I am requesting help with the following problem.
Below all rings are commutative with unit, and for a ring $R$, we define an $R$-algebra to be a ring $R'$ with ring homomorphism $f: R \to R'$. This induces a $R$-module structure on $R'$ by defining scalar multiplication as $ab := f(a)b$.
Let $R$ be ring, $A$ an $R$-module, and $S$ a $R$-algebra. For every positive integer $k$ construct an isomorphism $(\bigwedge^{k} A) \otimes_{R} S = \bigwedge^{k} (A \otimes_{R} S)$.
I know that I should use the universal properties of the tensor and exterior products. But I am stuck.
The notation below will look as though I'm working with left modules instead of right, but in fact, in working in the commutative ring case, every left module $M$ can be rendered as a right module by the definition $mr = rm$, and in fact in this way all (left) modules can be rendered as bimodules.
The following is a somewhat "high-level" categorical explanation.
To prove this, it suffices to observe that the functor
$$S \otimes_R -: R\textbf{-Mod} \to S\textbf{-Mod}$$ preserves colimits and is a symmetric monoidal functor. The latter means that the canonical map
$$(S \otimes_R M) \otimes_S (S \otimes_R N) \to S \otimes_R (M \otimes_R N)$$ defined by the assignment $$(s \otimes m) \otimes (t \otimes n) \mapsto st \otimes (m \otimes n)$$ is an isomorphism that is suitably compatible with the associativity and symmetry isomorphisms pertaining to the tensor products $\otimes_R$ and $\otimes_S$ (see the nLab, starting here and following links as needed). The inverse is defined by $s \otimes (m \otimes n) \mapsto (s \otimes m) \otimes (1_S \otimes n)$, where $1_S$ denotes the multiplicative identity of $S$. It is well-known that $S \otimes_R -$ preserves colimits, because it is left adjoint (see the discussion here) to the functor
$$S\textbf{-Mod} \to R\textbf{-Mod}$$ which takes an $S$-module $A$ to the $R$-module with scalar multiplication given by the definition $r \cdot a = f(r)a$, where $f: R \to S$ is the given ring homomorphism that gives $S$ its $R$-algebra structure.
The reason this suffices is that the exterior power $\Lambda^k M$ is expressed directly in terms of colimits and tensor products with their symmetric monoidal structure. Consider for example how $\Lambda^k$ is constructed in $R\textbf{-Mod}$: it is a coequalizer of two maps
$$\bigoplus_{\sigma \in S_k} M^{\otimes_R k} \rightrightarrows M^{\otimes_R k}$$ where the first map, evaluated at the summand indexed by a group element $\sigma$, is the map $M^{\otimes_R k} \to M^{\otimes_R k}$ obtained by permuting tensor factors according to the permutation $\sigma$, and the second map evaluated at the same summand is multiplication by $\mathrm{sign}(\sigma)$.