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? 



Can the failure of the multiplicativity of Euler factors at bad primes be corrected?
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!)


