Semiprimary rings: Sharpest bound for chain lengths of principal ideals A ring with unity $A$ is called semiprimary, if $\mathfrak r:=\mathrm{Jac}(A)$ is nilpotent and ${A}/{\mathfrak r}$ is semisimple (Artinian). I'm trying to find a proof (or counterexample) for:

When $A$ is semiprimary and $\mathfrak r^n=(0)$ while ${A}/{\mathfrak r}$ is of length $l$ as an $A$-module, then every sequence of principal left (or right) ideals in $A$ has at most $ln$ proper inclusions.

I stumbled upon this, while studying characterizations of semiprimary (and other families of) rings. I found the statement in the paper [1 (Björk), Section 0], but Björk claims it without proof. For commutative $A$, it's definitely true (see proof below), but I can't get a hold on the noncommutative case. If it's true, it holds in particular for one-sided Artinian rings - it might be easier to tackle those first, but I'm not sure if this actually simplifies the problem.
I'd be happy to get any help in proving the sharp bound or any idea for a counterexample.
Edit: Jeremy solved it by giving a nice counterexample in his answer below. As a follow-up question: Does anyone know the sharpest general bound $b(l,n)$ in the noncommutative case? End of edit
There is a weaker qualitative version:

A ring is semiprimary if and only if there is an upper bound for the lengths of proper chains of principal left (or right) ideals.

Up to now, I only found one other place where this is discussed in the literature: [2 (Rowen's book), Theorem 2.7.7]. Rowen gives a proof (see below for a sketch) of the qualitative characterization with the weaker general bound $l^{n+1}-1$ (when $l>1$). I think, one can actually obtain $l+l^2+\ldots+l^n$ from his proof, but that's still very far from Björk's bound.
Btw one could deduce a more general result as a corollary:

For an arbitrary $A$-module, any chain of submodules all having at most $r$ generators, has at most $b(lr,n)$ proper inclusions.

Sketch of proof: Such a proper chain can be lifted to a proper chain of $r$-generated submodules of the left (or right) module $A^r$ and this corresponds to a proper chain of principal left (or right) ideals of the matrix ring $C:=M_r(A)$. Now, $C$ is also semiprimary with $(\mathrm{Jac}(C))^n=(0)$ while ${C}/{\mathrm{Jac}(C)}$ has length $lr$ over $C$, so we're done by the $r=1$ case.

Coming back to Björk's claim about principal ideals, here's what I've achieved / tried so far:
Proof for the cases $n=1$ and $l=1$, respectively
$n=1$ is trivial. $l=1$ means $(A,\mathfrak r)$ is local, so if $\mathfrak a_0\subsetneq\ldots\subsetneq\mathfrak a_m$ is a proper chain of principal left ideals in $A$, then $\mathfrak a_{m-1}\subseteq\mathfrak r$, hence $\mathfrak a_{m-1}$ is an ${A}/{\mathfrak r^{n-1}}$-module with a proper chain of cyclic submodules of length $m-1$. This lifts to a proper chain of principal left ideals in ${A}/{\mathfrak r^{n-1}}$ of length $m-1$. Using induction, $m-1\leq n-1$, so $m\leq n$.
Proof for the commutative case
If $A$ is commutative, then $A\simeq A_1\times\ldots\times A_l$ with local rings $(A_i,\mathfrak m_i)$ with $\mathfrak m_i^n=(0)$ (the smallest exponent could actually be $n_i\leq n$ for some of the $\mathfrak m_i$) and the chain of principal ideals in $A$ gives corresponding chains of principal ideals in each $A_i$ (multiply with the $i$-th idempotent $e_i$). The chains in the $A_i$ have at most $n$ (even $n_i$) proper inclusions each, by the $l=1$ case. Hence, the original chain in $A$ can have at most $ln$ (even $n_1+\ldots+n_l$) proper inclusions.
What I've tried in the noncommutative case
Relation to perfect rings
If $\mathfrak r$ is nilpotent, it's especially $T$-nilpotent (on both sides), so every semiprimary ring is left (and right) perfect. Those rings have DCC on right (left) principal ideals (the converse holds as well). Is there a "direct constructive" proof for this implication? If so, it may contain helpful arguments for the above problem, but I only know "indirect" proofs.
First approach - Adapt from the commutative case
I tried induction on $l$. For $l=1$ see above. Now, let $l>1$. As $A$ is semiperfect, we can write $A=Ae_1\oplus\ldots\oplus Ae_l$ with pairwise orthogonal, local idempotents $e_i$ with $e_1+\ldots+e_l=1$. If $\mathfrak a_0\subseteq\ldots\subseteq\mathfrak a_m$ is a chain of principal left ideals $\mathfrak a_j=Aa_j$ in $A$, then, for every $i$, we obtain a chain $(Aa_je_i)_j$ of cyclic submodules of $Ae_i$. Now, the following 2 steps don't work:

