Lower bounds on the degrees of representatives of $u^n$ as $n \to \infty$ - MathOverflow

Lower bounds on the degrees of representatives of $u^n$ as $n \to \infty$

2010-09-01T15:55:56Z

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.

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 $n \to \infty$?

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.

Answer by Andreas Thom for Lower bounds on the degrees of representatives of $u^n$ as $n \to \infty$

2010-09-01T18:49:37Z

Consider $\phi: {\mathbb C}[x] \to A={\mathbb C}[x]/(x^2=1)$. Then $x \in A$ is a unit with $\deg(x)=1$ and $\deg(x^n)$ is clearly bounded.

Answer by Felipe Voloch for Lower bounds on the degrees of representatives of $u^n$ as $n \to \infty$

2010-09-01T19:04:35Z

If $A$ is the coordinate ring of an irreducible affine curve then the answer is yes. The unit $u$ has a pole at some point at infinity (since it is nonconstant) so the order of pole at this point of $u^n$ grows with $n$. On the other hand, the order of a pole can be bounded above in terms of the degree.

You might be able to do the general case, with $A$ a domain, by taking generic hyperplane sections and inducting on dimension. The hypothesis on $A$ is necessary as per Andreas's example.

Answer by Charles Staats for Lower bounds on the degrees of representatives of $u^n$ as $n \to \infty$

2010-09-18T18:54:00Z

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).

Answer by Ed Formanek for Lower bounds on the degrees of representatives of $u^n$ as $n \to \infty$

2010-09-19T04:40:04Z

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.