Can there be more ideals than elements of a ring?

Question

Can there be more ideals than elements of a ring? This is related to my other question https://math.stackexchange.com/questions/713610/having-elements-as-ideals . At first glance, it seems obvious that there would be less ideals than elements of a ring. But the ideals are not necessarily mutually disjoint, and they form a subset of the power set of the ring, which is strictly larger than the ring itself. I know fields only have trivial ideals, so maybe there is an "opposite" structure that has a maximal proliferation of ideals relative to elements? A (weaker) related question is if there can be more subrings of than elements of a ring.

Answer 1

For a finite example, consider $\mathbb{Z}[x,y,z]/(2,xx,xy,xz,yy,yz,zz)$ with 16 elements and 17 ideals.

The elements are all $a+bx+cy+dz$ for $a,b,c,d\in \{0,1\}$. The ideals are the whole ring and the 16 subspaces of $(x,y,z)$.

More generally, take $R_n = \mathbb{Z}[x_1,\ldots,x_n]/( (2) + (x_1,\ldots,x_n)^2 )$ with $2^{n+1}$ elements and $\displaystyle 1+\sum_{k=0}^n \prod_{i=1}^{n-k} \frac{2^{i+k} - 1}{2^{i\phantom{{}+k}} - 1} \approx 7.3 \cdot 2^{n^2/4}$ ideals. For example $R_4$ has 32 elements and 67 ideals, $R_5$ has 64 elements and 375 ideals, $R_6$ has 128 elements and 2826 ideals, and $R_7$ has 256 elements and 29213 ideals.

The motivation or intuition is that a vector space contain many more subspaces than elements, much like sets have many more subsets than elements. A “zero ring” is a ring in which all multiplications are 0, and so all subgroups are ideals. However, many people get confused by zero rings, so I tend to adjoin one before mentioning them. $(x_1,\cdots,x_n)/(x_1,\cdots,x_n)^2$ is a zero ring with $2^n$ elements and about $2^{n^2/4}$ ideals, and adjoining 1 (to get $R_n$) only doubles the number of elements (and gives one extra ideal).

Answer 2

Start with the polynomial ring, over your favorite countable field $K$, generated by countably many indeterminates $X_q$ indexed by the rational numbers $q$. Let $R$ be the quotient of this ring obtained by imposing the equations $X_qX_r=X_q$ for all $q<r$. (In fancier language, this is the semigroup algebra over $K$ of the semigroup consisting of the rational numbers with the operation $\min$.) Then $R$ is countably infinite, but every real number $\alpha$ gives rise to an ideal $I_\alpha$ generated by all the $X_q$ with $q<\alpha$. So this countable ring has continuum many ideals.

Actually, it wasn't necessary to pass from the polynomial ring $K[X_q:q\in\mathbb Q]$ to the quotient ring $R$. $K[X_q:q\in\mathbb Q]$ itself has continuum many ideals, namely the pre-images of the $I_\alpha$'s (and lots more besides).