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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!

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    $\begingroup$ I think that if $a$ is finite then $c \ge 2^a$. If $R$ is a ring with $a$ maximal ideals then quotienting by the Jacobson radical would only decrease $c$ and give a product of $a$ fields which has $2^a$ ideals. So $(3,3,5)$ is impossible $\endgroup$ – Jay Oct 13 '16 at 0:17
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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.


  1. 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 $t(R/M^2) = (1,1,\lambda)$.

  1. 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 $b(S)=n+1$.

  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 $(x+fy)$, $f\in \mathbb F$.]

  1. 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 $\lambda$.

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  • $\begingroup$ Thank you very much for this answer! I will read it carefully! $\endgroup$ – Watson Oct 15 '16 at 8:08
  • $\begingroup$ Since your answer is quite complete, I accepted it. I will maybe ask a new question for the specific case IV'. $\endgroup$ – Watson Oct 15 '16 at 17:43
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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.

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