let $x\in X$ be a general point of $D_i$ and $y = f(x)$. Since $Y$ is smooth, $D$ has a local defining equation at $y$, that is, $D = \{\varphi = 0\}$ for some irreducible function $\varphi\in \mathcal{O}_{Y,y}$. This $\varphi$ pulls back to a function $\varphi\circ f\in \mathcal{O}_{X,x}$ which will define $D_i$ near $x$. However, $\varphi\circ f$ will not necessarily be irreducible, it might be $\varphi\circ f = \psi^{n_i}$, where $\psi$ is an irreducible local defining equation for $D_i$ at $x$. This integer $n_i$ should be the multiplicity of $D_i$ when pulling back divisors --- it doesn't depend on the choice of $x$, so long at $x$ is chosen generic. I think $f^*D = \sum_i n_iD_i$ is how you would pull back $D$. Of course, you can extend $f^*$ linearly to all divisors $D$, not just prime divisors.

When $f\colon X\to Y$ is finite surjective like we had considered before, the integers $n_i$ won't be the multiplicities $m_f$, but should be the multiplicities $v_f$. That is, if $D$ is a prime divisor of $Y$ with generic point $y$, then $f^{-1}(y) = \{x_1,\ldots, x_n\}$ consists of the generic points of the components $D_1,\ldots, D_n$ of $f^{-1}(D)$, and $n_i = v_f(x_i)$. Of course, if both $X$ and $Y$ are curves, then divisors are closed points, and $m_f = v_f$ on closed points, so in the special case of curves the $n_i$ should be given by $m_f$.

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Wow that is great, i wanted to ask something about finite morphisms as well but i didnt since the question was long enough already. SO this really helps. It raises another question though, that i've added as an edit. See the post above. Thanks again! –  Joachim Aug 14 '12 at 20:11
    
+1 very nice explanations! –  Ehsan M. Kermani Aug 14 '12 at 20:38
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@Joachim: Added a section of pulling back divisors. Hope it's right (I'm not an algebraic geometer, seriously...) and helpful. There are way too many notions of multiplicity! –  froggie Aug 14 '12 at 20:39
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You claim not to be an algebraic geometer, but this is a suspiciously nice answer! ;-) One nit-picky point, that nevertheless has caused me confusion in the past, is that to pull back divisors we have to ensure that no component of $X$ maps into the divisor $D,$ otherwise pulling back does not give a (pure) codimension $1$ object. –  Andrew Aug 14 '12 at 21:30
    
@Andrew, i guess the assumption of $X$ being irreducible and the map being surjective makes sure this does not happen. –  Joachim Aug 15 '12 at 9:16

Well, in the first case, I think the simple answer is: Congratulations! You've discovered one of the fundamental reasons to work with schemes, rather than simply with varieties! Non-reduced points and spaces play a fundamental role in many parts of algebraic geometry, e.g. deformation theory and singularity theory to name a couple, and are one of the main reasons for the substantial shift in language and technique a la Grothendieck and company. In the category of varieties, the preimage of $(0,q)$ in your first example really is just $(0,q)$, and somehow we have lost the information of multiplicity present on other nearby fibres.

In your second question, you're right that the definitions agree. Remember that the map $z\mapsto z^n, \mathbb A^1_{\mathbb C}\to\mathbb A^1_{\mathbb C},$ which you mention in your second example, is regular (i.e. algebraic). We can compute the fibres of this map exactly as you have done in your examples, to find that $z\neq 0$ has a fibre with $n$ distinct points closed points, while over $z=0$ all the points collapse to a single non-reduced point (the scheme theoretic fibre keeps track of the ramification).

I don't think the fact that in one of the examples the fibres can be thought of as divisors really comes into it. After all, we could consider higher codimensional subspaces as abstract objects in a Chow ring, or some such. We really just want to keep track of the multiplicity, which can be done via the scheme structure.

I've always thought naively about fibres in the category of varieties, so I don't have a great answer re. computing fibres specifically in this category. I imagine that putting the reduced induced closed subscheme structure on the fibre computed in the category of schemes would work in many cases.

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Thanks! Haha you know that i recently asked for a good reason to work with schemes on MSE? =) Your answers raised another question, i've added an edit with it to the post, have a look if you want. Thanks! –  Joachim Aug 14 '12 at 20:14

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