Cokernel of map of étale sheaves

Question

Let $p:\mathbb{G}_m\to \operatorname{Spec} k$ be the structure map, and let $T$ be an algebraic $k$-torus viewed as an étale sheaf over $k$. Why is the cokernel of the canonical map $T\to p_*p^*T$ canonically isomorphic to the cocharacter lattice $L$ of $T$?

If $\operatorname{Spec}A$ is affine scheme, it seems to me that $p^*T=T\times \mathbb{G}_m$, so $$p_*p^*T(A)=p_*((T\times \mathbb{G}_m))(A)=T(A[t^{\pm 1}])\times \mathbb{G}_m(A[t^{\pm 1}]).$$ Say that $A=K$ is a field. Then $$p_*p^*T(K)=L\oplus\mathbb{Z},$$ but why is the image of $T(K)$ equal to $\mathbb{Z}$?

Answer 1

There is a small issue with your computation related to $\mathbb G_m$-linearity. By the $\operatorname{Gal}(k^{\text{sep}}/k)$-module view on étale sheaves on $\operatorname{Spec} k$, it suffices to understand the $k^{\text{sep}}$-points. We compute \begin{align*} p_*p^*T(k^{\text{sep}}) &= p^*T(\mathbb G_{m,k^{\text{sep}}}) = \operatorname{Hom}_{\mathbb G_m}(\mathbb G_{m,k^{\text{sep}}}, T \times \mathbb G_m) \\ &= \operatorname{Hom}_k(\mathbb G_{m,k^{\text{sep}}}, T) = \operatorname{Hom}_{k^{\text{sep}}}(\mathbb G_{m,k^{\text{sep}}}, T_{k^{\text{sep}}})\\ &=\operatorname{Hom}_{k^{\text{sep}}}(k^{\text{sep}}[T], k^{\text{sep}}[x,x^{-1}]). \end{align*} The image of the unit $T \to p_*p^*T$ will be the subgroup $\operatorname{Hom}_{k^{\text{sep}}}(k^{\text{sep}}[T],k^{\text{sep}})$. If¹ $T_{k^{\text{sep}}} \cong \mathbb G_m^n$, then we get \begin{align*} \operatorname{Hom}_k(k[T],k^{\text{sep}}[x,x^{-1}]) &= \operatorname{Hom}_{k^{\text{sep}}}(k^{\text{sep}}[t_1^{\pm 1}, \ldots, t_n^{\pm 1}],k^{\text{sep}}[x,x^{-1}])\\ &= \Big((k^{\text{sep}})^\times \times \mathbb Z\Big)^n,\\ \operatorname{Hom}_k(k[T],k^{\text{sep}}) &= \operatorname{Hom}_{k^{\text{sep}}}(k^{\text{sep}}[t_1^{\pm 1}, \ldots, t_n^{\pm 1}],k^{\text{sep}}) \\ &= \Big((k^{\text{sep}})^\times\Big)^n, \end{align*} so the quotient is just $\mathbb Z^n$. The Galois structure is that of the cocharacter lattice of $T$. $\square$

(In the above, all $\operatorname{Hom}$ sets indicate morphisms of rings or schemes, not of group schemes.)

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¹Every torus splits over $k^{\text{sep}}$. See for example Lemma B.1.5 of Conrad's notes.