Polynomial maps of finite-dimensional vector spaces

Question

I'm currently reading "General linear and functor cohomology over finite fields" by Franjou, Friedlander, Scorichenko and Suslin. There they define a polynomial map $T : V \rightarrow W$ of finite-dimensional space to be a morphism of the affine schemes $$ \text{Spec}(S^*(V^\#)) \rightarrow \text{Spec}(S^*(W^\#)) $$ where $S^*$ is the symmetric algebra and $-^{\#}$ is the standard $k$-duality. Moreover they say that this is equivalent to an element in $S^*(V^\#) \otimes W$. It is this isomorphism that I don't see. I see that for a finite-dimensional vector spaces we have that $\text{Hom}(V,W) \cong V^{\#} \otimes W$ using tensor-hom adjunction, but I can't see to find a similar argument for the symmetric algebra in this case.

Thank you very much for any help!

Answer 1

$\operatorname{Hom}(V,W)$ refers to the $linear$ maps between vector spaces $V,W$. Although linear maps are always polynomial maps, polynomial maps needn't be linear maps.

In the case that $W=k$ is the base field, the ring of polynomial functions is the ring generated by $V^\lor=\operatorname{Hom}(V,W)$ in the set of all functions on $V$. This is the symmetric/polynomial algebra, $S(V^\lor)$.

If you take bases $\{v_i\}$ and $\{w_j\}$ for the vector spaces, then the maps $V\to W$ are equivalent to maps $k^n\to k^m$ (assuming $V,W$ finite dimensional). Now, a polynomial map $T:k^n\to k^m$ is just a map of the form

$$T:(x_1,...,x_n)\mapsto (f_1(x_1,...,x_n),...,f_m(x_1,...,x_n))$$

where all the $f_i$ are polynomials in $n$-variables.

Ok, now putting this back into abtract form, a polynomial function on $V$ is an element of $S(V^\lor)$ (in my notation) and an element of $S(V^\lor)\otimes W$ is a sum of things of the form $f\otimes w$ for $f\in S(V^\lor)$ and $w\in W$, for instance, $\sum_{j=1}^m f_j\otimes w_j$, which is just the abstract way to write the coordinate definition above.