About simple motives I'm reading through Jannsen's paper Motives, numerical equivalence, and semi-simplicity and I'd like to pose two questions.
Suppose all motives are $F$-linear, for some characteristic zero field $F$, and $M$ is a simple numerical motive of weight zero over a fixed finite field $\mathbf{F}_q$.

If $\operatorname{End}(M)=F$, does $M$ have to be the trivial motive $1=(\operatorname{Spec}(\mathbf{F}_q),\mathrm{identity},0)$?

Moreover, in the very definition of motives as in the incipit of the paper, say $X$ is a smooth projective $\mathbf{F}_q$-variety, irreducible of dimension $d$:

for a motive $M=(X,p,j)$ does the idempotent $p\in A^d(X\times X,F)$ need to be central?

I expect not, although if not then I find the $F$-algebra structure on $\operatorname{End}(M)=pA^d(X\times X, F)p$ confusing. How is it a sub-$F$-algebra of $A^d(X\times X, F)$? (the answer could well be: "it's not, and that's fine".) What is its identity?
Note that for the first question, since we are considering motives over $\mathbf{F}_q$, the condition $\text{End}(M)=F$ means that the Frobenius element $\pi_M$ acts as a rational number, and since $M$ has weight zero, $\pi_M$ acts as the identity. It would be a consequence of the Tate conjecture that $M$ is isomorphic to $1$. However, what I'm asking assumes an a priori stronger condition than $\pi_M=\text{id}$: I'm assuming that we have $\text{End}(M)=F$. For example, this condition implies that the center of $\text{End}(M)$ is $F[\pi_M]$ (which is $F=\text{End}(M)$ itself), while if we merely assume $\pi_M=\text{id}$ we would know nothing about the center of $\text{End}(M)$ without assuming the Tate conjecture. I'm hoping the stronger condition $\text{End}(M)=F$ is sufficient to prove $M=1$ unconditionally, and that's what the question is about.
As a final remark, since both $M$ and $1$ are indecomposable, it would be enough to find a nonzero morphism between $M$ and $1$, and it would have to be an isomorphism.
 A: Your second question is easier to answer. The idempotent does not need to be central.
The subset $p A^d(X \times X, F) p $ of $A^d(X\times X, F)$ is closed under addition, additive inverses, and multiplication, and contains the additive identity, but not (usually) the multiplicative identity. However, $p1p=p$ is itself a multiplicative identity for $p A^d(X \times X, F) p $, so that $p A^d(X \times X, F) p $ is indeed an algebra (whether it is a subalgebra is a notational question, I guess).
Even if $p$ is central, $1$ is not contained in $p A^d(X \times X, F) p $  unless in fact $p=1$.
For the first, I think the answer is no because I am not sure I agree with your logic that Frobenius must act as a rational number. Why can't it act as an arbitrary algebraic number in $F$?
If $F= \mathbb Q$ I agree the answer should be yes by the Tate conjecture, but I don't think this should necessarily be any easier than the Tate conjecture itself (i.e. it may turn out to be easier in an eventual proof, but not at the current state of knowledge). Consider an algebraic variety $X$ where $H^{2i}(X, \mathbb Q_\ell)$ has an $n$-dimensional subspace where Frobenius acts by $q^i$, and suppose that the Tate conjecture is false and $X$ only has an $n-1$-dimensional space of codimension-$i$ cycles up to $\ell$-adic homological equivalence. Suppose also enough standard conjectures that the Lefschetz pairing on these cycles is nondegenerate. Then isn't the Tate twist by $i$ of the orthogonal complement of the space generated by algebraic cycles inside the  subspace of $H^{2i}(X)$ where Frobenius acts by $q^{i}$ a motive satisfying all your assumptions but not your conclusion?
I don't see why this case of the Tate conjecture should be easier than the general case.
