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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.

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Sorry, I made a mistake. I was thinking about $\operatorname{End}(M)$.

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    $\begingroup$ I think you meant to say that $f \mapsto \mu_f$ is a homomorphism to $\operatorname{End}_A(M)$, so that your question should really be about $\operatorname{End}_A(M)$ rather than $\operatorname{End}_A(A)$. Am I correct, or did you actually mean to ask a question about $\operatorname{End}_A(A)$? $\endgroup$ – Omnomnomnom Dec 15 '19 at 13:09
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    $\begingroup$ This seems to rather define a ring homomorphism $A\to\operatorname{End}_A(M)$. Do they specialize to $M=A$ (in which case $\operatorname{End}_A(A)$, the $A$-module endomorphisms of $A$, is still not the same as $\operatorname{End}(A)$, the ring endomorphisms of $A$. $\endgroup$ – Hagen von Eitzen Dec 15 '19 at 13:10
  • $\begingroup$ Sorry, I made a typo. I was considering the case about $End M$ $\endgroup$ – roro Dec 15 '19 at 13:21
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So, am I correct to think that in this case $End M$ is a noncommutative ring because $A$ is not commutative?

No, not necessarily. There isn’t a connection.

You can have $End(M)$ noncommutative and $A$ commutative ($A=\mathbb Z$ and $M=C_2\times C_2$)

You can also have $A$ noncommutative and $End(M)$ 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 $M=A/I$.)

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  • $\begingroup$ In this particular case, if we consider $R = \{ \mu_f : f \in A \}$, and if we consider the multiplication in $R$ as composition of function, then $\mu_f$ and $\mu_g$ commute if A is a commutative ring. So could we consider $R$ in this case as a commutative ring? $\endgroup$ – roro Dec 15 '19 at 14:50
  • $\begingroup$ @roro yes, if $A$ is commutative, its homomorphic image will be commutative... but $End(M)$ may still be not commutative. $\endgroup$ – rschwieb Dec 15 '19 at 15:46
  • $\begingroup$ That makes sense. I think I misunderstand the text. I do understand now that $End (M)$ is in general not commutative. Thanks. $\endgroup$ – roro Dec 15 '19 at 19:49
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    $\begingroup$ @roro The phrase “noncommutative ring” should be interpreted as “not necessarily commutative”, not “not commutative”. Most results in noncommutative ring theory are stated so that they apply to cases of commutative rings as well. $\endgroup$ – rschwieb Dec 15 '19 at 19:52
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It is true that $f \mapsto \mu_f$ defines a ring homomorphism $$\begin{align*}\mu : A &\to \operatorname{End}_{\mathbb Z}(M) \\ f & \mapsto \mu_f \end{align*}$$ Thus when $A$ is noncommutative and $\mu$ is injective, $\operatorname{End}_{\mathbb Z}(M)$ is noncommutative. The point is that $\mu$ is not necessarily injective.

The module $M$ is called "faithful" when $\mu$ is injective. So, a priori, there could exist examples of non-faithful modules over noncommutative rings with commutative endomorphism ring.

A silly example is the following: take any ring $A'$ and an $A'$-module $M'$ with commutative endomorphism ring. Take any noncommutative ring $B$. Define $A = A' \times B$ and define the $A$-module structure on $M$ by restricting scalars through the projection $A' \times B \to A'$. Then $A$ is noncommutative, even though $\operatorname{End}_{\mathbb Z}(M)$ is commutative.

Concretely, take e.g. $A = \mathbb Z \times M_2(\mathbb Z)$ and $M = \mathbb Z$, where $(a, b) \in A$ acts on $n \in M$ by $(a, b)n = an$. Then $\operatorname{End}_{\mathbb Z}(M) \cong \mathbb Z$ is commutative.


Conversely, for modules with noncommutative endomorphism rings over commutative rings: There are examples which you've certainly encountered before: let $k$ be a field, and $M$ an $n$-dimensional $k$-vector space. The endomorphism ring $\operatorname{End}_{\mathbb Z}(M)$ contains the ring $\operatorname{End}_{k}(M) \cong M_n(k)$. So if $\operatorname{End}_{\mathbb Z}(M)$ is commutative, so is the subring $M_n(k)$. However, the latter is not commutative for $n > 1$.

Thus $\operatorname{End}_{\mathbb Z}(M)$ is not commutative for $n > 1$, even though $k$ is commutative.

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You can start by proving that:

$$ \mu_{g}\circ\mu_{f} = \mu_{gf},\ \forall\, f,g\in A $$

and use that $M$ is a $A$-module to argue that:

$$ (a_{1}a_{2})m=a_{1}(a_{2}m),\ \forall\,m\in M,\forall\, a_{1},a_{2}\in A$$

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For a specific counterexample, take $k$ to be any field, $A = k\{x,y\}$ (polynomials over $k$ in two non-commuting variables), and $M = k$, considered as an $A$-module, where for $a \in k$ and $p \in k\{x,y\}$, we take $pa$ to be $ap(0,0)$. Then $A$ is certainly non-commutative, but you can show $\mathrm{End}(M) \cong k$, which is commutative.

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