Chow motives of quadratic fields Let us write $CM_k$ for the category of effective Chow motives up to rational equivalence over $k$.
Let $k = \mathbb{Q}$.
We consider for different primes $p,q$ the Varieties $X = \mathrm{Spec}(k(\sqrt{p}))$ and $Y =\mathrm{Spec}(k(\sqrt{q}))$. Are the motives $h(X)$ and $h(Y)$ isomorphic in $CM_k$?
To prove that we need to find correspondences $\alpha,\beta \in CH^0(X \times Y)$ such that $\alpha \circ \beta = \Delta_X$ and $\beta \circ \alpha = \Delta_Y$ while $\Delta$ denotes the diagonal morphism of $X,Y$ respectively.
Id say this is impossible to happen because zero cylces representing the classes of $\sqrt{p}$ and $\sqrt{q}$ are not rational equivalent and lie in different classes. (Why?)
Maybe this case is too trivial.
 A: Remark that $X\times_k X = X \sqcup X$ while $X \times_k Y = Y \times_k X = W$ where $W=\mathrm{Spec}(\mathbf Q(\sqrt p, \sqrt q))$. Thus, in the obvious way we can identify: $$\mathrm{CH}^0(X\times X) = \mathbf Z^2,$$ $$\mathrm{CH}^0(X\times Y) = \mathbf Z,$$ $$\mathrm{CH}^0(Y\times X) = \mathbf Z.$$
Remark also that the the diagonal of $X\times X$ corresponds to one of the components $X \subseteq X \sqcup X$ in the decomposition $X\times_k X = X \sqcup X$. This is because this decomposition comes from $a\otimes b \mapsto (ab, ab^\sigma)$, where $\sigma$ is the nontrivial automorphism of $X$; on the other hand, the diagonal corresponds to $a\otimes b \mapsto ab$, and this is just the projection $(ab, ab^\sigma) \mapsto ab$. (This might look a bit weird because it is not symmetrical, but neither is the choice of $a\otimes b \mapsto (ab, ab^\sigma)$ over $a\otimes b \mapsto (ab^\sigma, ab)$.)
Let $\alpha \in \mathrm{CH}^0(X\times Y)$ and $\beta \in \mathrm{CH}^0(Y\times X)$ be the generators. I claim that $ \beta \circ \alpha = (1,1) \in \mathbf Z^2 = \mathrm{CH}^0(X\times X).$ We have
$$ \beta\times_Y \alpha= \mathrm{Spec}(\mathbf Q(\sqrt p, \sqrt q) \otimes_{\mathbf Q(\sqrt q)}\mathbf Q(\sqrt p, \sqrt q))$$ $$= \mathrm{Spec}(\mathbf Q(\sqrt p, \sqrt q) \oplus \mathbf Q(\sqrt p, \sqrt q))$$ $$= \mathrm{Spec}(\mathbf Q(\sqrt p, \sqrt q)) \sqcup \mathrm{Spec}(\mathbf Q(\sqrt p, \sqrt q)).$$
The two morphisms $\beta\times_Y \alpha \to X$ are given by inclusion of $\mathbf Q(\sqrt p)$ in either component.  The image of the resulting map $\beta\times_Y \alpha \to X\times X$ in $\mathrm{CH}^0(X\times X)$ is $\beta \circ \alpha$, by definition; it consists of both points of $X\times X$, hence $\beta \circ \alpha = (1,1)$. Thus the composition map
$$\mathrm{CH}^0(Y\times X) \times \mathrm{CH}^0(X\times Y) \to \mathrm{CH}^0(X\times X)$$
has image the diagonal $\mathbf Z \hookrightarrow \mathbf Z \times \mathbf Z$. Since the diagonal of $X \times X$ corresponds to $(1,0) \in \mathrm{CH}^0(X\times X)$, this shows that we cannot have $\beta \circ \alpha = \Delta_X$ for any $\alpha, \beta$.
