The n-Category Café

That’s a nice question! We’d get that if the category of commutative rings had a factorization system which let us factor any morphism as an epimorphism followed by some other sort of morphism in a canonical way. This is different than the usual (regular epi, mono) factorization system we’ve got for commutative rings.

By the way, all rings in my post are assumed unital. That’s why they’re rings and not mere rngs.

If solid rings are closed under all colimits, can’t you just take the colimit of the diagram consisting of all solid rings that map to $R$?

Allan wrote:

Say I have a […] 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 $R$?

Yes, it’s the ‘core’ of $R$, a concept discussed here and less strenuously here.

I think the concept of regular image is relevant. In some categories we can factor any morphism $f \colon c \to d$ as an epimorphism $p \colon c \to im(f)$ followed by a regular monomorphism $i \colon im(f) \to d$ in a canonical way. Then $im(f)$ is called the regular image of $f$.

The nLab article core of a ring says the regular image of the unique morphism $f \colon \mathbb{Z} \to R$ is the core of $R$.

Beware: the ordinary set-theoretic image lets us factor any homomorphism as a regular epimorphism (that is, a quotient map) followed by a monomorphism. So, that’s giving us a (regular epi, mono) factorization. The regular image, when it exists, is giving us an (epi, regular mono) factorization!

I don’t really know if all homomorphisms of commutative rings have regular images, giving an (epi, regular mono) factorization system.

But I bet it does. And I bet to get the regular image you take the ‘dominion’ of the ordinary set-theoretic image. That’s another concept people talk about,which I’m having some trouble digging up information on.

There’s not a proper class of isomorphism classes of solid rings: the explicit classification of them here makes it clear there’s just a set:

• John Baez, When is the ring homomorphism $\mathbb{Z} \to R$ an epimorphism?.

I guess when you heard “in the category of commutative rings, any colimit of solid rings is solid” you thought we’d get a proper class of them — but we don’t!

It may help to remember that the coproduct of commutative rings is usually called their “tensor product”. So, the coproduct of $\mathbb{Z}$ and $\mathbb{Z}$ in the category of commutative rings is just $\mathbb{Z}$ again. Similarly the coproduct of two copies of $\mathbb{Q}$ is just $\mathbb{Q}$. The coproduct of $\mathbb{Z}/n$ and $\mathbb{Q}$ is the trivial ring! And so on.

So, you shouldn’t think of coproducts of solid rings as becoming arbitrarily “big”: there’s just a set’s worth of different choices (up to isomorphism). And then the colimits are just quotients of those.

This means that the category of solid rings is a cocomplete small category. Fortunately, it is also a preorder (like the category of epimorphisms out of any fixed object in any category), so it doesn’t run afoul of Freyd’s theorem. (-:

However, this does imply that it is a complete lattice, and so in particular it has all limits as well. Thus any family of solid rings has a “co-union”: a universal solid ring that maps to all of them. Of course, unlike the colimits, this is only universal among other solid rings, not all rings.

This increases my distaste for Hungerford’s book, which is inflicted on our graduate students at U.C. Riverside. He could have said “surjective homomorphism” to be clear and simple, or “regular epimorphism” to show off his knowledge of category theory, but he wanted to sound cool while not paying attention to the actual meaning of the word “epimorphism”, so he chose something that’s just wrong.

A regular epimorphism $f: y \to z$ is a morphism that’s the coequalizer of a pair of morphisms $g,g': x \to y$ for some $x$. In the category of rings this amounts to saying $z$ is the quotient of $y$ by some ideal, and $f$ is the quotient map. This is much stronger than saying $f$ is an epimorphism.

This generalizes. In any category of models of any algebraic theory, the regular epimorphisms are the same as the morphisms whose underlying map of sets is surjective. That’s why for sets equipped with any sort of “purely algebraic” structure, surjective homomorphisms are the same as regular epimorphisms. But of course Hungerford didn’t know or care.

