ac.commutative algebra - Lower bounds on the degrees of representatives of $u^n$ as $n to infty$

The following is inspired by/based on Felipe Voloch's excellent partial answer. It gives an affirmative answer, under a slightly stronger hypothesis of normality than that given in the question.

Note that the homomorphism $phi colon k[x_1, dotsc, x_n] to A$ that I assume given in the question is equivalent to giving a closed immersion $X = mathrm{Spec} A to mathbb{A}^n$. I am going to assume, not only that $A$ is integral and normal, but that the closure $overline{X}$ of $X$ in $mathbb{P}^n$ is normal. Although this does not quite answer the question I was asking, Donu Arapura's answer here shows that if we are given the freedom to choose $phi$, we can ensure that this condition is met. On the other hand, the proof does not require that $u$ be a unit, only that it be nonconstant.

Let $u in A$ be nonconstant. Then $u$ is a rational function on $overline{X}$. Moreover, any poles of $u$ must lie in $Y := overline{X} smallsetminus X$. Let $b(u)$ denote the order of the greatest-order pole of $u$ on $overline{X}$. If $u$ had no poles, then since $overline{X}$ is normal, $u$ could be extended to a regular function on $overline{X}$. Since $u$ is nonconstant, this is impossible, so $b(u) geq 1$.

Give $mathbb{P}^n$ homogeneous coordinates $T_0, dotsc, T_n$, where our embedding $mathbb{A}^n hookrightarrow mathbb{P}^n$ is given as $D_+(T_0)$. Then $T_0$ represents a global section of $mathcal{O}(1)$ on $overline{X}$. The set-theoretic union of the zeros of $T_0$ is $Y$. Let $c$ be the order of the highest-order zero of $T_0$ on $overline{X}$; clearly, $c geq 1$.

Claim: $b(u) leq c cdot deg(u)$.

If $deg(u) = d$, then $u = T_0^{-d} u'$ for some global section $u'$ of $mathcal{O}(d)$. Since $u'$ has no poles, the claim follows immediately.

Thus, we have

$$deg(u^n) geq frac{1}{c} b(u^n) = frac{n}{c} b(u) to infty$$
as $n to infty$.

If am, of course, quite interested to see if anyone can find a way around the normality hypothesis (or show that it is necessary).

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