# Homogeneous polynomials between vector spaces

Consider $\mathbb{C}$ vector space $V$ = span$(e_1,\cdots,e_n)$. Consider the following algebra embedding

$$\mathbb{C}[X_1,\cdots,X_n]\hookrightarrow F(V,\mathbb{C})$$ where $f\mapsto(\sum_i a_i e_i\mapsto f(a_1,\cdots,a_n))$. Denote the image as $O(V)$, call its elements polynomials on $V$. Define a polynomial to be homogeneous of degree $p$ if $f(tv) = t^p f(v)$ for all $v\in V$ and $t\in\mathbb{C}$.

I'm thinking about a different way of characterizing homogenous polynomials of degree $p$. Consider the map $i_p:V\to S^p(V)$ sending $v\mapsto \underbrace{v\otimes\cdots\otimes v}_{p}$, if we have any linear map $g:S^p(V)\to \mathbb{C}$, then I want to show

1) $g\circ i_p$ is indeed in $O(V)$ (and clearly homogeneous of degree p)
2) Every homogeneous polynomial can be obtained this way.

Is the above true, and how to show that?

If indeed true, then I can generalize and define homogeneous polynomials of degree p from $V$ to a vector space $W$ using the second way. Define a function $f$ to be polynomial of degree $p$ from $V$ to $W$ if there is linear map $g:S^p(V)\to W$ such that $f = g\circ i_p$.

Is this the same as all function from $V$ to $W$ such that $f(tv) = t^pf(v)$ for all $v\in V$ and $t\in\mathbb{C}$? And is this a good way to define the space of homogeneous polynomials between two vector spaces?

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I dont know if that's the answer you're looking for, but what you say is almost true. If V is a vector space over $\mathbb C$, then the (homogenous) algebra of polynomials (of degree p) over $V$ is given naturally by the (p-th part) of the symetric algebra on the dual of $V$.
You can prove it in the following fahsion you have a natural pairing $Sym^p(V^*)\times V$ which comes from the map $V^*\times ...\times V^* \times V$, given by $(\phi_1,...,\phi_p, v)\mapsto \phi_1(v)...\phi_p(v)$. This does factor through $Sym^p(V^*)\times V$, and identifies $Sym^p(V^*)$ with what you call $O(V)$, well its homogenous p-th piece.
Now it may seem nitpicking because if $V$ is $\mathbb C^n$, then is dual is too... but the natural (i.e functorial) map is the one i defined, and does not require to choose a basis of $V$ like you did.