Solid Rings
Posted by John Baez
“Solid ring” sounds self-contradictory, since a ring should have a hole in it. But mathematicians use words in funny ways.
Epimorphisms of rings aren’t always onto. For example the inclusion of the integers $\mathbb{Z}$ in the rationals $\mathbb{Q}$ is an epimorphism.
This makes it potentially interesting to ask: what are all the ring epimorphisms $f \colon \mathbb{Z} \to R$?
But $\mathbb{Z}$ is initial in the category of rings: there’s always exactly one ring homomorphism from $\mathbb{Z}$ to any ring $R$. So people say a ring is solid if the unique ring homomorphism $f \colon \mathbb{Z} \to R$ is an epimorphism.
My question then becomes: what are all the solid rings?
The most obvious examples are the quotient rings $\mathbb{Z}/n$. The homomorphism $f \colon \mathbb{Z} \to \mathbb{Z}/n$ is not just an epimorphism, it’s what we call a regular epimorphism: basically a quotient map. The rings $\mathbb{Z}/n$ are the only rings where the unique morphism from $\mathbb{Z}$ is a regular epimorphism.
But $\mathbb{Q}$ is also solid, since once you know what a ring homomorphism $g$ out of $\mathbb{Q}$ does to the element $1$, you know it completely, since $g(m/n) = g(m)g(n)^{-1}$. For the same reason, any subring of $\mathbb{Q}$ is solid.
And there are lots of these: in fact, a continuum of them. Take any set of prime numbers. Start with $\mathbb{Z}$, and throw in the inverses of these primes. You get a subring of $\mathbb{Q}$. So, there are at least $2^{\aleph_0}$ subrings of $\mathbb{Q}$, and clearly there can’t be more.
(In fact I’ve just told you how to get all the subrings of $\mathbb{Q}$.)
These are the ‘obvious’ solid rings. But there are lots more! Bousfeld and Kan classified all the commutative solid rings in 1972, in this paper freely available from the evil giant Elsevier:
- A. K. Bousfield and D. M. Kan, The core of a ring, Journal of Pure and Applied Algebra 2 (1972), 73–81.
For example, let $\mathbb{Z}[1/2]$ be the ring of dyadic rationals: the subring of $\mathbb{Q}$ generated by the number $1/2$. This is solid. So is $\mathbb{Z}/2$. We knew this already. But the product $\mathbb{Z}[1/2] \times \mathbb{Z}/2$ is also solid!
More generally, suppose $R$ is a subring of $\mathbb{Q}$. $R \times \mathbb{Z}/n$ is solid if:
- $R$ consists of fractions whose denominators are divisible only by primes in some set $P$,
- all the prime factors of $n$ lie in $P$.
But there are even more solid rings. Bousfeld and Kan show that in the category of commutative rings, any colimit of solid rings is again a solid ring! And they show any commutative solid ring is a colimit of the three kinds I’ve told you about so far. Using this, they get a precise classification of all commutative solid rings, whch I summarized here:
Here’s one reason this is interesting: the category of affine schemes is the opposite of $\mathsf{CommRing}$, so commutative solid rings give subobjects of the terminal affine scheme.
Subobjects of the terminal object in a category are called subterminal objects, and they’re cool because they look ‘no bigger than a point’. The category $\mathsf{Set}$ has just two, up to isomorphism: the 1-point set, and the empty set. These play the role of truth values.
But a general topos has lots of subterminal objects. These play the role of ‘external’ truth values — truth values as seen from outside, which form a set rather than an object in the topos. Now we’ve seen that the category of affine schemes has lots of subterminal objects too. These should be important in algebraic geometry somehow, but I have no idea how. Do you?
For that matter, what are the subterminal objects in the category of schemes?
One year after Bousfeld and Kan’s paper, Storrer showed that all solid rings are commutative:
- H. H. Storrer, Epimorphic extensions of non-commutative rings, Commentarii Mathematici Helvetici 48 (1973), 72–86.
So, we now know the complete classification of solid rings! But I feel there should be more that we can do with them. More than I know now, that is.
Images
Say I have a unital ring $R$, so $\mathbb{Z} \to R$. Is there a universal solid ring $S$ in between them, a sort of image of $\mathbb{Z}$ into $R$?
In terms of geometry, this would be about having a family defined over $Spec\ \mathbb{Z}$, and observing that some of its fibers are empty, so it’s really defined over some subscheme.