I was reading the "Undergraduate Commutative Algebra". It formalises the definition of Module.
Consider M, an A-module where A is a ring.
It defines $\mu_f : M \to M$ for the map $m \mapsto fm $, where $f \in A$.
Then the text claims that $f \mapsto \mu_f$ is a ring homomorphism $A \to \operatorname{End}(M)$ from A to the noncommutative ring of endomorphisms of M.
So, am I correct to think that in this case $\operatorname{End}(M)$ is a noncommutative ring because $A$ is not commutative?
$\operatorname{End}(M)$ seems to commutative if $A$ is commutative.
–update
Sorry, I made a mistake. I was thinking about $\operatorname{End}(M)$.
Best Answer
No, not necessarily. There isn’t a connection.
You can have $\operatorname{End}(M)$ noncommutative and $A$ commutative ($A=\mathbb Z$ and $M=C_2\times C_2$)
You can also have $A$ noncommutative and $\operatorname{End}(M)$ commutative (for this you can take a ring $A$ which isn’t commutative, but which has a unique maximal right ideal $I$ such that $A/I$ is commutative, and let $M=A/I$.)