*

*$M_i:=Ae_i$ is a (cyclic) $A$-module and ${M_i}/{\mathfrak rM_i}$ has length $l_i=1$. I would like to deduce (e.g. by the stronger statement ($\star$) below) that the chain $(Aa_je_i)_j$ of cyclic submodules has at most $nl_i=n$ proper inclusions.

*If our original chain has at least $ln+1$ inclusions and if 1. is true, there is at least one $j$ with $\mathfrak a_je_i=\mathfrak a_{j+1}e_i$ for all $i$. Then
$$
\mathfrak a_{j+1}\subseteq\mathfrak a_{j+1}e_1+\ldots+\mathfrak a_{j+1}e_r=\mathfrak a_je_1+\ldots+\mathfrak a_je_r,
$$
and we'd be done if this was contained in $\mathfrak a_j$. But this need not necessarily be true, as $\mathfrak a_j$ is only a left ideal.

The more general result mentioned in 1. is:

($\star$) If $M$ is a cyclic $A$-module with $\ell({M}/{\mathfrak rM})=1$ (or even $=k$), then every chain of cyclic submodules has at most $n$ (or $kn$) proper inclusions.

I'm not sure, if this is true in general. In the commutative case, one can again reduce to $A$ being local, so $l=1$. Then $k=1$ and the chain can be lifted to $A$, which reduces to the case $l=1$ from the beginning.
Second approach - Reduction modulo $\mathrm{Jac}(A)$
We can reduce modulo $\mathfrak r$, i.e. pass to ${A}/{\mathfrak r}$, i.e. look at $\mathfrak a_j+\mathfrak r$. There are at most $n$ proper inclusions there. If, by chance,
$$
\mathfrak r=\mathfrak a_0+\mathfrak r=\ldots=\mathfrak a_{m-l}+\mathfrak r\subsetneq\ldots\subsetneq\mathfrak a_m+\mathfrak r=A,
$$
then $\mathfrak a_{m-l}$ would be contained in $\mathfrak r$, so it's a ${A}/{\mathfrak r^{n-1}}$-module. By lifting the chain of length $m-l$ to ${A}/{\mathfrak r^{n-1}}$ (as in the proof of the case $l=1$), we would get $m-l\leq (n-1)l$ by induction and be done.
However, the distribution of the $l$ proper inclusions after reducing to ${A}/{\mathfrak r}$ can be very arbitrary. I tried different things, but wasn't able to "connect" the inclusions or manipulate the chain (keeping the length and properness) to shift the inclusions around. Somehow, the problem is that the chain can lie very "skew" to the chain $(0)=\mathfrak r^n\subseteq\ldots\subseteq\mathfrak r\subseteq A$.
One could also try to reduce modulo $\mathfrak r^{n-1}$ and use induction, but I faced similar problems there (no control of where the proper inclusions occur).
To get an idea how to proceed, I considered the next simplest case $n=l=2$ and tried to deduce a contradiction from a proper chain $(0)=\mathfrak a_0\subsetneq\ldots\subsetneq\mathfrak a_5=A$. But even there I couldn't settle all cases of distributions of the proper inclusions after reducing mod $\mathfrak r$.
Third approach - Cutting off factors from the right
Every ("maximal") proper chain of principal left ideals is of the form $(0)=Aa_0\cdots a_{m-1}\subsetneq Aa_1\cdots a_{m-1}\subsetneq\ldots\subsetneq Aa_{m-1}\subsetneq A$. Now, if we have equalities modulo $\mathfrak r$ from $i$ to $j>i$, i.e.
$$
Aa_i\cdots a_{m-1}+\mathfrak r=\ldots=Aa_j\cdots a_{m-1}+\mathfrak r,
$$
one can take a look at the chain
$$
Aa_i\cdots a_{j-1}\subsetneq\ldots\subsetneq Aa_{j-1}\subsetneq A.\qquad (\star\star)
$$
This is still proper. If it's long enough, it will again yield equalities after reducing modulo $\mathfrak r$. Rowen uses exactly this method in his proof (mentioned above) and just chooses the bound large enough such that, after doing $n$ recursion steps, he gets a chain
$$
Aa_i\cdots a_{i+n}\subsetneq\ldots\subsetneq Aa_{i+n}
$$
with $Aa_ja_{j+1}+\mathfrak r=Aa_{j+1}+\mathfrak r$ for $i\leq j<i+n$. Then, by another argument, $Aa_{i+1}\cdots a_{i+n}\subseteq\mathfrak r^n+Aa_i\cdots a_{i+n}=Aa_i\cdots a_{i+n}$, a contradiction.
However, as said above, this only works with the very large bound. I don't know, if parts of the arguments might be of help to get a proof for Björk's bound. I have the feeling, that some transition argument like ($\star\star$) should be used and might even be crucial.

