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.

  • $\begingroup$ If $\mathrm{dim}(V) = n$ and $\mathrm{dim}(W) = m$, then $S(V \otimes W)$ is a polynomial ring in $nm$ variables, because $\mathrm{dim}(V \otimes W) = nm$. $\endgroup$ – Nick Nov 21 '18 at 23:08
  • $\begingroup$ 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 $k \text{-alg}$. $\endgroup$ – Dean Young Nov 21 '18 at 23:19

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