Question

This is a revised and more sensible version of the original question, thanks to the kind help of Anthony Quas and J. C. Ottem.

Fix polynomials $f_{1},\ldots, f_{n} \in \mathbb{C}[t]$. Let $M_{k}$ be the set of polynomials $h\in \mathbb{Z}[x_{1},\ldots,x_{n}]$ such that the $t$-degree of $h(f_{1}(t),\ldots f_{n}(t))$ does not exceed $k$. Under addition $M_{k}$ is an abelian group. Questions:

1. Suppose that $M_0$ contains only constant polynomials. Must $M_k$ be finitely generated for all $k$?

2. If the answer to Question 1 is YES (which is my guess) then can anyone explain how to calculate, given $k$ and the $f_i$, an upper bound on the total degrees of the polynomials in $M_k$?

Remark: The requirement that $M_0$ contain only constant polynomials is clearly necessary if $M_k$ is to be finitely generated for all $k$, since if $h$ is a non-constant polynomial in $M_0$ then $h^2$, $h^3$, etc. are all in $M_0$.

Remark: The word GIVEN in Question 2 may seem troublesome since the $f_i$ have arbitrary complex coefficients, but actually this is not a problem. If $c_1, c_2,\ldots$ are all the coefficients of all the polynomials $f_i$, then all we need to know to calculate the $t$-degrees of compositions of the form $h(f_{1}(t),\ldots f_{n}(t))$ are all polynomial relations among the $c_i$. All such relations can be specified by giving a finite basis for the vanishing ideal of $c_1, c_2,\ldots$ in the polynomial ring $\mathbb{Q}[x_1,x_2,\ldots]$.

Example: Let $r$ be a transcendental, let $c=\sqrt{2}$, and let $f_1, f_2= t, ct+r$. It is simple to check that $M_0$ contains only constant polynomials. Note that $M_1$ contains, in addition to $x_1$ and $x_2$, the polynomial $2x_1^2-x_2^2$, and I suspect that these three polynomials and 1 generate $M_1$. Can anyone prove even in this case, that each of the $M_k$ are finitely generated?

Answer 1

Assuming we're talking about alg. indept. over $\mathbb C$ the answer is: yes it's finitely generated.

Let $\Phi(h)$ denote $h(f_1(t),\ldots,f_n(t))$. $\Phi$ is clearly a homomorphism.

Consider the set $\bar M_k$ of $h$ in $\mathbb C[x_1,\ldots,x_n]$ satisfying the condition that the degree of $\Phi(h)$ is at most $k$. Suppose that $\bar M_k$ is not finitely generated. $\bar M_k$ must then contain elements $\Phi(h)$ for $h$ of arbitrarily high degree. Since the range of $\Phi$ is a $(k+1)$-dimensional vector space, taking any $(k+2)$ elements of $M_k$ of different degrees: $h_1,\ldots,h_{k+2}$, there is a linear dependence amongst $\Phi(h_1),\ldots,\Phi(h_{k+2})$, but this gives an algebraic relation between $f_1,\ldots,f_n$, which contradicts algebraic independence of the polynomials over $\mathbb C$.

The set $M_k$ is then the intersection of $\bar M_k$ with a lattice, which is therefore finitely generated.

Answer 2

OK. Let's try this version which at least (I think) answers the special case.

Consider the map $\Psi_k$ sending a polynomial $h$ to $D^k(\Phi(h))$ (where $D$ represents differentiation with respect to $t$). We have that $M_{k-1}$ is precisely the kernel of $\Psi_k$. Notice that $\Psi_k$ is a group homomorphism (but not a ring homomorphism). In the case where $f_1$ and $f_2$ are linear functions, they satisfy the important property $D^2f_i=0$. Notice that $Df_1=1$ and $Df_2=\sqrt{2}$.

Now I claim that $$ D^k \Phi(h) = \sum_{i+j=k}\binom{k}{i\ j}\frac{\partial^k h}{\partial x_1^i\partial x_2^j}(f_1(t),f_2(t)). (Df_1)^i(Df_2)^j. $$ This can be rewritten as $$ \sum_{i+j=k}\binom{k}{i\ j}\Phi\left(\frac{\partial^k h}{\partial x_1^i\partial x_2^j}\right)\sqrt{2}^j. $$ Since $\Phi$ is a $\mathbb C$-linear group homomorphism, this is the same as $$ \Phi\left(\sum_{i\le k}\binom {k}{i} \sqrt{2}^{k-i}\frac{\partial^k h}{\partial x_1^i\partial x_2^{k-i}}\right) $$ Let $A_k(h)$ denote the element of $\mathbb Q[\sqrt 2][x_1,x_2]$ inside the parentheses. This may be succinctly written as $$ A_k(h)=\left(\frac{\partial}{\partial x_1}+\sqrt{2}\frac{\partial}{\partial x_2}\right)^kh $$

I claim that $f_1$ and $f_2$ are algebraically independent over $\mathbb Q[\sqrt 2]$. To see this consider the lexicographic ordering on monomials $x_1^ax_2^b$ where $x_1^ax_2^b\succ x_1^{a'}x_2^{b'}$ if $b>b'$ or $b=b'$ and $a>a'$. If $p\in \mathbb Q[\sqrt 2][x_1,x_2]$ is a relation on $f_1$ and $f_2$: $p(f_1,f_2)=0$, consider the lexicographically maximal term in $p$. This yields a term in $p(f_1,f_2)$ of the form $cr^at^b$ where $c\in \mathbb Q[\sqrt 2]$. This cannot be cancelled by any lower term in $p$.

We showed above that $D^k\Phi(h)=\Phi(A_k(h))$ so since the kernel of $\Phi$ is trivial we see that $M_k$ is equal to the kernel of $A_k$ lying inside $\mathbb Z[x_1,x_2]$. To compute the kernel of $A_k$, we define new monomials $u=x_1+x_2/\sqrt{2}$ and $v=x_1-x_2/\sqrt 2$. We calculate $\frac{\partial}{\partial u}=\frac12(\frac\partial{\partial x_1}+\sqrt 2\frac\partial{\partial x_2})$ and $\frac{\partial}{\partial v}=\frac12(\frac\partial{\partial x_1}-\sqrt 2\frac\partial{\partial x_2})$. The operator $A_k$ is precisely $2^k\partial^k/\partial u^k$ so we see that the kernel of $A_k$ (lying in $\mathbb Q[\sqrt 2][x_1,x_2]$) is generated by the elements of the form $u^iv^j$ for $0\le i < k$ and $j$ arbitrary and with coefficients in $\mathbb Q[\sqrt 2]$. We now need the elements of this group that lie in $\mathbb Z[x_1,x_2]$.

Let $P(u,v)=\sum_{i,j}a_{i,j}u^iv^j$ be a polynomial in $u$ and $v$ (with coefficients in $\mathbb Q[\sqrt 2]$. It lies in $\mathbb Q[x_1,x_2]$ if and only if it is fixed by the field automorphism $\theta$ of $\mathbb Q[\sqrt 2]$ switching $\sqrt 2$ and $-\sqrt 2$. Since $\theta(u)=v$ and $\theta(v)=u$, this means that we are requiring $a_{j,i}=\theta(a_{i,j})$. The elements of $M_{k-1}$ lying in $\mathbb Q[x_1,x_2]$ are spanned by $\sqrt{2}(u^iv^j-u^jv^i)$ for $i$ and $j$ satisfying $0\le i < j < k$ and $u^iv^j+u^jv^i$ for $i$ and $j$ satisfying $0\le i\le j < k$. It follows that $M_{k-1}$ is $k^2$-dimensional.