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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?

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Let $f_1=f_2=t$ and consider the polynomials $h(x_1,x_2)=x_1^n-x_2^n$. Then $M_k$ is not finitely generated for any $k$. –  J.C. Ottem Jan 24 '11 at 14:52
    
Do you mean that the polynomials are algebraically independent over $\mathbb C$ or over $\mathbb Q$? –  Anthony Quas Jan 27 '11 at 5:44
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3 Answers 3

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.

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Thanks Anthony... But actually the prototype for my question is the (belated) example at the end, in which, of course, the $f_i$ are algebraically independent over $\mathbb{Q}$ but not over $\mathbb{C}$. Sorry for the confusion. –  SJR Jan 27 '11 at 6:29
    
I think there's still some problem with the formulation. The polynomials $f_1(t)=t$ and $f_2(t)=t+e$ are algebraically independent, but $M_k$ is huge (containing all powers of $x_2-x_1$). Maybe the correct formulation is that the ideal generated by the polynomials over $\mathbb Q$ does not contain any constants? –  Anthony Quas Jan 27 '11 at 6:43
    
Yes, note the new, modified form of Item 1. I should have thought of all this before I posted... But now I do believe that the question is properly stated. –  SJR Jan 27 '11 at 12:21
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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.

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Anthony, note that I've edited the example given in my question to include the polynomial 1 in M_1. Your proof looks good, and the simple result $k^2$ is lovely. I will ponder all the details and see if I can generalize, but I must say that this (process of generalization) seems daunting at the moment. Thanks! –  SJR Feb 18 '11 at 6:10
    
I would be scared of the generalization to non-linear polynomials, but I think there are probably interesting things to be said about the case of more linear polys, or ones with different coefficients. –  Anthony Quas Feb 18 '11 at 16:19
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Second try deals with a very particular case of the problem for $n=2$ but, at least, it includes the baby example (solved in the other answer). First, I would like to slightly reformulate the question.

Problem. Let $f:{\mathbb A}_{\mathbb C}^1\to{\mathbb A}_{\mathbb C}^n$ be a polynomial map, let $I\subset A:={\mathbb C}[x_1,\dots,x_n]$ be the ideal of all polynomials vanishing on the image of $f$, and let $V\subset A$ be a finite-dimensional ${\mathbb C}$-linear subspace. Denote $A_0:={\mathbb Q}[x_1,\dots,x_n]$. One has to prove that $\dim_{\mathbb Q}(I+V)\cap A_0=\infty$ implies $\dim_{\mathbb Q}(I+{\mathbb C})\cap A_0>1$ (and, therefore, $\dim_{\mathbb Q}(I+{\mathbb C})\cap A_0=\infty$).

This is quite an intriguing problem, in particular, because there is a feeling that there are no developed tools to solve it. (Though, I think that there is a counter-example.)

When $n=2$, the ideal $I$ is generated by an irreducible polynomial $g$ (in the baby example, $g=x_2-\sqrt2x_1-r$ with a transcendent $r$) whose leading homogeneous part has the form $g_0:=x_1^{m-k}\displaystyle\prod_{i=1}^k(x_2-c_ix_1)$; we say that the $c_i$'s and $\infty$ (the latter, with multiplicity $m-k$) are roots of $g_0$. If one of the $c_i$'s is not algebraic over ${\mathbb Q}$, we are done because the leading homogeneous part of $gh$ can belong to $A_0$ only if $h=0$. On the other hand, if all coefficients of $g$ are algebraic over ${\mathbb Q}$, then $\overline x_1,\overline x_2\in\overline{\mathbb Q}[x_1,x_2]/I$ are algebraically dependent over $\overline{\mathbb Q}$ and, hence, over ${\mathbb Q}$, implying $I\cap A_0\ne0$.

Here, we consider a first nontrivial particular case of $g=g_0-r$ with transcendent $r\in{\mathbb C}$. Suppose that $\displaystyle\sum_{i=0}^{m+l}q_i=(g_0-r)\sum_{i=0}^la_i$, where $a_i,q_i$ are homogeneous of degree $i$, $a_l,q_{m+l}\ne0$, and $q_i\in A_0$ for all $i\ge s$. It suffices to show that $(m+1)(s-1)\ge l$.

