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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?

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up vote 3 down vote accepted

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.

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Thanks. I guess I have to study abstract category theory further to fully understand this answer, but it's still good to know the functor is a right adjoint. – Cihan Apr 13 '12 at 15:04
As I've indicated in the end, you don't need to know any theorems in order to construct $F$ (but they motivate it, as my answer shows): On free rings, we have $F(\mathbb{Z} \langle \{y_t\}_{t \in T} \rangle) = \mathbb{Z} \langle \{x_{ijt}\}_{1 \leq i,j \leq n, t \in T} \rangle$. In general, write a ring $R$ as the coequalizer of two maps $\mathbb{Z} \langle \{y_t\}_{t \in T} \rangle \to \mathbb{Z} \langle \{y_t\}_{t \in S} \rangle$ (this is a presentation of $R$ in terms of generators and relations), ... – Martin Brandenburg Apr 13 '12 at 15:15
... then $F(R)$ is is the coequalizer of two associated maps $\mathbb{Z} \langle \{x_{ijt}\}_{1 \leq i,j \leq n, t \in T} \rangle \to \mathbb{Z} \langle \{x_{ijt}\}_{1 \leq i,j \leq n, t \in S} \rangle$. – Martin Brandenburg Apr 13 '12 at 15:15

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