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Warning: This one of those does-anyone-know-how-to-fix-this-vague-problem questions, and not an actual mathematics question at all.

If $X$ is a scheme of finite type over a finite field, then the zeta function $Z(X,t)$ lies in $1+t\mathbf{Z}[[t]]$. We can calculate the zeta function of a disjoint union by the formula $Z(X\amalg Y,t)=Z(X,t)Z(Y,t)$. There is also a formula for $Z(X\times Y,t)$ in terms of $Z(X,t)$ and $Z(Y,t)$, but this is slightly more complicated. In fact, these two formulas are precisely the standard big Witt vector addition and multiplication law on the set $1+t\mathbf{Z}[[t]]$. (Actually, there's more than one standard normalization, so you have to get the right one. I believe this ring structure was first written down by Grothendieck in his appendix to Borel-Serre, but I don't know who first made the connection with the ring of Witt vectors as defined earlier by Witt.) If we let $K_0$ be the Grothendieck group on the isomorphism classes of such schemes, where addition is disjoint union and multiplication is cartesian product, then we get a ring map $K_0\to 1+t\mathbf{Z}[[t]]$. We could also do all this with the L-factor $L(X,s)=Z(X,q^{-s})$ (where $q$ is the cardinality of the finite field) instead of the zeta function. This is because they determine each other.

This is all good. The problem I have is when there is bad reduction. So now let $X$ be a scheme of finite type over $\mathbf{Q}$ (say). Then the L-factor $L_p(X,s)$ is defined by $$L_p(X,s)=\mathrm{det}(1-F_p p^{-s}|H(X,\mathbf{Q}_{\ell})^{I_p}),$$ where $I_p$ is the inertia group at $p$. (Sorry, I'm not going to explain the rest of the notation.) If $I$ acts trivially (in which case one might say $X$ has good reduction), then taking invariants under $I$ does nothing, and so as above, the L-factor of a product and sum of varieties is determined by the individual L-factors. If $I$ does not act trivially, then the L-factor of a sum is again the product of the individual L-factors, but for products there is no such formula! (The following should be an example showing this. Take $X=\mathrm{Spec}\ \mathbf{Q}(i)$, $Y=\mathrm{Spec}\ \mathbf{Q}(\sqrt{2})$. The we have the following Euler factors at 2: $L_2(X,s)=L_2(Y,s)=L_2(X\times Y,s)=1-2^{-s}$ and $L_2(X\times X,s)=(1-2^{-s})^2$. So the L-factors of two schemes do not determine that of the product.) Therefore the usual Euler factor cannot possibly give a ring map defined on the Grothendieck ring of varieties over $\mathbf{Q}$.

So, is there a way of fixing this problem? I would guess the answer is No, because while some people might allow you to scale Euler factors by numbers, I don't think anyone will let you change them by anything else. But maybe there is some "refined L-factor" that determines the usual one (and maybe incorporates the higher cohomology of the inertia group?) Assuming there is no known way of repairing things, I have a follow-up question: Is there some general formalism that handles this failure? And if so, how does that work?

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2 Answers 2

James: given that no-one else answered this yet, let me just make some naive comments that you probably know already.

Of course the problem is that if $I_p$ isn't acting trivially, then "taking $I_p$-invariants" isn't as functorial as you'd like it to be. For example consider two 1-dimensional ($\ell$-adic, or even complex) representations $\rho_1$ and $\rho_2$ of $D_p$, a decomposition group at $p$, each with $I_p$ acting non-trivially, and with $I_p$ also acting non-trivially on $\rho_1\otimes\rho_2$. Then the local $L$-factors of $\rho_1$, $\rho_2$ and $\rho_1\otimes\rho_2$ are all just 1, but the local $L$-factor of $\rho_1\otimes(\rho_1^{-1})$ is clearly not 1. In some sense this example is even simpler than the example you give.

But what the example is supposed to stress is the underlying basic problem (which presumably you know already): if $G$ is group acting on f.d. vector spaces $M_1$ and $M_2$, then $M_1^G\otimes M_2^G$ can easily be strictly smaller than $(M_1\otimes M_2)^G$. Hence the moment one considers the local Euler factor (which depends only on the $G$-invariants, where here $G$ is the inertia subgroup) one has lost too much information.

