# Symmetric Algebra over Tensor Product

Let $$k$$ be a field. I am interested in the symmetric algebra functor $$S : k \text{-vect} \rightarrow k \text{-alg}$$ taking a $$k$$-vector space $$V$$ to the symmetric algebra $$S(V)$$ over $$V$$, which is a quotient of the tensor algebra $$T(V) / I$$, where $$I$$ consists of the ideal generated by tensors of the form $$x \otimes y - y \otimes x$$ for $$x, y \in V$$. $$S$$ is a left adjoint functor.

$$S(V)$$ is sort of like an exponential map, in my intuition. I have far reaching but informal reasons for saying this. For the space I'm given it's hard to develop the analogy, though maybe someone here can make this formal.

1) $$S$$ sends $$0$$ to $$k$$. (like how the exponential map sends $$0$$ to $$1$$.)

2) $$S$$ sends $$k$$ to $$k(x)$$. (like how the exponential map sends $$0$$ to $$e$$).

3) $$S$$ sends $$k^n$$ to $$k(x_1) \otimes_k \cdots \otimes_k k(x_n)$$. (like how the exponential map sends $$n$$ to $$e^n$$).

4) $$S$$ sends $$V \oplus W$$ to $$S(V) \otimes_k S(W)$$ (like how the exponential map sends $$a+b$$ to $$e^a \cdot e^b$$).

Now my question:

5) $$S$$ sends $$V \otimes_k W$$ to what? In other words, what is a natural choice of a functor $$k \text{-alg} \times k \text{-alg} \rightarrow k \text{-alg}$$ sends $$(S(V) ,S(W))$$ to $$S(V \otimes W)$$. This would sort of be analogous to $$(e^a, e^b) \mapsto e^{ab}$$. We know to define it on free algebras. Maybe that means we can define it on quotients of free algebras using this formula.

• If $\mathrm{dim}(V) = n$ and $\mathrm{dim}(W) = m$, then $S(V \otimes W)$ is a polynomial ring in $nm$ variables, because $\mathrm{dim}(V \otimes W) = nm$. – Nick Nov 21 '18 at 23:08
• Yes, but of course this is not the complete picture since not all $k$-algebras are free. i.e. we want a natural choice of functor whose domain is $k \text{-alg} \times k \text{-alg}$ and whose codomain is $k \text{-alg}$. – Dean Young Nov 21 '18 at 23:19