Let $P_n$ be the space of polynomials in $n$ variables over a field of characteristic 0. I'm very sure that the space spanned by powers of linear functions is the whole space $P_n$. Anyone can come up with a proof, or indicate a reference for it? Thanks a lot.

2$\begingroup$ What is the field/ring? This is certainly not true if $n$ is a prime and the field is of characteristic $n$. Nor is it true over the integers, even if $n=2$: there's no way to get $xy$. $\endgroup$ – Alex Degtyarev Jan 19 '14 at 22:19

$\begingroup$ Sorry, I meant the degree, not $n$. Anyway, the statement is wrong. $\endgroup$ – Alex Degtyarev Jan 19 '14 at 22:25

1$\begingroup$ Sorry, I ask for the field of charactertic 0 $\endgroup$ – JJH Jan 19 '14 at 22:45

2$\begingroup$ If $n=2$, homogeneous polynomials of degree $d$ are generated by $(x+iy)^d$, $i=0,\ldots,d$ (Vandermonde determinant $\ne0$). I think a similar proof can be elaborated for any $n$. $\endgroup$ – Alex Degtyarev Jan 20 '14 at 0:01
Yes, actually, the space of homogeneous degree $d$ polynomials in $(n+1)$ variable is generated by $$(x_0+i_1x_1+\ldots+i_nx_n)^d,\quad 0\le i_1,\ldots,i_n\le d.$$ For proof, write it as a binomial $$[(x_0+i_1x_1+\ldots+i_{n1}x_{n1})+i_nx_n]^d$$ and, using Vandermonde, get all products $(x_0+i_1x_1+\ldots+i_{n1}x_{n1})^px_n^q$, $p+q=d$. Then, use induction.