My question is: why is the group of units in a local Artin ring with algebraically closed residue field $q$-divisible (i.e. the multiplication by $q$ endomorphism is surjective) for $q$ prime to char $k$?
In the following I fix the notation:
$A$ is our ring with unique maximal ideal $\mathfrak{m}$, which, since $A$ is Artinian, must be nilpotent, i.e. we have $\mathfrak{m}^N = 0$ for some $N\geq 1$. Let $p = $ char $k$, and take some integer $q$ not divisible by $p$.
My attempt is as follows. For each $n$, we have an exact sequence:$$0\to \mathfrak{m}^{n+1} \to \mathfrak{m}^n\to \mathfrak{m}^n/\mathfrak{m}^{n+1}\to 0$$ and since the final term is a $k$-vector space, it is uniquely $q$-divisible for $q$ prime to $p$. Thus, from the exact sequence: $$\mathfrak{m}^{n+1}/q\mathfrak{m}^{n+1}\to \mathfrak{m}^{n}/q\mathfrak{m}^{n}\to \frac{(\mathfrak{m}^{n}/\mathfrak{m}^{n+1})}{q(\mathfrak{m}^{n}/\mathfrak{m}^{n+1})}=0$$where the final $0$ comes from the $q$-divisibility of $\mathfrak{m}^{n}/\mathfrak{m}^{n+1}$. Thus we see that $\mathfrak{m}^{n+1}$ being $q$-divisible implies that $\mathfrak{m}^{n}$ is, and thus by induction starting with $n = N-1$ where the statement obviously holds, we get $\mathfrak{m}$ is $q$-divisible. Now we have an exact sequence:$$1\to 1+\mathfrak{m}\to A^*\to k^*\to 1$$with $1+\mathfrak{m}\cong \mathfrak{m}$ and hence $q$-divisible, and $k^*$ is $q$-divisible as well since $k$ is closed under taking $q^{th}$ roots, so we get that $A^*$ is $q$-divisible as desired. I'm wondering (a) if the above argument works, and if so, (b) if the hypothesis that $k$ is algebraically closed is necessary for any of the above steps besides $k^*$ $q$-divisible—any tips are appreciated! Thank you!