The example I originally posted doesn't work.
Here are a couple of This is way too addictive, so I'm going to try to quit, and I'll just leave my thoughts here in case they're useful for other addicts. This is based on the ideas in the previous non-answer that might lead to an example that's not a Boolean ringI posted last year, so I'm just editing that answer. I'll denote
Denote the number of ideals of a ring $A$$R$ by $n(A)$$n(R)$, and define the Alexeq Quotient $q(R)$ to be $\frac{n(R)}{\vert R\vert}$, so the question asks about rings $R$ with $q(R)=1$.
$q(R)$ is multiplicative on direct products (1i.e., $q(R\times S)=q(R)q(S)$) Suppose you, so one approach to finding rings with $q(R)=1$ is to look for examples with fairly simple fractions as $q(R)$ and take products.
For example, $q(\mathbb{F}_p)=\frac{2}{p}$, and $q(\mathbb{F}_4)=\frac{1}{2}$, so if we could find a ring with $A$ of order$q(R)=\frac{m}{2^k}$, where $2^k$ such that the number of$m$ has at least $k$ prime factors (counted with multiplicity) of $n(A)$ is at least $k$. Suppose $$n(A)=p_1\dots p_l,$$ with $l\geq k$, say.
Then $$A\times\mathbb{F}_{p_1}\times\dots\times\mathbb{F}_{p_l}\times\mathbb{F}_4^{l-k}$$ hasthen by taking a suitable direct product with fields we could achieve $2^{2l-k}p_1\dots p_l$ elements and ideals$q(R)=1$.
It seems quiteIt's not hard to produce examples. I triedfind $A=\mathbb{F}_2\oplus V$$\mathbb{F}_2$-algebras with $q(R)>1$. For example, take $R$ to be the $d$-dimensional algebra $\mathbb{F}_2\oplus V$ where $V$ is ana $\mathbb{F}_2$$(d-1)$-vector space and adimensional square zero-zero ideal infor $A$$d>3$.
However, so the proper ideals of $A$ are$n(R)$ is bounded by the number of subspaces of $V$$R$, but towhich behaves roughly as a constant times $2^{d^2/4}$ for large values of $d=\dim(R)$. But the limitnumber of my calculationsprime factors of a "typical" large number $n(A)$$n$ is roughly $\log\log n$, giving roughly $\log d$, so unless we can carefully design $R$ so that $n(R)$ has far too fewmany prime factors, and very rough heuristics suggest that this will continuewe probably need to be lucky to find an $\mathbb{F}_2$-algebra where $n(R)$ has as many prime factors as $\vert R\vert$.
(2) Here's a way of producing a ring $B$ It's possible to find rings where the number of$n(R)$ has more prime factors ofthan $n(B)$$\vert R\vert$. For example, $R=\mathbb{F}_q[x,y]/(x^2,y^2)$ has $n(R)=q+5$ and so $q(R)=\frac{q+5}{q^4}$. So if $q$ is a reasonably large compared to the numberprime of the form $2^k-5$ (e.g., $q=59$) then $n(R)$ has $2^k$ prime factors of, but $|B|$$\vert R\vert$ has only $4$.
Let $p$ be aUnfortunately, I don't know any way to do this without introducing large prime such that $p+5=2^t$ is a powerfactors in the denominator of $2$$q(R)$, andwhich are hard to get rid of: it's difficult to design $t>4$. Then$S$ so that $B=\mathbb{F}_p[x,y]/(x^2,y^2)$ has$n(S)$ is divisible by a particular prime, but $|B|=p^4$ and$n(S)$ doesn't have many fewer prime factors than $n(B)=2^t$$\vert S\vert$.
If youA weakening of the original question, that I don't think I've found an answer to, is:
Question: Is there a finite commutative ring as in$R$ (1apart from finite boolean rings) but with, where $l$ slightly smaller$n(R)\geq \vert R\vert$ and $n(R)$ has at least as many prime factors than $k$$\vert R\vert$?
I found one example that might be useful. If $R=\mathbb{F}_2[x]/(x^5)\oplus \mathbb{F}_2^4$, and where $p^4$ divides $n(A)$$\mathbb{F}_2^4$ is a square zero ideal, then considering $A\times B$$n(R)=1296=2^43^4$, and continuing as inso (1) might work$q(R)=\frac{3^4}{2^5}$. This means that if we could find a $d$-dimensional $\mathbb{F}_3$-algebra $S$ where $n(S)$ has at least $\frac{5d}{4}$ prime factors then we could construct $T$ with $q(T)=1$.