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.
Best Answer
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 $M_n(R)$.