Special cases
It might be helpful to tackle special cases to get proof ideas or even hints for counterexamples.
Product of rings
If $A\simeq A_1\times\ldots\times A_r$, it's easy to reduce from $A$ to all $A_i$ by adopting the proof of the commutative case. However, in general $A$ need not have a product decomposition at all - in particular the Artin-Wedderburn decomposition $A/{\mathfrak r}\simeq M_{l_1}(K_1)\times\ldots\times M_{l_r}(K_r)$ need not lift to $A$. E.g. $A=\begin{pmatrix} \mathbb{Q} & \mathbb{Q}^{(\mathbb{N})} \\ 0 & \mathbb{Q} \end{pmatrix}$ is a semiprimary, non-Artinian ring, which is not a product ring (it does not contain a pair of nontrivial orthogonal central idempotents).
Simple reduction case - Matrix ring over a local semiprimary ring
If $A/{\mathfrak r}\simeq M_l(K)$, then $A\simeq M_l(D)$ for a local ring $(D,\mathfrak m)$ with ${D}/{\mathfrak m}\simeq K$ (see [2 (Rowen), Proposition 2.7.21]), and (proper) chains of principal right ideals in $A$ correspond to (proper) chains of $D$-submodules of $D^r$ with at most $r$ generators each. Apart from that, I didn't get any further in this special case either.

Citations:
[1] Björk: "Noetherian and artinian chain conditions of associative rings." Arch. Math. 24 (1973), 366–378.
doi:10.1007/bf01228225
[2] L. H. Rowen: "Ring Theory. Volume 1." Academic Press, San Diego (1988).
https://www.elsevier.com/books/ring-theory-v1/rowen/978-0-12-599841-3
 A: Here is an artinian counterexample.
I'll describe a ring $A$ explicitly, but in the language of quivers
with relations, if $Q$ is a quiver with two vertices, a loop at each
vertex, and an arrow from vertex $1$ to vertex $2$, then $A$ is the
path algebra of $Q$ subject to relations making all paths of length
two equal to zero.
Let $A$ be the algebra over a field $k$ with basis $\{e_{1},e_{2},a,b,c\}$,
with all products of two basis elements equal to zero except:
$$e_{1}^{2}=e_{1},\quad
e_{2}^{2}=e_{2},\quad
e_{1}a=a=ae_{1},\quad
e_{1}b=b=be_{2},\quad
e_{2}c=c=ce_{2}.$$
Then

*

*$A$ is a five-dimensional associative algebra with unity
$1=e_{1}+e_{2}$.

*The Jacobson radical $\mathfrak{r}$ is spanned by $\{a,b,c\}$,
and $\mathfrak{r}^{2}=0$. so in the notation of the question, $n=2$.

*$A/\mathfrak{r}\cong k\times k$, so in the notation of the
question, $l=2$.

*So $ln=4$.

But there is a properly descending chain of principal left ideals
$$A > A(a+e_{2}) > Ae_{2} > A(b+c) > Ac > 0$$
with bases
$$\{e_{1},e_{2},a,b,c\}\supset\{e_{2},a,b,c\}\supset\{e_{2},b,c\}
\supset\{b,c\}\supset\{c\}\supset\emptyset.$$
