## September 16, 2021

### 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:

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.

Posted at September 16, 2021 9:51 PM UTC

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### 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.

Posted by: Allen Knutson on September 17, 2021 12:45 AM | Permalink | Reply to this

### Re: Images

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.

Posted by: John Baez on September 17, 2021 4:54 AM | Permalink | Reply to this

### Re: Images

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 $R$?

Posted by: Mike Shulman on September 17, 2021 7:36 AM | Permalink | Reply to this

### Re: Images

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.

Posted by: John Baez on September 17, 2021 6:23 PM | Permalink | Reply to this

### Re: Solid Rings

A question: BK showed that all colimits of solid rings of one of the three basic types accounts for all solid rings. Even if we allow for isomorphism classes, it seems that we have a class worth of them. So it is difficult for me to see how these correspond to truth values as sub-objects of the terminal object.

Posted by: Swill Stroganoff on September 17, 2021 1:02 AM | Permalink | Reply to this

### Re: Solid Rings

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:

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.

Posted by: John Baez on September 17, 2021 4:31 AM | Permalink | Reply to this

### Re: Solid Rings

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.

Posted by: Mike Shulman on September 17, 2021 7:35 AM | Permalink | Reply to this

### Re: Solid Rings

Nice! So I guess that the poset structure of solid rings can be defined in two equivalent ways:

• $R \le S$ iff $R \otimes S \cong S$

• $R \le S$ iff there is an epimorphism from $R$ to $S$.

The bottom element of this poset is $\mathbb{Z}$, but its top element may be more esoteric. Based on Bousfeld and Kan’s classification of solid rings, I will wildly guess that it’s

$c\big(\mathbb{Q} \times \prod_{p \in \mathcal{P}} \mathbb{Z}/p\big)$

where $\mathcal{P}$ is the set of all primes and $c(R) \subseteq R$ is what Bousfeld and Kan call the core of the commutative ring $R$:

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

The core of any ring is solid, and a ring is solid iff it’s own core. I don’t really understand what effect taking the core has on $\mathbb{Q} \times \prod_{p \in \mathcal{P}} \mathbb{Z}/p$, but Bousfeld and Kan do take the core.

Come to think of it,

$c\big(\mathbb{Q} \times \prod_{p \in \mathcal{P}} \mathbb{Z}/p\big)$

reminds me of the adeles. I don’t know if there’s an interesting relationship.

Posted by: John Baez on September 17, 2021 3:56 PM | Permalink | Reply to this

### Re: Solid Rings

John wrote:

$c\big(\mathbb{Q} \times \prod_{p \in \mathcal{P}} \mathbb{Z}/p\big)$

My ears prick up whenever I hear mention of $\mathbb{Q} \times \prod \mathbb{Z}/p$. If $T$ denotes the codensity monad of the inclusion $\mathbf{Field} \hookrightarrow \mathbf{Ring}$, then

$\mathbb{Q} \times \prod_{p \in \mathcal{P}} \mathbb{Z}/p = T(\mathbb{Z}).$

In other words, $\mathbb{Q} \times \prod \mathbb{Z}/p$ is the closest approximation to the “free field on $\mathbb{Z}$”, given that free fields on rings don’t exist.

Posted by: Tom Leinster on September 21, 2021 10:33 AM | Permalink | Reply to this

### Re: Solid Rings

Interesting! I’m conjecturing the core of $\mathbb{Q} \times \prod_{p \in \mathcal{P}} \mathbb{Z}/p$ is the terminal solid ring.

I don’t know what taking the core actually does to it! It would be funny if it did nothing. Then there might be some really nice relation to what you just said.

Posted by: John Baez on September 21, 2021 8:55 PM | Permalink | Reply to this

### Re: Solid Rings

I’m conjecturing the core of $\mathbb{Q} \times \prod_{p \in \mathcal{P}} \mathbb{Z}/p$ is the terminal solid ring.

Wouldn’t the terminal solid ring simply be the trivial ring?

### Re: Solid Rings

Gosh, I guess so! I don’t recall Bousfeld and Kan talking about that one: they talk about $\mathbb{Z}/p$ for $p$ prime, but not I think $\mathbb{Z}/1$.

So, maybe the solid ring I’m talking about is the ‘preterminal’ solid ring: terminal if we remove the terminal ring.

Posted by: John Baez on September 27, 2021 12:38 AM | Permalink | Reply to this

### Re: Solid Rings

It would be funny if it did nothing.

I think this is indeed true.

Taking the core does nothing on any of $\mathbb{Z}/p$’s or $\mathbb{Q}$. This is basically the same as saying that $k \otimes_{\mathbb{Z}} k$ is isomorphic to $k$, and this is true for a field $k$ if (and in fact, which relates I believe to both your guess and to Tom’s comment, only if) $k$ is $\mathbb{Q}$ or a $\mathbb{Z}/p$.

