an interesting result from the course on rings and modules
I’m currently taking a course on Rings and Modules at UFMG, and this week I came across an interesting result that I’ve decided to share here.
Let \(R\) be a ring with unity. We wish to show that if \(R\) is commutative and if \(M\) is a simple \(R\)-module, then the ring of \(R\)-endomorphisms \(\text{End}_R(M)\) is a field.
We will first prove an auxiliary result that turns out to be quite useful.
Lemma 1: Let \(R\) be a unitary ring, and let \(M,N\) be two isomorphic \(R\)-modules. Then \(\text{End}_R(M)\) and \(\text{End}_R(N)\) are isomorphic as rings.
Proof:
Let \(\phi:M\mapsto N\) be the \(R\)-isomorphism between \(M\) and \(N\). Define \(\psi:\text{End}_R(M)\mapsto \text{End}_R(N)\), where \(\psi(f)(n) = \phi(f(\phi^{-1}(n)))\), for all \(f \in \text{End}_R(M),n \in N\). We claim that \(\psi\) is a ring isomorphism.
If \(f,g \in \text{End}_R(M)\), then
\[\psi(f+g)(n)=\phi((f+g)(\phi^{-1}(n)))\] \[=\phi(f(\phi^{-1}(n))+g(\phi^{-1}(n)))\] \[= \phi(f(\phi^{-1}(n)))+\phi(g(\phi^{-1}(n)))\] \[= \psi(f)(n) + \psi(g)(n),\forall n \in N\]since \(\phi\) is itself an endomorphism, and thus additive over \(M\).
Also
\[\psi(f\circ g)(n) = \phi(f(g(\phi^{-1}(n))))\] \[=\phi(f(\phi^{-1}(\phi(g(\phi^{-1}(n))))))\] \[=\phi(f(\phi^{-1}(\psi(g)(n))))\] \[=\psi(f)(\psi(g)(n)) = (\psi(f) \circ \psi(g))(n),\forall n \in N\]therefore \(\psi\) is a ring homomorphism.
If \(f \in \ker(\psi)\), we have that \(\forall n \in N: \psi(f)(n) = 0 \Rightarrow \phi(f(\phi^{-1}(n))) = 0\). Since \(\phi\) is a \(R\)-module isomorphism, this implies that \(\forall n \in N: f(\phi^{-1}(n)) = 0\), but again \(\phi\) is an isomorphism, thus its inverse is also surjective, which implies that \(\forall m \in M: f(m) = 0\), and so \(f = 0 \Rightarrow \ker(\psi)=\{0\}\), i.e., \(\psi\) is injective.
Given \(g \in \text{End}_R(N)\), define \(f(m) = \phi^{-1}(g(\phi(m)))\). Note that \(f\) is an endomorphism, since \(\phi\) and \(g\) both are. Also, note that
\[\psi(f)(n) = \phi(\phi^{-1}(g(\phi(\phi^{-1}(n))))) = g(n),\forall n \in N\]thus \(\psi(f) = g\), and so \(\psi\) is surjective. The previous points show that \(\psi\) is indeed a ring isomorphism, which concludes the proof.
\(\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\blacksquare\)
We now look at the ring of \(R\)-homomorphisms of a commutative unitary ring \(R\).
Lemma 2: Let \(R\) be a commutative ring with unity. The following statements hold:
Proof:
For the first statement, let \(f,g \in \text{End}_R(R)\), and note that
\[\phi(f+g)=(f+g)(1)=f(1)+g(1)=\phi(f)+\phi(g)\] \[\phi(f\circ g) = f(g(1)) = f(1\cdot g(1))=f(1)g(1)=\phi(f)\cdot\phi(g)\]thus \(\phi\) is a ring homomorphism. If \(f \in \ker(\phi)\), we have that \(f(1) = 0\), but note that \(\forall a \in R:f(a) = a\cdot f(1)\), since \(f\) is a \(R\)-homomorphism, thus if \(f(1)=0\), this implies that \(f = 0\), and so \(\phi\) is injective. Moreover, given any \(a \in R\), we can define \(f(1)=a\), and \(f(b)=b\cdot a\), which yields that \(\phi(f)=a\) and that \(f\) is a \(R\)-homomorphism. Therefore \(\phi\) is a ring isomorphism.
For the second statement, note that since \(R\) is commutative and \(I\) is a bilateral ideal of \(R\), \(R/I\) is a ring, and thus \(R/I\) can be viewed as an \(R/I\)-module. On the other hand, since \(I\) is an ideal and \(I\) is an \(R\)-module, \(R/I\) is also an \(R\)-module. Thus, if \(f \in \text{End}_{R/I}(R/I)\), note that
\[f(a \cdot (b+I))=f((a+I)(b+I)) = (a+I)f(b+I)=a\cdot f(b+I)\]thus \(f \in \text{End}_{R}(R/I)\). Conversely, if \(f \in \text{End}_{R}(R/I)\), we have that
\[f((a+I)(b+I))=f(ab+I)=f(a(b+I))=af(b+I)=(a+I)f(b+I)\]and so \(f \in \text{End}_{R}(R/I)\), implying that \(\text{End}_{R/I}(R/I) = \text{End}_{R}(R/I)\). Using this fact and the previous result, we conclude that \(R/I\) and \(\text{End}_{R}(R/I)\) are isomorphic rings.
\[\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\blacksquare\]Recall that an \(R\)-module \(M\) is called simple if it has no proper non-trivial submodules, i.e., its only submodules are \(\{0\}\) and \(M\). We know that this is equivalent to two things: first that \(M\) must be cyclic, and every non-zero element of \(M\) must be a generator, and second that \(M\) is \(R\)-isomorphic to \(R/I\), where \(I\) is some maximal left-ideal of \(R\). We are now ready to prove our main result.
Theorem: Let \(R\) be a commutative ring with unity, and \(M\) be a simple \(R\)-module. Then \(\text{End}_{R}(M)\) is a field.
Proof:
Since \(M\) is simple, it follows that it is \(R\)-isomorphic to \(R/I\), where \(I\) is a maximal left-ideal of \(R\), but since \(R\) is commutative, this implies that \(I\) is a maximal bilateral ideal. From Lemma 1, it follows that \(\text{End}_{R}(M)\) and \(\text{End}_{R}(R/I)\) are isomorphic rings, and from Lemma 2 it follows that \(\text{End}_{R}(R/I)\) is isomorphic to \(R/I\) as rings, so \(\text{End}_{R}(M)\) is isomorphic to \(R/I\) as rings. However, note that since \(I\) is a maximal bilateral ideal, \(R/I\) is a field, and thus \(\text{End}_{R}(M)\) is itself a field.
\[\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\blacksquare\]The proof of this result is quite simple due to it being mainly a series of straightforward algebraic manipulations. However, I still think that it portrays some interesting techniques.
Lemma 1 essentially provides a way of converting a module homomorphism between two modules to a ring homomorphism between two rings that are closely related to the original modules. This surely has limited applications, however in our case it is precisely what we needed: we start by assuming that \(M\) is simple, and thus it is isomorphic to \(R/I\) as a module, however due to the fact that \(I\) is maximal we also know that \(R/I\) is a field. We then need a way of connecting this module isomorphism to a ring isomorphism, and this is precisely what Lemma 1 does. Lemma 2 on the other hand is just a bundle of simple observations that follow from basic properties of unitary commutative rings, and so it itself isn’t of much interest, however combined with Lemma 1 it yields the desired result. As for the main theorem itself, it yields a more powerful version of Schur’s Lemma: function composition for simple modules over commutative rings is itself commutative.