On those integers $n>1$ such that there exists a commutative ring with identity with exactly $n$ ideals

Question

Let $n>1$ be an integer; we call $n$ a "ring number" if there exists a commutative ring $R$, with identity, having exactly $n$ ideals (including $\{0\}$ and $R$); now since for every $n>1$, $\mathbb Z_n$ is a commutative ring with unity having exactly $d(n)$ (no. positive divisors of $n$) ideals and $d(n)$ is unbounded as $n$ grows large, so we know that there are infinitely many ring numbers.

My question:

Is it true that all ring numbers are of the form $d(n)$ for some $n$ ? If not all ring numbers are of the form $d(n)$ then how many exceptions are there; finitely many or infinitely many? Are there infinitely many positive integers which 'are not' ring numbers? Can we characterize ring numbers in any way? Is it true that if $n$ is a ring number then there is a commutative ring with identity $R$ which is also a PIR, having exactly $n$ ideals?

Answer or reference to any of the questions would be highly appreciated. Please help. Thanks in advance.

Answer 1

Yes, $d(n)$ can attain any positive integer for some $n$, consider $n=2^k$. Hence all ring numbers are of the form $d(n)$.

Answer 2

Take $R = \mathbb{F}_2[X]/(X^n)$. Then $R$ is finite and has exactly $n+1$ ideals.

Indeed, ideals of $R$ are in canonical bijection with ideals of $\mathbb{F}_2[X]$ containing $X^n$, ie with polynomials dividing $X^n$ : these are the $X^k$ for $0\leqslant k\leqslant n$.