# Noncommutative rings, matrices and homomorphisms of free modules

In Blyth's book "Module Theory: An Approach to Linear Algebra" matrix theory is developed generally over a noncommutative ring $$R$$ with $$1$$. However, it seems that there is a mistake in way that an important property doesn't carry over to noncommutative rings, namely the isomorphic of a ring of $$n\times n$$ matrices with coefficients in $$R$$ and the endomorphism ring of a free module over $$R$$

Given an a homomorphism $$\phi\colon M\to N$$ of free $$R$$-modules with respective bases $$(a_i)_m$$ and $$(b_i)_n$$, the matrix $$\mathrm{Mat}(\phi,(b_i)_n,(a_i)_m)$$ of this homomorphism with respect to said bases is a matrix $$(r_{ij})$$ such that $$r_{ij}$$ is the unique element of $$R$$ so that $$\phi(a_i) = s_{1i}b_1 + ... + r_{ij}b_j + ... + s_{ni}$$.

One can prove that, for the respective bases, there is an isomorphism $$\vartheta\colon\mathrm{Hom}_R(M,N)\to\mathrm{Mat}_{n\times n}(R), \phi \mapsto \mathrm{Mat}(\phi,(b_i)_n,(a_i)_m)$$.

But am I right to assume that this is not a ring homomorphism? It seems modules over noncommutative rings lack the multiplicative property $$\mathrm{Mat}(\psi\circ\phi, (c_i)_p,(a_i)_n) = \mathrm{Mat}(\psi,(c_i)_p,(b_i)_m)\mathrm{Mat}(\phi,(b_i)_m,(a_i)_n)$$ for free modules $$M,N,P$$ with respective bases $$(a_i)_n, (b_i)_m$$ and $$(c_i)_p$$ and their homomorphisms $$\phi\colon M\to N, \psi\colon N\to P$$.

Am I right or there is a mistake there? I was doing all the matrix theory for commutative rings before now, but encountered a proof in the aforementioned book which uses (probably wrong) ring isomorphism.

Let $$\mathrm{Mat}(\phi,(b_i)_m,(a_i)_n) = (r_{ij})$$ and $$\mathrm{Mat}(\psi,(c_i)_p, (b_i)_m) = (s_{ij})$$. Then we have $$\mathrm{Mat}(\psi,(c_i)_p, (b_i)_m)\mathrm{Mat}(\phi,(b_i)_m,(a_i)_n) = (t_{ik})$$ where $$t_{ik} = \sum_{j = 1}^m s_{ij}r_{jk}$$. Also, $$(\psi\circ\phi)(a_i) = \psi\left(\sum_{i = 1}^n r_{ij}b_i\right) \\ = \sum_{i = 1}^n r_{ij}\psi(b_i) \\ = \sum_{i = 1}^n r_{ij}\left(\sum_{k = 1}^p s_{ki}c_k\right) \\ = \sum_{k = 1}^p \left(\sum_{i = 1}^n r_{ij}s_{ki}\right)c_k \\ \neq \sum_{k = 1}^p \left(\sum_{i = 1}^n s_{ki}r_{ij}\right)c_k$$ generally.

You're right. The correct statement is that composition of maps between free (left) $$R$$-modules corresponds to multiplication of matrices with entries in $$R^{op}$$, i.e. the ring obtained by reversing the order of multiplication in $$R$$. The most basic case of this is $$1\times 1$$ matrices, where you find that the endomorphism ring of $$R$$ as a left $$R$$-module is $$R^{op}$$, not $$R$$, because an endomorphism is given by right multiplication by an element of $$R$$ which reverses the relationship between multiplication and composition. Similarly, the endomorphism ring of $$R^n$$ as a left $$R$$-module is $$M_n(R^{op})$$, not $$M_n(R)$$.