2
$\begingroup$

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.

$\endgroup$
4
  • 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 $xy$. $\endgroup$ Jan 19, 2014 at 22:19
  • $\begingroup$ Sorry, I meant the degree, not $n$. Anyway, the statement is wrong. $\endgroup$ Jan 19, 2014 at 22:25
  • 1
    $\begingroup$ Sorry, I ask for the field of charactertic 0 $\endgroup$
    – JJH
    Jan 19, 2014 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 $n$. $\endgroup$ Jan 20, 2014 at 0:01

1 Answer 1

8
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.