# End(M) for cyclic R-module M is a commutative ring where R is a PID.

Show that End(M) for cyclic R-module M is a commutative ring where R is a PID.

So I know the module M is an abelian group so by a previous result, End(M) is a ring under pointwise addition and function composition.

What remains to show is that the function composition is commutative. To that end, let $$\phi, \alpha \in End(M)$$. We want to show the ring End(M) is commutative so we must show $$\phi(\alpha(m)) = \alpha(\phi(m)), \forall m \in M$$.

We are given M is a cyclic R-module (R is a PID) so $$M = Rx$$ for some $$x \in M$$. Then for any $$m \in M$$ we have that $$m = rx$$ for some $$r \in R$$.

Then, $$\phi(\alpha(m)) = \phi(\alpha(rx)) = \phi(r\alpha(x)) = r\phi(\alpha(x))$$

Likewise, $$\alpha(\phi(m)) = r\alpha(\phi(x))$$

So, we just need to show $$\alpha(\phi(x)) = \phi(\alpha(x))$$ but I'm not sure what to do from here? Any hints?

Notice $$\phi(x)=r_\phi x$$ for some $$r_\phi\in R$$, since $$\phi(x)\in M=Rx$$; similarly $$\alpha(x)=r_\alpha x$$ for some $$r_\alpha\in R$$. Then we can calculate $$\alpha(\phi(x))=\alpha(r_\phi x)=r_\phi\,\alpha(x)=r_\phi(r_\alpha x)=r_\alpha(r_\phi x)=r_\alpha\,\phi(x)=\phi(r_\alpha x)=\phi(\alpha(x))$$

• How can we say that $r_{\alpha} (r_{\phi}x) = r_{\phi} (r_{\alpha}x)$? What justifies this? – jmac Dec 19 '19 at 5:09
• commutativity of $R$ and the "associativity" axiom for a module; in more detail $r_\alpha(r_\phi x)=(r_\alpha r_\phi)x=(r_\phi r_\alpha)x=r_\phi(r_\alpha x)$ – Alex Mathers Dec 19 '19 at 5:10
• Ah yes, I totally blanked that R was commutative. That makes sense. – jmac Dec 19 '19 at 5:12

Since $$M$$ is cyclic, there exists a surjective homomorphism of $$R$$-modules $$\psi:R\rightarrow M$$.

The kernel of $$\psi$$ is an ideal of $$R$$, call it $$I$$.

We then have $$M \simeq R/I$$ as $$R$$-modules. Hence $$\operatorname{End}_R(M)$$ is isomorphic to $$\operatorname{End}_R(R/I)$$.

However, an $$R$$-linear endomorphism of $$R/I$$ is the same as an $$R/I$$-linear endomorphism of $$R/I$$, hence we have $$\operatorname{End}_R(R/I) = \operatorname{End}_{R/I}(R/I)$$.

But for any commutative ring $$S$$, the endomorphism ring $$\operatorname{End}_S(S)$$ is canonically isomorphic to $$S$$ itself: we have maps $$u:\operatorname{End}_S(S)\rightarrow S$$ sending any $$\phi$$ to $$\phi(1)$$, and $$v:S\rightarrow \operatorname{End}_S(S)$$ sending any $$s$$ to the endomorphism $$x\mapsto sx$$.

In particular, $$\operatorname{End}_{R/I}(R/I) \simeq R/I$$ is a commutative ring.

The condition "$$R$$ is PID" is not needed.