Why is the Category of Correspondences not pseudo abelian?

i've just gotten into the theory of motives.I understand the construction of the Karoubian envelope (pseudo-abelian completion) to ensure that morphisms have kernels and images in order to get certain decompositions of the diagonal of Motives in $\mathcal{M}_k$.But i don't see why the morphisms in $Corr_k$ which are elements of $CH(X \times Y)$ for the objects $X,Y$ don't have kernels and/or images in general.

A second one: In the category of effective Chow-Motives, do you know an example of non isomorphic varieties,that have the same motive,except for a totally split quadric and projective spaces?

• It would probably help if you defined the category of correspondences. There are probably people here who can answer your question if given such a definition but who aren't familiar with motives. – Qiaochu Yuan Nov 3 '13 at 19:42
• Its totally clear on en.wikipedia.org/wiki/Motive_(algebraic_geometry). Besides: When i talk about effective Motives i mean the Karoubian evelope of the category of degree zero correspondences. – Jason Pioneer Nov 3 '13 at 19:50
• I second Qiaochu Yuan. Also specify your equivalence relation: I assume it is rational equivalence. Further, not all morphisms have kernels and cokernels, only idempotent morphisms. This is the difference between a pseudo-abelian category, and a preabelian category. – jmc Nov 5 '13 at 10:53
• I should add that if one uses numerical equivalence, the category of motives is abelian; in particular all morphisms have kernels and cokernels. One of the Standard Conjectures says that numerical equivalence is homological equivalence, implying that homology theories factor through this category of numerical motives. – jmc Nov 5 '13 at 14:59
• Yes, I mean rational equivalence.Thanks for pointing out these differences. – Jason Pioneer Nov 6 '13 at 13:57

Answer 1. Take the projective line $\mathbb{P}^{1}_{k}$, and let $p$ be the correspondence projecting to a point. Then this has an image, namely $\mathrm{Spec}(k)$. However, it does not have a kernel. One can see this by looking at cohomological realisations. If the kernel existed, it would only have an $\mathrm{H}^{2}$; since the $\mathrm{H}^{0}$ is accounted for by $\textrm{Spec}(k)$.
 I realise I am made this more difficult then necessary. Suppose that all idempotents have kernels. Then the category of correspondences is equivalent to its Karoubian envelope. In the Karoubian envelope we have a decomposition $\mathbb{P}^{1}_{k} = \textrm{ker}(p) \oplus \textrm{ker}(1 - p)$. The direct sum in this category is given by the disjoint union of the underlying schemes and correspondences. Under our assumption of equivalence with the category of correspondences, this exhibits the connected scheme $\mathbb{P}^{1}_{k}$ as the disjoint union of $\textrm{Spec}(k)$ and some other scheme. Contradiction. [/Edit]
Answer 2. Since the Hom-sets have $\mathbb{Q}$-coefficients, every isogeny becomes an isomorphism. But maybe you find this answer a bit cheating.
• @ Answer 1. You mean we would have an exact cohomoloy-sequence $\mathbb{Z} \xrightarrow{p*} \mathbb{Z} \xrightarrow{i*} ker(p)$ ,which then obviously forces $ker(p)$ to vanish? – Jason Pioneer Nov 6 '13 at 17:25
• @JasonPioneer, well, $p$ is idempotent, hence so is $1 - p$. Then $\mathbb{P}^{1}_{k}$ decomposes as $\textrm{ker}(p) \oplus \textrm{ker}(1 - p)$. You then have to check that the realisations respect direct sums to arrive at a contradiction. (I admit that my answer is a bit sketchy...) – jmc Nov 6 '13 at 18:18