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Let $k$ be a finite field with $q$ elements, $U$ an integral scheme of finite type over $k$ and $f: U \to \mathbb{A}_k^d$ a finite dominant morphism (obtained by Noether normalization, $d=\dim U$). If $K$ is an extension of $k$ of degree $n$, how exactly does one get the inequality $|U(K)| \leq deg(f)\cdot q^{n\cdot d}$?

This seems intuitive to me but I failed trying to explain it to myself. $U(K)$ is the set of $K$-rational points of $U$.

Thanks in advance for your help!

user1728
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  • Before anyone attempts this, let me explain why the obvious approach doesn't work. The obvious approach says that the number of $K$-points of $\mathbb A^d_k$ is $q^{nd}$, and the fibres of the map all have degree $\deg f$. But the second statement is only true if $f$ is flat, which we may not in general assume. In fact, I'm not even convinced (yet) that the result you're trying to prove is true! – Remy Feb 18 '16 at 06:05
  • @Remy I apologize if I'm being silly, but isn't that precisely why there is an inequality? Namely, consider the fiber $f^{-1}(y)\to \text{Spec}(k(y))$ for any point $y\in\mathbb{A}^n_k$. Then, this map is finite (of the same degree as $f$) and so, in particular $f^{-1}(y)=\text{Spec}(A)$ for some $k(y)$-algebra of dimension $\deg(f)$. But, then, the $K$-points of $f^{-1}(Y)$ correspond to the maps $A\to K$ which there are patently less than $\dim(A)=\deg(f)$. Thus, we see that for a given $K$-point $y$ in $\mathbb{A}^d_k$ that $#(f^{-1}(y)(K))\leqslant \deg f$. – Alex Youcis Feb 18 '16 at 13:57
  • And then notices that $#(U(K))$ is certainly bounded by $\displaystyle \sum_{y\in \mathbb{A}^d_k(K)}#(f^{-1}(y)(K))$ (with possible strict inequality because an $L$-point of might map to $y$ for some other extension $L/K$). But, the above tells us that $#(f^{-1}(y)(K))\leqslant \deg(f)$ so that we can conclude that $#(U(K))\leqslant \deg(f)#(\mathbb{A}^d_k(K))=\deg(f)q^{dn}$.

    Please point any silly mistakes I've made!

     – Alex Youcis Feb 18 '16 at 13:59
  • @AlexYoucis: Can you explain why $f^{-1}(y) \to \operatorname{Spec} (k(y))$ is finite of degree $\deg f$ (the degree of the function field extension)? I can only prove this when $f$ is flat, and I think it is false in general (but maybe it is at most $\deg f$?). – Remy Feb 18 '16 at 18:28
  • @Remy Ah, yes! I knew that I was being silly about something. I was, in the proof in my head, assuming it was locally free! :) I'll have to think about that, but yes, I agree, now it seems a little scary. Two trivial things I can offer to the OP in this case. First, this is true when $U$ is a smooth curve just because our map is guaranteed to be flat. – Alex Youcis Feb 18 '16 at 21:15
  • Secondly, if the OP is more just curious about bounding the rough side of $U(\mathbb{F}_q)$, say to prove convergence of $\zeta(U,s)$ ins some half-plane, one can use the above argument to get that $U(\mathbb{F}_q)$ should be $O(q^{dn})$. Indeed, take our map $U\to\mathbb{A}^d_k$ and apply our argument to the flat locus. Then, proceed by Noetherian induction! Hopefully I haven't made a mistake there as well. – Alex Youcis Feb 18 '16 at 21:15
  • Thank you so much for your comments Alex and Remy... I was trying to figure out why the degree of the induced map on the fibers is the same as the degree of $f$. I am indeed interested in proving the estimate above but can not use the flat locus of the scheme or reduce the claim to the case that $U$ is a curve :( the proof that I'm trying to understand can be found on the second page of: (http://www.math.lsa.umich.edu/~mityab/beilinson/SamREU07.pdf) – user1728 Feb 18 '16 at 21:49

2 Answers2

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As noted in the comments (and, as I understand, what caused the OP to post this question in the first place), there is a bit of a problem with the obvious approach of bounding the number of $K$-points in the fibres.

However, for the real question you're interested in (the proof of Lemma 1.4 of these notes), we're allowed to pass to a dense open by the same induction argument. Thus, by generic freeness (Tag 051S), we can restrict to an open $V \subseteq \mathbb A^n_k$ such that $f \colon f^{-1}(V) \to V$ is finite flat. For such morphisms, the (scheme-theoretic) degree is constant, thus $$|f^{-1}(V)(K)| \leq \deg f \cdot |V(K)| \leq \deg f \cdot q^{nd}.$$ The same induction now applies.

