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Algebraic K-theory defines a functor K taking commutative rings to E_\infty ring spectra. I'm interested in which pushouts (tensor/smash products) K preserves. For example, if R is a regular noetherian ring then (I believe) K(R[t, t^{-1}]) = K(R) / ΣK(R) = K(R) /\K(Z) K(Z[t, t^{-1}]). On the other hand, K(Q) = K(Q ⊗ Q) is not the same as K(Q) /\K(Z) K(Q) as you can check by computing π1.

Are there useful conditions under which K-theory preserves pushouts?

Edit: I'm equally interested in more general positive answers and more geometric counterexamples. For example, what is an example of smooth schemes X and Y over Spec k such that K(X) /\K(k) K(Y) -> K(X xk Y) is not an equivalence?

Also, what if I only cared about K0? Is the product map more often an isomorphism then?

More generally, is there a spectral sequence to compute the K-theory of a fiber product of schemes?

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2 Answers 2

For smooth varieties $X$ and $Y$ over a field $k$, it is pretty rare to have that the product map $K_0(X)\otimes K_0(Y)\to K_0(X\times_kY)$ is an isomorphism (or surjective).

For instance, suppose $X$ is smooth and proper over $k$, and the product map above is surjective for $Y=X$. Consider the class of the diagonal $\Delta_X$ in $K_0(X\times_kX)$ (this makes sense because $K_0$ is the same for vector bundles and coherent sheaves). Then the class of $\Delta_X$ is expressible as a finite sum $\sum_ia_i\otimes b_i$ with $a_i,b_i\in K_0(X)$.

View elements $\alpha\in K_0(X\times_kX)$ as correspondences, i.e., as endomorphisms of $K_0(X)$, given by $x\mapsto (p_2)_*(p_1^*(x)\cdot\alpha)$, where $\cdot$ is the multiplication in $K_0$ (the direct image $(p_2)_*$ exists because $X$ is proper). The class of the diagonal gives the identity endomorphism, while the class of an element which is the image of $a\otimes b$ is rather special: it is of the form $x\mapsto \chi(x\cdot a)b$, where $\chi:K_0(X)\to {\mathbb Z}$ is the Euler characteristic ($\chi$ is the direct image map $K_0(X)\to K_0(k)$). Using this, one can deduce that $K_0(X)$ must be free abelian of finite rank, and $a_i$ (and also $b_i$) form a $\mathbb Z$-basis (and in fact, for the non-degenerate pairing $(x,y)\mapsto \chi(x\cdot y)$, they form dual bases). This argument, in some form, is well-known in some circles; this ``dual bases'' formulation appears in work of Ivan Panin.

Thus it is easy to give examples of such $X$ for which $K_0(X)\otimes K_0(X)\to K_0(X\times_kX)$ is not surjective, e.g., a smooth projective curve of positive genus over an algebraically closed field, or (more tricky) a complex Enriques surface ($K_0$ is finitely generated, but has torsion).

I do not know a counterexample to the following:

let $X$ be a smooth (say, proper) variety over a field $k$ for which $K_0(X)\otimes K_0(X)\to K_0(X\times_kX)$ is an isomorphism; then for any smooth variety $Y$, the product map $K_0(X)\otimes K_0(Y)\to K_0(X\times_kY)$ is an isomorphism.

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Welcome to MO ! –  Chandan Singh Dalawat Mar 15 '13 at 17:36

I don't have a complete answer to this. However, there is an argument (which I have not checked carefully, but I believe it works) to proves that K(XxY) = K(X) /\L K(Y) when X, Y are smooth schemes over k, and one of them (say Y) is a linear variety. Here /\L is the derived smash product over K(Spec k).

The class of linear varieties is the smallest class of quasi-projective varieties such that

  1. Affine spaces are linear,
  2. Let X be a variety, U an open subvariety and Y its closed complement. If Y and either U or X is linear, so is the other.

For example, any toric variety is linear.

Now using the localization exact triangle for the variety Y, the homotopy invariance of K-theory of smooth schemes (i.e. K(XxA^k) = K(X)) and the fact that derived-smashing with K(X) preserves exact triangles, I believe one can use an inductive five-lema to show that K(XxY) = K(X) /\L K(Y).

Maybe this argument can be extended to deal with more general fibre products over a general base S. But as it uses homotopy invariance of K-theory, which does not hold for singular schemes, and as XxSY may be singular, this might lead to trouble.

This is a very special case though (Y is very special). For a general Y this result will not be true.

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