If $I\subseteq R$ is a uniserial ideal of $R$ with finite length and $R/I$ is uniserial of finite length, then $R$ is uniserial of finite length

Question

A module is called uniserial if the lattice of its submodules is a chain, i.e., the set of all its submodules is linearly ordered by inclusion. A ring is called right (resp. left) uniserial if it is a right (resp. left) uniserial module over itself, i.e., the lattice of right ideals is linearly ordered.

Prove that if $I\subseteq R$ is a uniserial ideal of $R$ with finite length and $R/I$ is uniserial of finite length, then $R$ is uniserial of finite length.

This is my attempt.

If $I$ is uniserial of finite length, then $I\supseteq IJ(I)\supseteq IJ(I)^2\supseteq\ldots \supseteq IJ(I)^n=0$ where $J(I)$ is the Jacobson radical of $I$ in $R$.

Similarly, $R/I:=\overline{R}\supseteq \overline{R}J(\overline{R})\supseteq \overline{R}J(\overline{R})^2\supseteq\ldots \supseteq \overline{R}J(\overline{R})^n=\overline{0}=I$ where $J(\overline{R})$ is the Jacobson radical of $R/I$.

Thus, we have $$\overline{R}\supseteq J(\overline{R})\supseteq J(\overline{R})^2\supseteq\ldots \supseteq J(\overline{R})^n=\overline{0}=I\supseteq J(I)\supseteq J(I)^2\supseteq\ldots \supseteq J(I)^n=0.$$ We also notice that the composition serries of both $I$ and $R/I$ are unique. So $J(I)\supseteq J(I)^2\supseteq\ldots \supseteq J(I)^n=0$ are the only ideals (submodules) in $I$ and $J(\overline{R})\supseteq J(\overline{R})^2\supseteq\ldots \supseteq J(\overline{R})^n=\overline{0}=I$ are the only ideals (submodules) of $R/I$.

Howeve, I am failing to deduce that $R$ has a finite chain like $R\supseteq J(R)\supseteq J(R)^2\supseteq\ldots \supseteq J(R)^n=0.$

Answer 1

However, I am failing to deduce that $$ has a finite chain like ...

Hm, well, that seems to be the easiest part. If $0=A_0\subseteq A_1\subseteq\ldots A_n=I$ is a composition series for $I$ and $I/I=A_n/I\subseteq A_{n+1}/I\subseteq\ldots\subseteq A_{n+m}/I=R/I$ is a composition series for $R/I$, then $0=A_0\subseteq\ldots \subseteq A_{n+m}=R$ is a composition series for $R$.

then is uniserial of finite length.

Finite length, yeah, but the claim that it's uniserial seems to be bogus.

Take $R=F_2\times F_2$ for example. The ideal $I=F_2\times\{0\}$ is uniserial (it's simple) and the quotient $R/I$ is uniserial (it's simple), but of course the lattice of ideals of $R$ is a square.