This question is somewhat inspired by Kevin Buzzard's answer to What is the interpretation of complex multiplication in terms of Langlands? and somewhat from my own curiosity about such topics.

Let $X$ be a variety over $\mathbb{Q}$. This variety induces a pure motive of weight $1$ (it induces pure motives of other weights, but I will focus on the one with weight $1$). I understand that weight $1$ motives come (conjecturally, of course) from weight $2$ newforms.

Okay, now let's trace it back. Let's start with a weight $2$ newform. Then, as James D. Taylor alluded to (and from what I know from http://staff.science.uva.nl/~bmoonen/MTGps.pdf), the corresponding newform must be the pure motive of weight 1 that is induced by an abelian variety (this is special to the weight $2$ newforms).

If so, then it seems that this proves that any pure motive of weight 1 is equal to the pure motive of weight 1 of a motive coming from some abelian variety.

Put back in words that are not conjectural: Is it true that for every variety $X$ over $\mathbb{Q}$ there is an abelian variety over $\mathbb{Q}$, $A$, such that $H^1_{et}(X,\mathbb{Q}_l) \cong H^1 _{et} (A,\mathbb{Q}_l)$ as $Gal(\mathbb{Q})$-representations?

Or perhaps is the following weaker statement true (if I somehow managed to get something wrong in the above): For every variety $X$ over $\mathbb{Q}$, $L(X,s)$ (coming from the action on the pure motive of weight 1 -- which is well defined even without motives, since one can create it using $l$-adic cohomology) = $\prod_i L(A_i,s)$ where the $A_i$'s are (finitely many) abelian varieties over $\mathbb{Q}$ and the $L$'s are coming from their pure motives of weight $1$.

I would very much like to know, if the above is wrong, where exactly the fallacy was. But if everything above is right, then is this known without assuming crazy conjectures like the standard conjectures or forms of Langlands?

of dimension $2$, and with Hodge numbers $(1,0)$ and $(0,1)$come from weight $2$ newforms; higher rank pure weight $1$ motives (even those with Hodge number just of the form $(1,0)$ and $(0,1)$, such as those coming about as $H^1$ of a smooth projective variety) will be attached to automorphic forms on other groups. – Emerton Sep 8 '11 at 2:43