I’m not familiar with Hungerford’s book, but perhaps its usage of epimorphism shouldn’t be judged too harshly. In Categories for the Working Mathematician, Mac Lane uses epimorphism to mean surjective homomorphism and epi for the categorical notion. For example, in Section 1.5 he writes:

In $\mathbf{Set}$ the epi arrows are precisely the surjections (epimorphisms) in the usual sense

and in Section 1.7, talking about $\mathbf{Ab}$:

Also, an epimorphism (a homomorphism onto) is clearly epi.

Then a paragraph later, talking about the category of rings, he writes:

Every epimorphism of rings is epi as an arrow, but the inclusion $\mathbf{Z} \to \mathbf{Q}$ of $\mathbf{Z}$ in the field $\mathbf{Q}$ of rational numbers is epi, but not an epimorphism.

Obviously Mac Lane’s terminological distinction didn’t catch on, but Hungerford’s usage at least has precedent.

The absolute elements of a commutative ring form a subring. So does the core, which Bousfeld and Kan defined as follows: given a commutative ring $R$, its core is

$$c(R) = \{r \in R: \; r \otimes 1 = 1 \otimes r \, \in R \otimes_\mathbb{Z} R \}$$

What’s nice about the core is that we can check if something is in it just by checking an equation; the concept of ‘absolute element’ involves universal quantification over all homomorphisms out of your ring.

I will wildly guess that the core of a commutative ring is the same as its subring of absolute elements. If this is true, a lot is known about absolute elements, thanks to Bousfeld and Kan’s paper and later work by Storrer.

Here’s why I’ll make this guess. According to the Stacks Project, a commutative ring homomorphism $f : S \to R$ is an epimorphism iff the two ring homorphisms

$$f \otimes 1, 1 \otimes f \colon R \to R \otimes_S R$$

are equal. So, the unique homomorphism $f\colon \mathbb{Z} \to R$ is an epimorphism iff

$$r \otimes 1 = 1 \otimes r \in R \otimes_\mathbb{Z} R$$

for all $r \in R$. This is true iff the core of $R$ is all of $R$.

Now, if $R$ is any ring, the unique homomorphism

$$f \colon \mathbb{Z} \to R$$

corestricts to a homomorphism

$$f_c \colon \mathbb{Z} \to c(R)$$

$f_c$ is an epimorphism iff the core of $c(R)$ is all of $c(R)$. But you can easily check that $c(c(R)) = c(R)$. So it’s true: $f_c$ is an epimorphism!

Summarizing: for any commutative ring $R$, the unique homomorphism $f \colon \mathbb{Z} \to R$ has range contained in the core of $R$, and the resulting homomorphism $f_c \colon \mathbb{Z} \to c(R)$ is actually an epimorphism.

I haven’t carefully checked, but I’m willing to guess that the core of $R$ is the largest subring of $R$ to which $f \colon \mathbb{Z} \to R$ corestricts to an epimorphism.

And I haven’t checked at all, but I’m willing to wildly guess that the subring of absolute elements of $R$ is characterized by the exact same property!

If these guesses are right, the core consists exactly of the absolute elements of $R$.

So

Excellent! Thanks for solving this old mystery (at least, mystery to me). The absolute elements of a commutative ring are indeed the elements of its core, and I think there’s a much simpler argument than the one you give.

The tensor product of two commutative rings (and I’m happy to restrict to only commutative rings) is their coproduct. If $r \in R$ is absolute then the two coprojections from $R$ to $R \otimes R$ must take the same value on $r$ — explicitly, $r \otimes 1 = 1 \otimes r$. On the other hand, if the two coprojections take the same value on $r$ then by the universal property of coproducts, any two homomorphisms out of $R$ with the same codomain take the same value on $r$. And that does it.

When I was trying to understand absoluteness years ago, I’m pretty sure I realized that $r$ is absolute iff $r \otimes 1 = 1 \otimes r$, but what I didn’t know is that such elements $r$ have a name. So I couldn’t look up all the stuff that’s known about them. But thanks to your answer, now I can!

No, no relation.