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Let $V$ be a $\mathbb{C}$-vector space of finite dimension. Denote its $d$-th symmetric power by $V^{\odot d}$. I am looking for a proof that $V^{\odot d}$ is generated by the elements $v^{\odot d}$ for $v\in V$.

A different way to look at it is the following: Consider the polynomial ring $R=\mathbb{C}[x_1,\ldots,x_n]$ and $f$ a homogeneous polynomial of degree $d$: Then, I want to show that there are linear polynomials $h_1,\ldots,h_k$ such that $f$ is a linear combination of the $d$-th powers $h_i^d$.

In the case $d=2$, this follows from $2xy = (x+y)^2 - x^2 - y^2$. For higher $d$, I recall seeing a proof involving multinomial coefficients once, but I do not remember the details. I have tried to work it out again, but it seems a bit cumbersome, so I am asking whether you know any textbook where this result is proved. If you know an easy proof, I'd be very happy if you could outline it, though.

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Maybe will help. – KCd May 16 '12 at 8:58
up vote 2 down vote accepted

In my answer here I note that symmetric tensors, as multilinear functionals, descend to linear maps on the symmetric power of the underlying vector space. I then reason that if we could show that $\mathrm{Sym}^n V$ is generated by $n$th powers of elements from $V$ the question on tensors would then be answered in its general form decisively. I remark that this is formally equivalent to the elementary symmetric polynomials $e_n$ being expressible as sums of $n$th powers of homogeneous polynomials.

This was the subject of my question here, which received a correct answer (containing a proof of the claim) from user m_l. It was very combinatorial and indeed involved multinomial coefficients, though I'm not sure how related it is to what you've seen before. (Unfortunately, at this point in time I am the only person to have upvoted poor m_l.) It requires the characteristic of the base field be greater than the power $n$ in question (or zero, of course).

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Let $W$ be the (finite-dimensional) vector space generated by $d$-th powers of linear functions in $x_1,\dots,x_n$. Let $h_1,\dots,h_d$ be such linear functions. Consider the polynomial map $f:\mathbb{C}^d\to W$ given by $f(t_1,\dots,t_d)=(t_1h_1+\dots+t_d h_d)^d$. As $$\frac{\partial}{\partial t_1}\cdots\frac{\partial}{\partial t_d}f=d!\;h_1\dots h_d,$$ we have $h_1\dots h_d\in W$. This shows that any degree-$d$ monomial is in $W$, and therefore also any homogeneous degree-$d$ polynomial.

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I don't understand what the $h_i$'s are: surely there are more than $d$ $d$-th powers of linear functions? – Georges Elencwajg May 16 '12 at 9:24
@GeorgesElencwajg: I edited my last sentence to make it clearer. – user8268 May 16 '12 at 9:36
I don't quite understand your argument. Assuming that $f$ is holomorphic, which seems alright: How and why is $\partial/\partial t_1\cdots\partial/\partial t_d f$ an element of $W$? – Jesko Hüttenhain May 16 '12 at 11:24
@rattle: well, if $f:\mathbb{C}^n\to W$, then $\partial f/\partial t_1$ is again a function $\mathbb{C}^n\to W$, hence so is $\partial^2 f/\partial t_1\partial t_2$, etc. – user8268 May 16 '12 at 14:21
Dear user8268, I certainly didn't vote you down and I think there is absolutely no reason why you should take a month off from this site: I hope you will not keep that promise! – Georges Elencwajg May 16 '12 at 14:39

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