TITLE: Existence proof of the tensor product using the Adjoint functor theorem.
QUESTION [9 upvotes]: Can one prove the existence of the tensor product by the adjoint functor theorem? (of, say, modules over a commutative ring) If yes, how would one check the SSC (solution set condition) for the hom functor?

REPLY [11 votes]: If $R$ is a ring, $M$ is a right $R$-module and $N$ is a left $R$-module, then the functor of balanced maps $F : \mathsf{Ab} \to \mathsf{Set}, A \mapsto \{\beta:  |M| \times |N| \to |A| \text{ balanced}\}$ satisfies the assumptions of Freyd's criterion for representability: It is easy to check that it preserves limits. For the solution set condition, let $\beta : |M| \times |N| \to |A|$ be a balanced map. Let $A'$ be the subgroup of $A$ which is generated by all elements of the form $\beta(m,n)$ with $m \in M, n \in N$. Then $\# |A'| \leq \aleph_0 \cdot \# |M| \cdot \# |N|$. Hence, up to isomorphism, there is only a set of such $A'$s. This provides the solution set, since $\beta$ factors as a balanced map $|M| \times |N| \to |A'|$ followed by $|A'| \hookrightarrow |A|$. 
Hence, $F$ is representable, which means that the tensor product $M \otimes_R N$ exists. In fact, all classical universal constructions of basic algebra can be obtained with Freyd's criterion for representability - without any effort.
Now if $R$ is commutative, then $M \otimes_R N$ is actually an $R$-module which represents the functor of bilinear maps $\mathsf{Mod}(R) \to \mathsf{Set}$ on $|M| \times |N|$. This functor is isomorphic to $\hom_R(M,\underline{\hom}_R(N,-))$, so that in fact $\hom_R(M \otimes_R N,T) \cong \hom_R(M,\underline{\hom}_R(N,T))$, i.e. $- \otimes_R N$ is left adjoint to $\underline{\hom}_R(N,-)$. Now your question was actually how to use Freyd's adjoint functor theorem to obtain $- \otimes_R N$ this way:
Again it is clear that $\underline{\hom}_R(N,-) : \mathsf{Mod}(R) \to \mathsf{Mod}(R)$  preserves limits, so that we only have to verify the solution set condition. But this comes down to the proof above: If $M \to \underline{\hom}_R(N,T)$ is an $R$-linear map, it corresponds to an $R$-bilinear map $|M| \times |N| \to |T|$, hence factors through $|T'|$ for some submodule $T'$ whose cardinality can be bounded depending only on $M,N,R$.