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Suppose $F$ is a Frobenius map on a connected, reductive group $G$, and $T$ is an $F$-invariant torus. Let $X$ be the character group on $T$, so that $F$ acts on $X$ by $F(\chi)(t)=\chi(F(t))$, where $\chi\in X$ and $t\in T$. Why is $$ \#X/(F-1)X=|\det_{X\otimes\mathbb{Q}}(F-1)|? $$

The text (Carter, Finite Groups of Lie Type, Prop 3.2.3) I'm reading simply says is follows by "considering elementary divisors" of $F-1\colon X\to X$ without elaboration.

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    This is a question about matrix with entries in $\mathbb{Z}$ a PID. You need to use smith normal form https://en.wikipedia.org/wiki/Smith_normal_form . From there it should be easy. I am assuming that you already have the finiteness of LHS. – 2017-09-17

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