Is $S(E\otimes_AB)\cong S(E)\otimes_AB$?

Question

Suppose A is a commutative ring, $E$ is an $A$-module, $B$ is an $A$-algebra, ${S}$ is the symmetric $A$-algebra functor.

I try to use universial property, where ${S}$ is the left adjoint of the forgetful functor from $A-alg$ to $A-mod$.

Suppose we have $E\otimes_AB\to C$, $C$ an $A-alg$, we have $E\to C$ thus $S(E)\to C$, but the map $B\to E\otimes_AB \to C$ may not be an algebra homomorphism, since the second one is only an $A-mod$ map. I don't know how to deal with it. Can anyone help me? Thanks

You have to begin with a commutative $B$-algebra $C$, not just $A$-algebra. After all you want to prove an isomorphism of commutative $B$-algebras. — Martin Brandenburg, Jan 07 '14 at 15:53
If we denote $S_A$ and $S_B$ the symmetric A/B algebra functor, so we have $S_B(E\otimes_AB)\cong S_A(E)\otimes_AB$? — , Jan 07 '14 at 16:03
First comment: Yes this is exactly what you have been asking. Second comment: No (try examples!). — Martin Brandenburg, Jan 07 '14 at 16:04

Martin Brandenburg · Accepted Answer · 2014-01-07T16:07:48.917

This is true by abstract nonsense, namely the general rule that the left adjoint of a composition is the composition (vice versa) of the left adjoints. Consider the composition of forgetful functors

$\mathsf{CAlg}(B) \to \mathsf{Mod}(B) \to \mathsf{Mod}(A).$

The left adjoint maps $E \in \mathsf{Mod}(A)$ to $S_B(E \otimes_A B)$.

But the composition can also be seen as

$\mathsf{CAlg}(B) \to \mathsf{CAlg}(A) \to \mathsf{Mod}(A).$

The left adjoint therefore maps $E \in \mathsf{Mod}(A)$ to $S_A(E) \otimes_A B$.

Left adjoints are unique, hence $S_B(E \otimes_A B) \cong S_A(E) \otimes_A B$.

Is $S(E\otimes_AB)\cong S(E)\otimes_AB$?

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