But why not simply consider the ring of isomorphism classes of $\ell$-adic representations of $D_p$ instead? That's a perfectly good ring, and it has direct sums and tensor products, and presumably if you consider virtual representations too then maybe you can see maps from the ring you mention above to this ring (via Kunneth?). Perhaps this ring is "too coarse" for you? I'm not sure. But, if it is, and you're after some finer invariant, then surely the invariant will depend only on the action of $D_p$. So at least this observation somehow removes all the geometry from the question, which now perhaps is "give me a quotient ring of the ring of reps of $D_p$" that still distinguishes non-isomorphic unramified representations" or something.

EDIT: (major addition to answer). Overnight I realised that really all the answer above was saying was the following. You want to find a map from the Grothendieck ring to "zeta-functions". I am suggesting that we start by factoring this map into three pieces. I now realise that I am a bit hazy as to whether this can be done. First I am suggesting that we start by sending a variety to the corresponding motive. Now already I realise I might be in trouble, because I think a motive always has realisations, which are representations, and something in your Grothendieck ring might sometimes only have a virtual rep attached to it. But let me ignore this issue. Next I want to go from motives to their realisations (rep of $D_p$ on an $\ell$-adic vector space). Now I want to go from this representation to its $L$-function (defined as in the question), and I want to claim that your map (I am confused about your map though because I don't know whether you're fixing a degree of cohomology or looking at all degrees at once and taking an alternating sum) factors somehow as a composite of these 3 maps. Finally I am claiming that your complaints are only about the last map, so really your question is purely representation-theoretic. I am not sure whether I am right about this formalism, so I wanted to flag it explicitly.

Next, here's an exercise you might want to try. The group $I_p$ has a pro-$\ell$ component so it might have some higher cohomology on $\ell$-adic vector spaces. I suspect that $D_p/I_p$ acts on $H^i(I_p,M_\ell)$ for all $i$. Why not take the alternating product of these and then take char poly of Frobenius? Stab in the dark! Might get nowhere.

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Thanks, Kevin. I had considered looking at the representation ring, but I guess I was hoping for something zeta-like. It would be nice to have some non-tautological positive statement about Euler factors at the bad primes, especially since we have a very nice statement at the good primes. Also thanks for the simpler example. I agree that any reasonable solution would have work on the level of motives rather than just on the level of the Grothendieck ring of varieties. –  JBorger Feb 9 '10 at 21:28
Let me add a few more words, though I fear this question is beyond salvaging. It would be great to find a ring map from K_0 to some ring of zeta-like objects in analytic number theory. I suspect that that's not possible, in which case it would be nice to have some general formalism that controls its failure, perhaps like how Grothendieck's abstract Riemann-Roch formalism controls the failure of push-forward to commute with the Chern character. –  JBorger Feb 10 '10 at 1:09
James: I had some more comments to make but they wouldn't fit in the margin so I just appended them to my answer. You might want to clarify whether I have misunderstood things. –  Kevin Buzzard Feb 10 '10 at 7:32
Thanks. Yes, I agree with the first paragraph of your edit, or at least that solving the question at the level of Galois reps would be sufficient. As for your suggestion in the second paragraph, I like the spirit, but it looks to me that even when M is the trivial 1-dim rep, you get some H^1 and no higher H^i, so the Euler factors can change even when there is good reduction. I guess that's not a contradiction yet, but I can't be too confident that we'd be moving closer to any ultimate understanding of zeta-functions. I'll think about it some more, though. –  JBorger Feb 10 '10 at 10:24
up vote 3 down vote accepted

When I asked Niranjan Ramachandran this question a few days ago, he pointed out that you can indeed fix the problem if you work with integral models instead of varieties over $\mathbf{Q}$: Let $X$ be a scheme of finite type over $\mathbf{Z}$ and define the Euler factor $L_p(X,s)$ to be $P_p(X,p^{-s})$, where $P_p(X,t)$ is the Zeta-function of the fiber of $X$ over $p$. Then the multiplicativity of the product $\prod_p L_p(X,s)$ follows immediately from that of Zeta-functions for varieties over finite fields. Lo and behold, in the example I gave above, the products of the minimal integral models are different from the minimal integral models of the products. (This is all a little embarrassing, not just because the solution is really easy, but because I take a certain amount of pleasure in telling people that it's better to work directly with integral models! Also, apologies to the people who spent time thinking about this. I probably gave the impression that working with varieties over $\mathbf{Q}$ was non-negotiable!)

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In your counterexample above, does the inequality amount to the statement that the integral closure of Z in the tensor product of fields is not necessarily isomorphic (over 2) to the tensor product of integral closures? –  S. Carnahan Apr 18 '10 at 22:34
Yes. By 'minimal integral model of' I meant (and should have said) 'integral closure of Z in'. –  JBorger Apr 18 '10 at 22:40

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