Give an example of a ring with the right minimum condition, but not the left. Can you find such an example with no nonzero nilpotent ideals? Why?
Solution
1. Let \(K\subset L\) be a field extension with \([L:K]=\infty\), and set\[ R= \begin{pmatrix} K & L\\ 0 & L \end{pmatrix}. \]Then \(R\) satisfies the minimum condition on right ideals.
Let \[ e_1= \begin{pmatrix} 1&0\\ 0&0 \end{pmatrix}, \qquad e_2= \begin{pmatrix} 0&0\\ 0&1 \end{pmatrix}. \] As a right \(R\)-module, \[ R_R=e_1R\oplus e_2R. \] The module \(e_2R\) is simple: via \[ \begin{pmatrix} 0&0\\ 0&\ell \end{pmatrix} \longleftrightarrow \ell, \] its right action factors through the lower-right copy of the field \(L\), so it is one-dimensional over \(L\).
Inside \(e_1R\), let \[ N= \left\{ \begin{pmatrix} 0&m\\ 0&0 \end{pmatrix}:m\in L \right\}. \] This is a simple right \(R\)-submodule, again one-dimensional over the lower-right copy of \(L\). Moreover, \[ e_1R/N\cong K \] as a right module through the upper-left copy of \(K\), hence is simple. Thus \(e_1R\) has composition length \(2\), while \(e_2R\) has composition length \(1\). Therefore \(R_R\) has finite length, so it is Artinian. Equivalently, \(R\) satisfies the minimum condition on right ideals.
2. The ring \(R\) does not satisfy the minimum condition on left ideals.
For every \(K\)-subspace \(U\subseteq L\), define \[ I_U= \left\{ \begin{pmatrix} 0&u\\ 0&0 \end{pmatrix}:u\in U \right\}. \] For \[ r= \begin{pmatrix} k&m\\0&\ell\end{pmatrix} \quad\text{and}\quad x= \begin{pmatrix}0&u\\0&0\end{pmatrix}, \] we have \[ rx= \begin{pmatrix}0&ku\\0&0\end{pmatrix}. \] Hence \(I_U\) is a left ideal precisely because \(U\) is a \(K\)-subspace.
Since \([L:K]=\infty\), choose \(K\)-linearly independent elements \[ u_1,u_2,u_3,\ldots\in L. \] Set \[ U_n=\operatorname{span}_K\{u_n,u_{n+1},u_{n+2},\ldots\}. \] Then \[ U_1\supsetneq U_2\supsetneq U_3\supsetneq\cdots, \] so \[ I_{U_1}\supsetneq I_{U_2}\supsetneq I_{U_3}\supsetneq\cdots \] is an infinite strictly descending chain of left ideals. Thus \(R\) is not left Artinian.
3. This example necessarily has a nonzero nilpotent ideal.
The ideal \[ N= \left\{ \begin{pmatrix} 0&m\\ 0&0 \end{pmatrix}:m\in L \right\} \] is nonzero and satisfies \[ N^2=0. \]
4. There is no example with the right minimum condition, not the left minimum condition, and no nonzero nilpotent ideals.
Suppose \(A\) is right Artinian and has no nonzero nilpotent ideals. For every right-Artinian ring, the Jacobson radical \(J(A)\) is nilpotent. Hence the hypothesis forces \[ J(A)=0. \] Also, a right-Artinian ring modulo its Jacobson radical is semisimple Artinian. Therefore \[ A=A/J(A) \] is semisimple Artinian. By the Wedderburn–Artin theorem, a semisimple Artinian ring is a finite product of full matrix rings over division rings, and in particular is Artinian on both the right and the left. Thus a right-Artinian ring with no nonzero nilpotent ideals must also be left Artinian, so the requested stronger example cannot exist.