A rationality problem about $F$-points in tori Let $F$ be a finite field, and $T$ be a torus over $F$. Assume that $T_1,T_2$ are two $F$-subtori of $T$, such that $T_1 \times T_2 \to T,(t_1,t_2) \mapsto t_1 t_2$ is surjective with finite kernel $K$. I wonder whether $T_1(F)$ and $T_2(F)$ would generate $T(F)$.
Of course the case that $T$ is $F$-split is trivial. For general case, one could use Galois cohomology to describe it, namely that $1 \to K(F) \to T_1(F) \times T_2(F) \to T(F) \to H^1(\langle Frob \rangle,K) \to 1$. But I do not think the last cohomology group would vanish in general. Is the claim wrong or did I miss something?
 A: I think the F-points of the two subtori do not, in general, generate the group of F-points of the bigger torus.   Take $F=\mathbb{F}_3$ and $L=\mathbb{F}_9=\mathbb{F}_3[\sqrt{-1}]$.  Then let $T$ denote the Weil restriction from $L$ to $K$ of the split rank $1$ torus.  We can view $T$ as the group of invertible $2\times 2$ matrices over $F$ of the form $\begin{pmatrix} x & -y\\ y & x\end{pmatrix}$. 
Let $S$ denote the kernel of the determinant map from $T$ to $\mathbb{G}_m$,
and let $D$ denote the rank $1$ split torus viewed as the diagonal scalar
matrices in $T$.  Then the map $m:S\times D\to T$ given by $(s,d)\mapsto sd$ is surjective with kernel a diagonally embedded copy of $\mu_2=\{\pm 1\}$.  But, the image of the map $m(F):S(F)\times D(F)\to T(F)$ consists of the matrices with square determinant.  So the cokernel is cyclic of order $2$.  If you like you can also see that, as abstract groups, we have $S(F)=C_4, D(F)=C_2$ and $T(F)=C_8$.  So, since the kernel of $S(F)\times D(F)\to T(F)$ is a diagonally embedded copy of $C_2$, the map $m(F)$ cannot be onto.   You can see from this that your analysis of the cokernel, as $H^1(\mathrm{Gal}(F), K)$, is correct since that group is just, in this case, just the group of homomorphisms from $\hat{\mathbb{Z}}$ to $K$.  In other words, it is isomorphic to $K$ itself.
