I was wondering when the Kunneth formula holds for motivic cohomology: $$ H^p(X,A(\alpha)) = \bigoplus_{i+j=p;\beta+\gamma = \alpha} H^j(X,A(\beta)) \otimes H^i(X,A(\gamma)) $$ where $H^p(X,A(\alpha))$ is defined as you wish: by higher Chow groups, Hom groups in $DM(X)$, etc... The case I'm most interested in is $A= \mathbb{Q}$ and $M(X) \in DM(\mathbb{Q})_{\mathbb{Q}}$ in the thick subtriangagulated category generated by the $\mathbb{Q}(n)$, $n\in \mathbb{Z}$.
6 Answers
I now remember a nice argument, why there's no Kunneth formula for Chow groups of $X \times X$ unless $X$ has a Tate motive. Let $X$ be smooth projective of dimension $d$. We start with a decomposition of a diagonal: $$ [\Delta] = \sum_{i,j} \alpha^i_j \beta^{di}_j \in \oplus_i CH^i(X) \otimes CH^{di}(X) $$ We can assume $\alpha^i_j$ are linearly independent. In this case we can show that $\alpha^i_j$ form a basis of Chow groups and $\beta^{di}_j$ is the dual basis.
Indeed, as a correspondence $[\Delta]$ acts as identity on Chow groups, so for any class c, $$c = [\Delta]c = \sum_{i,j} \alpha^i_j deg(\beta^{di}_j \cup c),$$ and the claim follows if we substitute $c = \alpha^i_j$.
Now $CH_i(X) = Hom(\mathbb Z(i)[2i], M(X))$ and we can consider the set of $\alpha^i_j$ as a morphism of motives $$\oplus_{i,j}\mathbb Z(i)[2i] \to M(X).$$ A simple computation shows that it is an isomorphism with the inverse given by $\beta^i_j$.
And of course, on the other hand, if $X$ has a Tate motive, then Kunneth formula for Chow groups follows (it doesn't answer the question, since I only consider smooth projective varieties).
I don't think so; for example if X is Spec(Q) then this doesn't seem to be true. If you define, for any $Y$ in this thick subcategory, $\mathbb{H}(Y)$ to be the bigraded ring $\oplus H^s(Y, \mathbb{Q}(t))$, then I'd expect there to be a spectral sequence of the form $$ {\rm Tor}^{{\mathbb H}({\rm Spec} \mathbb{Q}))}(\mathbb{H}(X),\mathbb{H}(X)) \Rightarrow \mathbb{H}(X \times X) $$ instead, i.e. you should take the module structure over the motivic cohomology of a point into account on the righthand side of the formula you were proposing.
Correct me if I'm wrong. I think here is how it goes. For any $X$ and any Tate motive $M$, there is an isomorphism of modules over $H^{\*,\*}(Spec F)$: $$H^{\*,\*}(M(X) \otimes M) = H^{\*,\*}(X) \otimes_{H^{\*,\*}(Spec F)} H^{\*,\*}(M).$$
Remarks.
Right hand side makes sense, since $H^{\*,\*}(M)$ is indeed a module over $H^{\*,\*}(Spec F)$
When $M = M(\mathbf P^n)$, this is the the statement of the projective bundle theorem.
It's unclear to me how to write down the individual terms $H^{p,q}(M(X) \otimes M)$ it terms of cohomology $M(X)$ and $M$.
To prove the theorem in this form it is sufficient to consider the case $M = \mathbf Q(n)$. Note that $H^{\*,\*}(\mathbf Q(n))$ is $H^{\*,\*}(Spec F)$ with bidegree shifted by (0,n), therefore tensoring with this module is the same as shifting the bidegree by (0,n). And the same thing is in the lefthand side, according to Voevodsky's Cancellation Theorem.
There is a paper by Dugger & Isaksen called Motivic Cell Structures in which they establish a Künneth formula for various cohomology theories that are represented in the $\mathbb{A}^1$homotopy category, provided the object $X$ satisfies some sort of cellularity condition which is similar to the requirement of having a Tate motive.
Of course, as Tyler suggested this Künneth formula is of the form of a spectral sequence over the motivic cohomology of the ground field, to wit $\mathrm{Tor}_{H(\mathrm{spec}\; k)} (H(X) , H(Y)) \Rightarrow H(X \times Y)$.
In general, this spectral sequence fails, as can probably be seen by all the counterexamples already given, and certainly can be seen by considering $\mathrm{spec}\; \mathbb{C} \times_\mathbb{R} \mathrm{spec} \; \mathbb{C}$ with $\mathbb{Z}/2$coefficients, where the motivic cohomology rings are known in their entirities (These calculations also appear in papers of Dugger & Isaksen on the motivic Adams spectral sequence).
Very briefly, I believe the following is true: Motivic cohomology does not satisfy a Kunneth formula on the level of cohomology groups, but it does satisfy a kind of Kunneth formula on the level of some suitable derived category of sheaves. This should hold true in general for any BlochOgus cohomology theory, I think.
Did you mean $H^p(X \times X, A(\alpha) = \dots$?
I 'm pretty sure Kunneth formula doesn't hold for motivic cohomology in general. I think you'll get the wrong statement even when you specialize to the case of Picard group of a product of two curves of genus > 0.
However, for Tate motives over $\mathbf Q$ it probaby does hold. Seems like it's sufficient to check the case of $\mathbf{P^n}$ for which motivic cohomology "coincide" with the usual cohomology.

$\begingroup$ This is exactly what I had in mind. In the case of a mixed Tate motives, it feels like this is equivalent to Ext^2(A(0),A(n)) = 0 for the base field but I don't know how to turn this intuition into a proof. Any thoughts? $\endgroup$– AFKNov 25, 2009 at 21:09

$\begingroup$ I had a wrong impression, that $H^{p,q}(Spec(F), \mathbf Q)$ is nonzero only for $p=q=0$. However, since it is not the case I agree with Tyler Lawson, that the formula must have tensor product over $H^{*,*}(Spec F)$ or something like that. Otherwise even H^{,}(X \times Spec(F)) doesn't satisfy the decomposition. $\endgroup$ Nov 25, 2009 at 21:39