Let $k$ be an algebraically closed field and $A$ a finitely generated $k$-algebra, together with a specified surjective morphism $\phi \colon k[x\_1, \dotsc, x\_n] \to A$. For $f \in A$, define $\mathrm{deg}(f)$ to be the minimum of $\mathrm{deg}(g)$, where $g$ ranges over all polynomials in $k[x\_1, \dotsc, x\_n]$ such that $\phi(g) = f$. [Note: by $\mathrm{deg}(g)$, I mean the degree of the highest-degree monomial, where $\mathrm{deg}(x\_1^{i\_1} \dotsm x\_n^{i\_n}) = i\_1 + \dotsb + i\_n$.] If it is helpful, we can assume $A$ is an integral domain, even integrally closed if necessary.



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> Let $u \in A^\*$ be a unit such that $\mathrm{deg}(u) > 0$, or equivalently, $u \not\in k^\*$. Is it necessarily true that $\deg(u^n) \to \infty$ as $n \to \infty$?
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Thoughts: If we have a monomial order that respects degree (such as grlex or grevlex, but not lex), and take a Groebner basis of $\ker(\phi)$, then we see that powers of $u$ remain predictable as long as their leading terms fall outside the ideal generated by the leading terms of the groebner basis (aka, the initial ideal).


Motivation: I'm trying to prove a classical theorem using model theory, and the proof I have in mind would require the above to be true.



Lower bounds on the degrees of representatives of $u^n$ as $n \to \infty$
If $u \in A$ and $deg(u^t), t = 0,1,2, \dots$ is bounded by $d$, then the powers of $u$ lie
in $\varphi(V)$, where $V$ is the finite-dimensional $k$-vector space spanned by the 
monomials in $k[x\_1, \dots , x\_n]$ of degree $\leq d$. Hence the powers of $u$ are linearly
dependent over $k$, so $u$ is algebraic over $k$. If $A$ is a domain, this implies that
$u \in k$, since $k$ is algebraically closed.


