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.
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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 DegtyarevCommented Jan 19, 2014 at 22:19
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$\begingroup$ Sorry, I meant the degree, not $n$. Anyway, the statement is wrong. $\endgroup$– Alex DegtyarevCommented Jan 19, 2014 at 22:25
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1$\begingroup$ Sorry, I ask for the field of charactertic 0 $\endgroup$– JJHCommented Jan 19, 2014 at 22:45
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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 DegtyarevCommented Jan 20, 2014 at 0:01
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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 $$[(x_0+i_1x_1+\ldots+i_{n-1}x_{n-1})+i_nx_n]^d$$ and, using Vandermonde, get all products $(x_0+i_1x_1+\ldots+i_{n-1}x_{n-1})^px_n^q$, $p+q=d$. Then, use induction.
answered Jan 20, 2014 at 0:20