We have $$g_0a_l=q_{m+l},\qquad q_{l-im}=g_0a_{l-(i+1)m}-ra_{l-im},\quad0\le i\le\frac{l-s}m.$$

Since $g_0a_l=q_{m+l}\in A_0$, the roots of $a_l$ and, therefore, the coefficients of $a_l$ are algebraic ($\infty$ is considered as algebraic). Let us show that $g_0$ divides $a_l$. Indeed, denote by $d:=\text{gcd}(g_0,a_l)$ the greatest common divisor of $g_0$ and $a_l$, and let $c$ be a root of $\frac{g_0}d$. Substituting $x_2:=cx_1$ (if $c=\infty$, we substitute $x_1:=0$) in $\frac{q_l}d=\frac{g_0}da_{l-m}-r\frac{a_l}d$, we obtain $\frac{q_l}d|_{x_2:=cx_1}=-r\frac{a_l}d|_{x_2:=cx_1}$, where the polynomials $\frac{a_l}d|_{x_2:=cx_1}\ne0$ and $\frac{q_l}d|_{x_2:=cx_1}$ have algebraic coefficients. This contradicts the assumption that $r$ is transcendent. Consequently, $a_{l-m}=\frac{q_l}{g_0}+r\frac{q_{m+l}}{g_0^2}$.

By induction, suppose that $$a_{l-im}=\sum_{j=0}^ir^j\frac{q_{l+(j-i+1)m}}{g_0^{j+1}}$$ with $q_{l+(j-i+1)m}$ divisible by $g_0^{j+1}$ for all $0\le j\le i$, where $0\le i\le\frac{l-s}m$. From $q_{l-im}=g_0a_{l-(i+1)m}-ra_{l-im}$, we obtain $q_{l-im}=g_0a_{l-(i+1)m}-\displaystyle\sum_{j=0}^ir^{j+1}\textstyle\frac{q_{l+(j-i+1)m}}{g_0^{j+1}}$. If one of the polynomials $\frac{q_{l+(j-i+1)m}}{g_0^{j+1}}$, $0\le j\le i$, is not divisible by $g_0$, we divide the equality by the greatest common divisor $d$ of $g_0$ and these polynomials. As above, we arrive at a contradiction after substituting a root of $\frac{g_0}d$. Consequently, $a_{l-(i+1)m}=\frac{q_{l-im}}{g_0}+\displaystyle\sum_{j=0}^ir^{j+1}\textstyle\frac{q_{l+(j-i+1)m}}{g_0^{j+2}}$.

Thus, we conclude that $q_{l+m}$ is divisible by $g_0^{i_0+2}$, where $i_0:=[\frac{l-s}m]$.

There exists a homogeneous polynomial $g_1\in A$ such that $g_0g_1\in A_0$ and $g_0h\in A_0$ for a homogeneous $h\in A$ implies $h\in g_1A_0$. Such a minimal $g_1$ can be easily constructed in the form $g_1=\displaystyle\prod_{j=1}^t(x_2-c'_jx_1)$, where $c'_j$ are conjugated to suitable $c_i$'s so that the total collection of conjugated $c$'s in $g_0g_1$ is "complete" thus representing $g_0g_1$ as a product of polynomials of the form $x_1^{\deg p}p(\frac{x_2}{x_1})$, where $p$ is the minimal polynomial of some $c_i$. Note that $\deg g_1\ge1$ as, otherwise, $g_0\in A_0$ and ${\mathbb Q}+{\mathbb Q}g_0\subset(I+{\mathbb C})\cap A_0$.

It follows that $q_{l+m}$ is divisible by $(g_0g_1)^{i_0+2}$, so $l+m\ge(m+1)(i_0+2)>(m+1)(\frac{l-s}m+1)$ and $(m+1)(s-1)\ge l$.

First try (stupid and wrong). Let $f_1(t):=t+\pi$ and $f_2(t):=t$. As $f_1$ and $f_2$ are algebraically independent over ${\mathbb Q}$, the abelian group $M_0$ consists only of constants. On the other hand, the polynomials $h_m(x_1,x_2):=(x_1-x_2)^mx_2$, $m\in{\mathbb N}$, all belong to $M_1$. So, $M_1$ is not a finitely generated abelian group.

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The sequence of revisions in this post is confusing, and may have misled you, for which I am sorry. Note that in the example you give, the group $M_0$ contains the nonconstant polynomial $x_1-x_0$. –  SJR Feb 7 at 4:17
    
@SJR I am not sure if you noted that there is another try to solve a problem (in a particular case). Hope now it is less stupid. –  Sasha Anan'in Jun 1 at 4:44
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