Let \(A\) be a simple ring with identity element. Show that if \(A\) has a minimal right ideal, then \(A\) satisfies the minimum condition for right ideals.
Solution
1. Let \(I\) be a minimal nonzero right ideal of \(A\), and let \(S\) be the sum of all right ideals of \(A\) that are isomorphic to \(I\) as right \(A\)-modules.Then \(S\) is a nonzero two-sided ideal of \(A\).
By construction, \(S\) is a right ideal and \(I\subseteq S\), so \(S\ne0\).
Let \(J\subseteq A\) be a right ideal isomorphic to \(I\), and let \(a\in A\). Left multiplication by \(a\) defines a homomorphism of right \(A\)-modules \[ \lambda_a:J\longrightarrow A, \qquad x\longmapsto ax. \] Since \(J\) is simple, \(\ker\lambda_a\) is either \(J\) or \(0\). Thus \(aJ=0\), or else \(aJ\) is a simple right ideal isomorphic to \(J\), hence to \(I\). In either case \(aJ\subseteq S\). Because \(S\) is the sum of all such \(J\), this shows \(aS\subseteq S\) for every \(a\in A\). Hence \(S\) is also a left ideal, so it is two-sided.
2. The right \(A\)-module \(A_A\) is a sum of simple submodules.
The ring \(A\) is simple and \(S\) from 1 is a nonzero two-sided ideal. Therefore \[ S=A. \] Every summand used to define \(S\) is a minimal right ideal, hence a simple right \(A\)-module. Thus \(A_A\) is a sum of simple submodules.
3. A module that is a sum of simple submodules is a direct sum of simple submodules.
Let \(M\) be a sum of simple submodules. Choose, by Zorn’s lemma, a maximal family \(\{M_\lambda\}_{\lambda\in\Lambda}\) of simple submodules whose sum is direct, and put \[ N=\bigoplus_{\lambda\in\Lambda}M_\lambda. \] Suppose \(N\ne M\). Because \(M\) is a sum of simple submodules, there is a simple submodule \(T\subseteq M\) with \(T\nsubseteq N\). Now \(T\cap N\) is a submodule of the simple module \(T\), so it is either \(0\) or \(T\). It cannot be \(T\), because \(T\nsubseteq N\). Hence \(T\cap N=0\), and therefore \[ N\oplus T \] is a strictly larger direct sum of simple submodules, contradicting maximality. Thus \(N=M\).
4. The module \(A_A\) is a finite direct sum of simple right ideals.
By 2 and 3, write \[ A=\bigoplus_{\lambda\in\Lambda} I_\lambda, \] where every \(I_\lambda\) is a simple right ideal. Because an element of a direct sum has finite support, there is a finite set \(F\subseteq\Lambda\) such that \[ 1\in\bigoplus_{\lambda\in F}I_\lambda. \] The finite sum on the right is a right ideal, so \[ A=1A\subseteq\bigoplus_{\lambda\in F}I_\lambda\subseteq A. \] Hence \[ A=\bigoplus_{\lambda\in F}I_\lambda, \] a finite direct sum of simple right \(A\)-modules.
5. Therefore \(A\) satisfies the minimum condition for right ideals.
A simple module is Artinian: its only submodules are \(0\) and itself. A finite direct sum of Artinian modules is Artinian, so 4 implies that the right module \(A_A\) is Artinian. The submodules of \(A_A\) are exactly the right ideals of \(A\). Therefore every descending chain of right ideals stabilizes; equivalently, \(A\) satisfies the minimum condition for right ideals.