Remark. This is a little unsatisfying because it does not answer the literal question that the OP posted. I would be interested to know if there exists a counterexample, or whether the result is somehow true regardless of the failure of the argument suggested above.

Remy
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  • This is a much better argument as the one given in the notes where the inequality seems to fall from heaven ; ) i would also be interested in a counterexample for a finite morphism with more points in the fiber than its degree. I'll keep looking for a while... Thank you so much! – user1728 Feb 18 '16 at 23:20
  • @MartinArtin I guess the other thing which, might, make you happy, is to observe that if $U$ is non-singular (in fact even Cohen-Macaulay!) that $f$ is automatically flat (by miracle flatness), and so the argument we want to make works. – Alex Youcis Feb 20 '16 at 15:18
  • @AlexYoucis, thank you but I may not assume that $U$ is Cohen-Macaulay :( – user1728 Feb 20 '16 at 15:38
  • @MartinArtin If you're thinking about the Weil conjectures, you surely can. You're only going to be dealing with smooth schemes! – Alex Youcis Feb 20 '16 at 15:46
  • I want to have some general results about zeta functions for schemes of finite type even over the integers before I consider smooth varieties over finite fields. Serre has this beautiful proof for the convergence in the general setting but I'm stuck with this elementary lemma: http://math.stackexchange.com/questions/1664299/convergence-of-zeta-functions-for-schemes-of-finite-type-over-the-integers – user1728 Feb 20 '16 at 16:15
  • If you have any ideas @AlexYoucis, I would be very grateful! – user1728 Feb 20 '16 at 16:16
  • @MartinArtin Perhaps I'm being silly, but the question you linked to ONLY needs an inequality of the form $\leqslant \deg(f)q^{nd}+O(\deg(f)+q^{n\ell})$ for $\ell<d$. This follows from the Noetherian induction argument discussed several times throughout this thread. – Alex Youcis Feb 20 '16 at 17:25
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This is an answer to the question posed in Remy's answer (I don't have the points to leave a comment). If you have a finite (or quasi-finite) dominant morphism $f : X \to Y$ of varieties over a field $k$ and $Y$ is normal, then each fibre $f^{-1}(y)$ has at most $\deg(f)$ points, where $\deg(f)$ is defined as the degree of the extension $k(Y) \subset k(X)$ of function fields.

In the case of a finite morphism $f : X \to \mathbf{A}^d_k$ with $k$ infinite and $y$ a $k$-rational point you can prove it as follows. Choose a general line $L \subset \mathbf{A}^d_k$ through $y$. Then $L$ meets the locus where $f$ is flat hence $C = f^{-1}(L) \to L$ is generically flat of the same degree. By going down for integral over normal, we see that every point of $f^{-1}(y)$ is the specialization of a point of $C$ mapping to the generic point of $L$. Hence if $C' \subset C$ is the closed subscheme cut out by the torsion (sections supported in dim $0$) in $\mathcal{O}_C$, then $C' \to L$ and $C \to L$ have the same set of points over $y$. OK, and now the morphism $C' \to L$ is flat (as $L$ is a smooth curve) and this is the case you professed to be happy with. (Warning: $C'$ need not be irreducible and need not even be reduced, but it doesn't matter for the argument --- you can replace $C'$ by its reduction and even by its normalization which would only increase the number of points over $y$.)

If $k$ is finite, then you have to replace the line in the argument above by a higher degree curve to make sure it meets the flat locus.

This argument doesn't prove the general normal case because we don't know we can find a smooth curve through every point. To prove it in general I suggest using completions.

ajdj
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  • Do you mind explaining what $C'$ is—I'm not particularly good with the minutiae of non-reduced things. So, I assume you're trying to apply the result I mentioned in the comments above that if $X$ is integral than any surjection $X\to C$ with $C$ Dedekind is flat. I know that if you don't assume that $X$ is integral, just reduced, you need every irreducible component to surject (else you could do something stupid like take $C\sqcup C\to C$). So, given that you're trying to use this, I'm a little unclear how you're doing it. The gut reaction is to look at $X_\text{red}$, but, of course, that – Alex Youcis Feb 20 '16 at 15:35
  • might not be flat to $C$, and so the composition $X_\text{red}\to C$ might not be flat. So, can you explain to me what you mean by cutting out the torsion, and why that fixes things? I assume you're trying to use some souped up version of the statement I know—perhaps something like if you can control where associated points go—but I don't know this theorem, and I don't really understand what $C'$ is. Thanks! – Alex Youcis Feb 20 '16 at 15:36