Rings with a given number of (prime, maximal) ideals 
Given three cardinal numbers $1≤a≤b<c$, I would like to know if we can find a commutative ring $R$ (with unit) such that

*

*the number of ideals of $R$ is $c$

*the number of prime ideals of $R$ is $b$

*the number of maximal ideals of $R$ is $a$

For instance, for $a=b=c=\aleph_0$, the ring $R=\Bbb Z$ provides an example. I'm particularly interested in the case where $a,b,c$ may be infinite.
If $c$ is finite, then either $a=b=1$ and $c=2$ or $R$ is not an integral domain (see here). In any finite ring, prime ideals are maximal (because a finite integral domain is a field, see here), so that possible examples for $a \neq b$ must be infinite. If $c=3$, then $a=1$ (see here).
If $a=b$ and $c=2^a$ is finite, then I think that we can consider $R=\Bbb Z/p_1 \cdots p_a \Bbb Z$ where $p_1, \dots, p_a$ are distinct primes. I wasn't able to solve the case $c=5,a=b=3$ for instance. Maybe there are conditions on $a,b,c$ for such a ring to exist, at least when $c$ is finite.
Possibly related: (1), (2), (3).
Thank you very much for your help!
 A: I don't have a complete solution, but I will isolate out what appears
to be the hardest case, namely whether there is a commutative
unital ring with $1$ maximal ideal, $\lambda$-many prime ideals
($\lambda$ infinite), and $\mu$-many ideals, where $\mu>\lambda$
and $\mu\neq \lambda^{\nu}$ for any cardinal $\nu$.

Say that a triple $(a,b,c)$ is realizable if there is
a nontrivial commutative unital ring with
$a$ maximal ideals, $b$ prime ideals, and $c$ ideals.
If $R$ is a commutative unital ring write $t(R)$ for
its associated triple, $(a,b,c)$. I might write $a(R)$
for the number of maximal ideals of $R$, $b(R)$
for the number of prime ideals of $R$, and
$c(R)$ for the number of ideals of $R$.
We must have $1\leq a\leq b\leq c$, since nontrivial rings
have at least one maximal ideal, maximal ideals are prime,
and prime ideals are ideals. I will always consider triples
like this.
As Jay observed, if $c$ is finite, then a triple is realizable
iff $a=b$ and $c$ can be factored as $(e_1+1)\cdots (e_a+1)$ where
each $e_j$ is a positive integer.
To prove this, Let $R$ be a commutative ring realizing
$(a,b,c)$ with $c$ finite. $R$ must be Artinian, so it
is isomorphic to a product $L_1\times \cdots\times L_k$
of nontrivial local rings. Necessarily $k = a = b$. Each ideal
of $R$ has the form $I_1\times \cdots\times I_k$ with $I_j\lhd L_j$
for all $j$, since $R$ is unital, so $c(R) = c(L_1)\cdots c(L_k)$.
Each $c(L_j)\geq 2$, since a nontrivial ring has at least $2$ ideals,
hence $c(L_j)=e_j+1$ for some positive integer $e_j$. Thus
$c(R)$ factors as $(e_1+1)\cdots (e_a+1)$ where
$e_j+1=c(L_j)\geq 2$ for all $j$.
This shows that the conditions mentioned for triples
of finite numbers must hold.
Conversely, suppose the conditions for triples
of finite numbers hold, namely that
$a=b$ and $c = (e_1+1)\cdots (e_a+1)$ where $e_j\geq 1$ for each $j$.
Then $(a,b,c)$ is realized by the ring
$$
\mathbb Z_{p_1^{e_1}\cdots p_a^{e_a}}\cong 
\mathbb Z_{p_1^{e_1}}\times 
\cdots\times \mathbb Z_{p_a^{e_a}},
$$
where $p_1<p_2<\cdots<p_a$ are distinct primes.

Now I consider only those triples $(a,b,c)$ with $c$ infinite.
Background fact 1: There is a field of every infinite cardinality.



*

*Case $t(R) = (1,1,\lambda)$, $\lambda$ infinite.


Let $\mathbb F$ be a field of cardinality $\lambda$ and let
$R = \mathbb F[x,y]$. Let $M = (x,y)\lhd R$.
Then $M/M^2$ is the unique maximal and the unique
prime ideal of $R/M^2$. Since $|R|=\lambda$ and
$R$ is Noetherian, $R$ has at most $\lambda$-many ideals,
hence $R/M^2$ has at most $\lambda$-many ideals.
But the ideals $(x+fy)+M^2$, $f\in \mathbb F$, are
$\lambda$-many distinct ideals. Thus
$t(R/M^2) = (1,1,\lambda)$.


