# basis for space of polynomials

In the proof of Chevalley's theorem on invariants in the polynomials on a semisimple Lie algebra, one uses the following general fact: the polynomials $(\sum m_ix_i)^n$ with $m_i$ positive integers span the homogeneous polynomials of degree $n$ in the variables $x_i$ with coefficients in a field of characteristic zero. Does anyone know a proof of this fact, ideally without painful combinatorics?

-
It can't be true. At most they will span the homogenous polynomials of degree $n$. – Henning Makholm Oct 30 '11 at 1:40
Oops! This is a bit confusing: the standard use of "degree $n$ polynomial" translates into "polynomial with filtration $n$" in a more general standard terminology, which is not what I meant. Thanks. – Justin Campbell Oct 30 '11 at 2:52

Let $K$ be a field of characteristic $0$, and let $S^m$ be the space of homogeneous degree $m$ polynomials in $x_1,\dots,x_n$ with coefficients in $K$. Identify $S^1$ to $K^n$ in the obvious way. Let $Z$ be a Zariski dense subset of $K^n$.

We claim that the $m$-powers of elements of $Z$ generate $S^m$ over $K$.

Let $M$ be the set of length $m$ multi-indices $\alpha$ in $\mathbb N^n$, and equip $S^m$ with the symmetric bilinear form $(f,g)\mapsto\langle f\ |\ g\rangle$ for which the monomials $x^\alpha$, $\alpha\in M$, form an orthonormal basis.

Let $f\in S^m$ be orthogonal to $a^m$ for all $a$ in $Z$. It suffices to show $f=0$.

For $\alpha$ in $M$, define $\binom{m}{\alpha}$ by the expansion $$(x_1+\cdots+x_n)^m=\sum_{\alpha\in M}\ \binom{m}{\alpha}\ x^\alpha$$ in $\mathbb Z[x_1,\dots,x_n]$. The $\binom{m}{\alpha}$ are then positive integers; in particular, they are nonzero in $K$ (because $K$ is of characteristic $0$).

In $K[x_1,\dots,x_n]$ we have, for $a$ in $K^n$,
$$(a_1\,x_1+\cdots+a_n\,x_n)^m=\sum_{\alpha\in M}\ \binom{m}{\alpha}\ a^\alpha\ x^\alpha,$$ and thus $$\Big\langle(a_1\,x_1+\cdots+a_1\,x_n)^m\ \Big|\ x^\alpha\Big\rangle=\binom{m}{\alpha}\ a^\alpha,$$ which (in view of the above identification $S^1=K^n$) we can write $$\langle a^m\ |\ x^\alpha\rangle=\binom{m}{\alpha}\ a^\alpha.$$ Now, if $f=\sum b_\alpha\, x^\alpha$ is orthogonal to $a^m$ for all $a$ in $Z$, we get $$\sum_{\alpha\in M}\ \binom{m}{\alpha}\ b_\alpha\ a^\alpha=0$$ for all $a$ in $Z$, which implies indeed $f=0$.

Let $K$ be a field of characteristic $0$, and let $S^m$ be the space of homogeneous degree $m$ polynomials in $x_1,\dots,x_n$ with coefficients on $K$. Identify $S^1$ to $K^n$ in the obvious way. Let $Z$ be a Zariski dense subset of $K^n$.

We claim that the $m$-powers of elements of $Z$ generate $S^m$ over $K$.

Put $$\partial_i:=\frac{\partial}{\partial x_i}\quad,\quad\partial:=(\partial_1,\dots,\partial_n).$$ The formula $$(f\ |\ g):=f(\partial)\,g(x)$$ defines a non-degenerate symmetric bilinear form on $S^m$.

Let $\alpha$ be an $n$-multi-index of length $m$. The assumption that $K$ is characteristic $0$ implies, by a straightforward argument, the existence of a nonzero scalar $s(\alpha)$ such that $$(a^m\ |\ x^\alpha)=s(\alpha)\,a^\alpha$$ for all $a$ in $K^n$. [We're using standard multi-index notation.]

Let $f\in S^m$ be orthogonal to $a^m$ for all $a$ in $Z$. By the above observation, $f=0$. This proves the claim.

-
Very nice! And more general than my claim. I'm happy modulo this "straightforward argument" for the existence of $s(\alpha)$: can you give me a hint? How is the characteristic zero assumption used? – Justin Campbell Oct 31 '11 at 0:28
Dear @Justin: Thanks! We have $$\binom{m}{\alpha}=\frac{m!}{\alpha_1!\cdots\alpha_n!}\quad,$$ but this is not needed. – Pierre-Yves Gaillard Oct 31 '11 at 7:38

For brevity, define a basal polynomial to mean a positive integer combination of variables, $\sum_i m_ix_i$. Then the claim is equivalent to asserting that every monomial $x_1^{i_1}x_2^{i_2}\cdots x_k^{i_k}$ is a linear combination $a_1p_1^n+\cdots+a_mp_m^n$ where $n=i_1+\cdots +i_k$ and each $p_j$ is basal polynomial.

Proof by induction on $k$. The base case $k=1$ is immediate. The induction hypothesis tells us that $x_1^{i_1}\cdots x_{k-1}^{i_{k-1}}$ is a combination of $(n-i_k)$th powers of basal polynomials in $x_1$ through $x_k$, so it suffices to prove for an arbitrary such basal $p_0$ that $p_0^{n-i_k}x_k^{i_k}$ is a combination of $n$th powers of basal polynomials.

Consider the polynomials $v_j=\binom{n}{j}p_0^{n-j}x_k^j$ for $0\le j\le n$. They are linearly independent because each $v_j$ is a sum of monomials with degree $j$ in $x_k$. Moreover, for $1\le s\le n+1$, the polynomial $w_s=p_0+s x_k$ is basal, and binomial theorem gives $w_s = \sum_{j=0}^{n} s^j v_s$. Therefore each $w_s$ is in the span of the $v_j$s, and has coordinates $(1,s,\ldots,s^n)$ in the basis given by the $v_j$s; such vectors are well known to be linearly independent. Because there are as many $w_s$s as there are $v_j$s, the $w_s$s are a basis for the span of the $v_j$s.

Thus, in particular, $p_0^{n-i_k}x_k^{i_k} = \binom{n}{i_k}^{-1} v_{i_k}$ is a linear combination of $w_s$s, and because each $w_s$ is an $n$th power of a basal polynomial, this completes the proof.

(Edit: fixed major gap in the first proof given)

-