Exterior product respects base change of modules This is a homework assignment so I am only looking for a hint. Suppose that $f: R \rightarrow S$ is a homomorphism of commutative rings, and $M$ is an $R$-module. I am tasked with establishing an isomorphism of $S$-modules
$$\bigwedge^k\, (M\otimes_R S) \cong \bigl(\bigwedge^k M\bigr)\otimes_R S$$
I would like to use the most "concrete" approach possible, so I am trying to define two $S$-linear maps that are inverses of each. In one direction I defined $\psi$ to be the $R$-linear extension of the assignment on basic wedges,
$$\wedge_i (m_i \otimes s_i) \mapsto \wedge_i m_i \otimes \left(\prod s_i\right)$$
I would need to show that this gives a well-defined map, for which I'd like to show that elements $\wedge_i (m_i \otimes s_i)$ with $(m_j \otimes s_j) = (m_l \otimes s_l)$ for some $j\neq l$ are mapped to zero. In particular I'd like to show that $\psi$ takes such an element to an element $\wedge_i m_i \otimes s'$ with $m_j = m_l$. I don't know how to show this, so now I'm questioning if I picked the right definition for $\psi$.
Note: there's a related unanswered question on this site, which is looking for a hint given an approach using exact sequences/category theory, and this question is looking for a "concrete" treatment of this question.
 A: Suppose that $\omega = \wedge_i(m_i \otimes s_i)$ is a simple tensor which is 0. We can rewrite $\omega$ as $\omega=(\prod_i s_i)(\wedge_i m_i \otimes 1_S)$. This is zero in $\wedge (M \otimes_R S)$ if and only if either $(\prod_i s_i) =0$ in $S$ or $\exists j\neq l$ with $m_j \otimes 1_S = m_l \otimes 1_S$ (i.e. $m_j = m_l$).
In both cases the map from above $\psi: \wedge^k (M \otimes_R S) \rightarrow (\wedge^k M) \otimes_R S$ clearly contains $\omega$ in its kernel. Thus $\psi$ is well-defined.
Conversely, suppose that $u = \wedge_i m_i \otimes \prod s_i = 0$ in $(\wedge^k M) \otimes_R S$. THen either $\prod_is_i = 0$ or $\exists j\neq l$ with $m_j = m_l$. As before it is easy to see that the inverse map $\phi: (\wedge^k M) \otimes_R S \rightarrow \wedge^k (M \otimes_R S)$  contains $u$ in its kernel in both cases. Thus the isomorphism of $(M \otimes_R S)^{\otimes k} \simeq (M^{\otimes k})\otimes_R S$ "descends" to give the required isomorphism with the exterior power in place of the tensor power.