*Case $t(R) = (1,n+1,\lambda)$, $n\geq 1$ finite, $\lambda$ infinite.


Let $\mathbb F$ be a field of cardinality $\lambda$ and let
$R = \mathbb F[x_1,\ldots,x_n]$. Let $M = (x_1,\ldots,x_n)\lhd R$
and let $I$ be the ideal of $R$ generated by $\{x_ix_j\;|\;i\neq j\}$.
Let $S = (R/I)_{M/I}$, the localization of $R/I$ at the
maximal ideal $M/I$. This ring is local, so $a(S)=1$.
It can be seen to have $c(S) = \lambda$ via arguments
similar to the last case. To compute the number of prime ideals
note that in $S$ the (images of the) elements $x_i$ and $x_j$
have product zero, so any prime contains one or the other.
Hence any prime contains all except at most one of the $x_i$.
If $J\subseteq M_{M/I}$ contains all of the $x_i$, then $J$ is the maximal
ideal of $S$, which is prime. If $J$ is generated
by all except, say, $x_i$, then
$S/J\cong \mathbb F[x_i]_{(x_i)}$, which is a domain of dimension $1$.
So this locates all of the primes: those ideals of $S$ generated
by a subset of $\{x_1,\ldots,x_n\}$ containing at least $n-1$
of the elements. Thus $b(S)=n+1$.


*Case $t(R) = (1,\lambda,\lambda)$, $\lambda$ infinite.


Let $\mathbb F$ be a field of cardinality $\lambda$ and let
$R = \mathbb F[x,y]$. Let $M = (x,y)\lhd R$.
The localization $S=R_M$ of $R$ at $M$
satisfies $t(R_M)=(1,\lambda,\lambda)$.
[To see that $b(R_M)\geq\lambda$, consider
the ideals $(x+fy)$, $f\in \mathbb F$.]


*Case $t(R) = (\lambda,\lambda,\lambda)$, $\lambda$ infinite.


Let $\mathbb F$ be a field of cardinality $\lambda$ and let
$R = \mathbb F[x]$. Then $t(R)=(\lambda,\lambda,\lambda)$.

