# Endomorphisms of a projective (non-free) module

I've confused myself about endomorphisms of a non-free module. Here's the situation:

Let $$R$$ be a non-commutative ring, let $$M\subset R^2$$ be a projective left $$R$$-module which has two generators over $$R$$, say $$e_1,e_2\in R^2$$. Then it seems that any $$R$$-linear endomorphism $$\varphi:M\to M$$ is determined by where we send these two generators, since all elements of $$M$$ are of the form $$ae_1+be_2$$, and we have that $$\varphi(ae_1+be_2)=a\varphi(e_1)+b\varphi(e_2)$$. Then any endomorphism is of the form $$\varphi(e_1)=a_1e_1+a_3e_2$$, and $$\varphi(e_2)=a_2e_1+a_4e_2$$, which we can write as a matrix if we want: $$\begin{bmatrix}a_1&a_2\\a_3&a_4\end{bmatrix}\in M_2(R).$$

But then it seems that $$\text{End}_R(M)\cong M_2(R)\cong \text{End}_R(R^2)$$? This seems strange to me, so perhaps I've made an error somewhere? As a semi-related question, is the determinant of such a matrix defined the same way for this non-commutative $$R$$, i.e. $$a_1a_4-a_2a_3$$?

• Not every choice of $a_1,a_2,a_3,a_4$ yields a well-defined map, unless $M$ is free. – egreg Jan 21 at 23:00

Take your favorite ring $$A$$ and consider $$R=A\times A\times A$$. As projective module take $$M=A\times A\times\{0\}$$, with generators $$e_1=(1,0,0)$$ and $$e_2=(0,1,0)$$.
There is no endomorphism of $$M$$ sending $$e_1$$ to $$e_2$$, for instance. Indeed, if $$\varphi(e_1)=e_2$$, we have $$\varphi(e_1)=\varphi(e_1e_1)=e_1\varphi(e_1)=e_1e_2=0$$, a contradiction.
Actually, the endomorphism ring of $$M$$ is $$A^\mathrm{op}\times A^\mathrm{op}$$, which is very different from $$M_2(R^\mathrm{op})$$.
The problem in your reasoning is that you can't arbitrarily choose $$a_1,a_2,a_3,a_4$$. This is possible if $$M$$ is free with basis $$\{e_1,e_2\}$$.
Every endomorphism can be expressed as a matrix in the way you say. However, that hardly defines an isomorphism $$\text{End}_R(M)\cong M_2(R)$$. Indeed, it doesn't even define a function $$\text{End}_R(M)\to M_2(R)$$ (or $$M_2(R)\to\text{End}_R(M)$$). The representation of an endomorphism as a matrix is typically not unique, since the $$a_1$$ and $$a_3$$ such that $$\varphi(e_1)=a_1e_1+a_3e_2$$ are typically not unique (and similarly for $$a_2$$ and $$a_4$$). Moreover, there is no reason to think that an arbitrary matrix actually gives a well-defined endomorphism of $$M$$. At best, what you get is a surjective homomorphism from a subring of $$M_2(R^{op})$$ (those matrices that do give well-defined homomorphisms) to $$\text{End}_R(M)$$. (Note also that I wrote $$R^{op}$$ because composition of endomorphisms corresponds to matrix multiplication where you use the ring structure on $$R$$ with multiplication in reversed order; see Noncommutative rings, matrices and homomorphisms of free modules for more on this.)