Is the functor $R \mapsto \mathbb{M}_n(R)$ a right adjoint? Given a positive integer $n$, there is a functor $F: \mathsf{Ring} \rightarrow \mathsf{Ring}$ such that $F(R) = \mathbb{M}_n(R)$ on objects and the action of $F$ on morphisms are given entrywise. Is $F$ a right adjoint?
 A: Theorem 1.66 in Locally presentable categories by Adamek-Rosicky states that a functor between locally presentable categories is a right adjoint iff it preserves limits and $\lambda$-directed colimits for some regular cardinal. This is a consequence of Freyd's Adjoint Functor Theorem (which you can find in Mac Lane, V.6). We can apply this here with $\lambda=\aleph_0$:
The forgetful functor $U : \mathsf{Ring} \to \mathsf{Set}$ reflects (and also creates) limits and directed colimits. Thus it is enough to show that $U \circ M_n : \mathsf{Ring} \to \mathsf{Set}$ is a right adjoint. But this functor is even representable, namely by the (non-commutative) polynomial ring $P=\mathbb{Z} \langle \{x_{ij}\}_{1 \leq i,j \leq n} \rangle$, so that its left adjoint is given by the copower $T \mapsto \coprod_T P = \mathbb{Z} \langle \{x_{ijt}\}_{1 \leq i,j \leq n, t \in T} \rangle$.
So what is the left adjoint $F$ of $M_n$ explicitly? The left adjoint of $U$ maps a set $T$ to $\mathbb{Z} \langle \{y_t\}_{t \in T} \rangle$. It follows $F(\mathbb{Z} \langle \{y_t\}_{t \in T} \rangle) = \mathbb{Z} \langle \{x_{ijt}\}_{1 \leq i,j \leq n, t \in T} \rangle$. This enables us to compute $F$ of free rings. Since $F$ preserves colimits, it even shows how $F(R)$ looks like for an arbitrary ring $R$ when we represent $R$ in terms of generators and relations. This also allows us to argue directly - without all those general theorems - that $F$ exists at all.
I wonder if there is any description of $F(R)$ which doesn't need a presentation of $R$, but with tensor products and direct sums over $\mathbb{Z}$ allowed.
