If all residue fields $A/m$ have degree $\geq2$ then $A$ is generated by elements $a-a^q$

Question

I found the following claim which I haven't been able to prove.

Let $q$ be a power of a prime $p$. Let $A$ be a finitely generated $\mathbb{F}_q$-algebra and suppose that for every maximal ideal $m\subset A$ we have $[A/m:\mathbb{F}_q]\geq 2$. Then the set $\{a-a^q\}$ generates the unit ideal.

I was wondering if anyone could provide a hint (even for a simpler case, like $A$ being semilocal) so I can work out the details, because I haven't found a way to proceed.

Thanks in advance.

Answer 1

Hints:

1. Prove that for an extension $K/\Bbb F_{q}$ with $[K:\Bbb F_q]\geq 2$ there are elements $x\in K$ such that $x-x^q\ne0$.
2. By localizing reduce to the case where $A$ is a local ring and use 1.

Edit: Let $I$ be the ideal in $A$ generated by the elements of the form $a-a^q$ where $a\in A$. Let $m$ be a maximal ideal of $A$. It suffices to show that the localized ideal $I_m$ is the ideal $J$ of $A_m$ generated by the elements of the form $x-x^q$ where $x\in A_m$ as we may then apply the argument in the local ring case to conclude $I_m=J=A_m$ and hence $I=A$.
Obviously we have $I_m\subseteq J$. Let $x=\frac{a}{s}\in A_m$ with $s\in A\setminus m$. We have$$x-x^q=\frac{a}{s}-\frac{a^q}{s^q}=\frac{as^q-sa^q}{s^{q+1}}=\frac{as^q-as+as-sa^q}{s^{q+1}}=\frac{1}{s^{q+1}}\left(a\underbrace{(s^q-s)}_{\in I}+s\underbrace{(a-a^q)}_{\in I}\right)$$ Hence $x-x^q\in I_m$ and we get $I_m=J$.