Something implicit in the part of this post
concerning the case where $c(R)$ is finite is that,
for any $R$ and $S$, if
$t(R) = (a_1,b_1, c_1)$ and 
$t(S) = (a_2,b_2, c_2)$,
then $t(R\times S) = (a_1+b_1,a_2+b_2,c_1\cdot c_2)$.
Background fact 2: If at least one of the
nonzero cardinals $\alpha$ and $\beta$ is infinite,
then we have $\alpha+\beta=\alpha\beta = \max\{\alpha,\beta\}$.
Let's use this info to combine the examples above:
I. Case $(a,b,c) = (\kappa,\lambda,\mu)$, all infinite
and satisfying $\kappa\leq\lambda\leq\mu$:
Take the product of a ring of type $(1,1,\mu)$ from Construction 1
with a ring of type $(1,\lambda,\lambda)$ from Construction 3
and a ring
of type $(\kappa,\kappa,\kappa)$ from Construction 4.
II. Case $(a,b,c) = (m,n,\mu)$, $m\leq n$ finite, $\mu$ infinite:
If $m=1$, then use a ring of type $(1,n,\mu)$ from Construction 1 or 2.
Else $2\leq m\leq n$. In this case, take the product of a ring of type
$(1,1,\mu)$ with one of type $(m-1,n-1,\mu)$. [Latter exists
by induction.]
III. Case $(a,b,c)=(m,\lambda,\mu)$ where $2\leq m$, $m$ finite,
and $\lambda\leq\mu$ are infinite:
Take a product of a ring of type $(1,\lambda,\lambda)$ and one of type
$(m-1,m-1,\mu)$. The latter exists by Item II.]
IV.  Case $(a,b,c)=(1,\lambda,\mu)$ where $\lambda\leq\mu$ are infinite:
The case $\lambda=\mu$ is handled by Construction 3, so we may
restrict to THE HARD CASE:
IV.'  Case $(a,b,c)=(1,\lambda,\mu)$ where $\lambda <\mu$ are infinite:
I only see how to handle the subcase where
$\mu = \lambda^{\nu}$ for some infinite $\nu$.
The idea is to combine a Noetherian ring of type $(1,\lambda,\lambda)$
from Construction 3 with a new ring of type $(1,1,\lambda^{\nu})$
using a pullback instead of a product. I use pullback
instead of product to keep the $a$-value from increasing.
Let $R$ be a Noetherian ring of type $(1,\lambda,\lambda)$
from Construction 3 above. $R$ is local, so let $\alpha\colon R\to k$
be the natural map onto its residue field $k$.
(Actually $k=\mathbb F$ from that construction.)
Let $V$ be a $\nu$-dimensional
vector space over $k$, for some infinite $\nu$. Let $S = k\oplus V$
with multiplication defined by $(r,v)*(r',v') = (rr',rv'+r'v)$.
Then $S$ is a local ring with maximal ideal $V$
which squares to $0$.
Any proper ideal of $S$ is a $k$-subspace of $V$,
and any $k$-subspace is an ideal.
This yields $t(S)=(1,1,c)$,
where the third coordinate is $c(S) = 1+$ the number
of $k$-subspaces of $V$. When the dimension $\nu$
of $V$ is infinite, this number is $|k|^{\nu}$.
For our situation it is $\lambda^{\nu}$.
Let $\beta\colon S\to k$ be the first projection map.
The ring we take for the subcase $(1,\lambda,\lambda^{\nu})$
is $R\times_k S$, the pullback of $\alpha$ and $\beta$.
It may be viewed as the subring of $R\times S$ consisting of
all $(r,s)$ such that $\alpha(r)=\beta(s)$.
Suppose that $t(R)=(a_1,b_1,c_1)$ and
$t(S)=(a_2,b_2,c_2)$. Since $R$ and $S$ are local
with the same residue field $k$, and $R\times_k S$ is the pullback
of the maps onto $k$, then $R\times_k S$ is
also local with residue field $k$. Thus
$a(R\times_k S) = 1$. It is not hard to show that
$b(R\times_k S) = b_1+b_2-1$ for this type of construction,
so for us $t(R\times_k S) = (1,\lambda,c)$.
I claim that if $\lambda<\lambda^{\nu}$, then
$c = \lambda^{\nu}$.
We have to count the ideals of a local ring that is a pullback.
The proper ideals of our pullback are contained in
$M\times V$ where $M$ and $V$ are the maximal ideals of $R$
and $S$. If $I\subseteq M\times V$ is an ideal, then there
are a smallest product ideal $A\times B$ that contains $I$
and a largest product ideal $C\times D$ contained in $I$.
The interval in the ideal lattice of $R\times_k S$ from
$C\times D$ to $A\times B$ will be called a minimal
product interval.
I claim that the map $I\mapsto (A,B,C,D)$ maps onto a set
of size $\lambda^{\mu}$ and has fibers of size at most $\lambda$,
so the number of proper ideals of $R\times_k S$
is exactly $\lambda^{\nu}$.
There are at most $\lambda$-many choices for $A$ and $C$
and at most $\lambda^{\nu}$-many choices for $B$ and $D$,
so at most $\lambda^{\nu}$ many possible 4-tuples.
But this many are in the range of the assignment, since
there are $\lambda^{\nu}$-many ideals of the form
$I=\{0\}\times U$, $U\leq V$,
and for such an ideal we assign the 4-tuple $(0,U,0,U)$.
To measure the fiber size, argue that if $I$
is assigned $(A,B,C,D)$, then the $R$-module
$(A\times B)/(C\times D)$ is annihilated by
the maximal ideal of $R\times_k S$,
since it is a minimal product interval and 
the second factor $B/D$ is annihilated by the maximal ideal of $S$.
The module $(A\times B)/(C\times D)$ is
Noetherian, since it is a minimal product interval
and the first factor $A/B$ is Noetherian. The first of these
two claims tell us that the ideals in this interval
are just the $k$-subspaces, while the second
tells us that this module, as a $k$-space,
is finite dimensional. Thus the number
of ideals in $R\times_k S$ contained in
$A\times B$ and containing $C\times D$
is at most the the number of subspaces of some finite dimensional
$k$-space, i.e., at most $\lambda$.
A: I found a characterization of the possible triples $(a,b,c)$ when $c$ is finite:
If $c < \infty$, then $(a,b,c)$ occurs iff $a = b$ and $c$ is a product of $a$ integers all of which are $\ge 2$.
Proof: Let $c(R)$ be the number of ideals of $R$ and take a ring with $c(R) < \infty$. Then $R$ is Artinian because any descending chain of ideals stabilizes. This implies $a = b$ since $\dim R = 0$. Also $R = \prod_{i=1}^a R_i$ is a product of $a$ Artinian local rings so $c(R) = \prod_{i=1}^a c(R_i)$. But $c(R_i)$ can be any positive integer $\ge 2$ since e.g. $k[x]/(x^{c-1})$ has exactly $c$ ideals. 
This shows that e.g. $(3,3,5), (3,3,9)$ are impossible but $(3,3,8), (3,3,12)$ are possible.