Once one has this, one just observes that one can work levelwise.

Posted by: Richard Williamson on September 22, 2021 12:33 PM | Permalink | Reply to this

### Re: Solid Rings

I was just reasoning directly from the definitions, but realised later that essentially everything in my second paragraph was already noted at one of the links earlier in the thread; apologies for that!

I also was too hasty with regard to reasoning levelwise. What I had in mind was that the proof that $\mathbb{Z} / p$ is solid, for instance, is fairly trivial:

\begin{aligned} 1 \otimes x &= x \cdot x^{-1} \otimes x \\ &= \underbrace{x + \ldots + x}_{x^{-1}} \otimes x \\ &= \underbrace{x \otimes x + \ldots + x \otimes x}_{x^{-1}} \\ &= x \otimes \underbrace{x + \ldots + x}_{x^{-1}} \\ &= x \otimes x \cdot x^{-1} = x \otimes 1. \end{aligned}

The same kind of argument works for $\mathbb{Q}$.

Given some $(q,r_1,r_2, \ldots)$ where $r_i \in \mathbb{Z} / p_i$ for $p_i$ the $i$-th prime, we can carry out the same kind of argument one component at a time. However, one of course does need to stop at some point, since the axioms of a tensor product only involve pairwise (i.e. finite) additions; I was too quick, and omitted to take this into consideration in my initial comment.

I now no longer think it is true that the core does nothing on $\mathbb{Q} \times \prod_{p \in \mathcal{P}} \mathbb{Z} / p$. I think it might be that the core consists of elements of the form $(q, r_1, r_2, \ldots,)$ where only finitely many of the $r_i's$ are not $1$. This matches well the fact that Bousfield and Kan observe that one has to be able to build the core as a colimit from the basic examples of solid rings, and infinite coproducts of rings have this kind of ‘only finitely many not 1’ construction.

One might try to construct some elements in the core not of this form by for example letting $r_i = \frac{p_i + 1}{2}$ for all $i \gt 1$ (so that $2$ is a multiplicative inverse of $r_i$ for all $i \gt 1$), which almost allows one to carry out the same kind of argument as above for all components simultaneously, except that there is nothing one can set $r_1$ to for which it will go through. I can’t think of any other way that one could do it, although I’m not sure how one could rigorously prove that it is not possible, except possibly by some kind of appeal to the fact about coproducts that I mentioned in the previous paragraph.

Posted by: Richard Williamson on September 23, 2021 11:58 PM | Permalink | Reply to this

### Re: Solid Rings

Here’s a proof that the ring

$R = \mathbb{Q} \times \prod_{p \in \mathcal{P}} \mathbb{Z}/p$

is not solid, i.e. $c(R) \neq R$. It’s more elaborate than I’d like, but here goes anyway.

For convenience, I’ll write $\mathbb{Q}$ as $\mathbb{Z}/\infty$, so that

$R = \prod_{p \in \mathcal{P} \cup \{\infty\}} \mathbb{Z}/p.$

Since there are infinitely many primes $\equiv 1 \pmod{4}$, we can choose a nonprincipal ultrafilter $\mathcal{U}$ on $\mathcal{P} \cup \{\infty\}$ containing the set of such primes.

Now take the ultraproduct

$F = \bigg(\prod_{p \in \mathcal{P} \cup \{\infty\}} \mathbb{Z}/p\bigg)/\mathcal{U}.$

In other words, $F$ is the quotient of $R$ by the equivalence relation that identifies two sequences when the set of indices where they’re equal belongs to $\mathcal{U}$. It inherits a ring structure from $R$. In fact, it’s a field, since each of the factors $\mathbb{Z}/p$ is.

The field $F$ has characteristic $0$, since whenever $p$ is a prime, only one coordinate of $p \cdot (1, 1, \ldots)$ is zero (and $\mathcal{U}$ is nonprincipal).

The field $F$ also contains a square root of $-1$, since $\mathbb{Z}/p$ has one for all primes $p \equiv 1 \pmod{4}$ (e.g. $((p - 1)/2)!$), and the set of such primes is in $\mathcal{U}$.

As a preview of where we’re going, we’re going to consider maps

$R \twoheadrightarrow F \hookrightarrow \overline{F} \hookrightarrow \mathbb{C} \rightrightarrows \mathbb{C},$

as I’ll explain.

The algebraic closure $\overline{F}$ of $F$ has the same cardinality as $F$, since $F$ is infinite. I don’t know what that cardinality is (anyone?), but certainly it’s at most the cardinality of $R$, which is $2^{\aleph_0}$. I’ll now show that whatever its cardinality may be, $\overline{F}$ embeds in $\mathbb{C}$.

• Suppose that $F$ has cardinality $2^{\aleph_0}$. Then so does $\overline{F}$. It’s a theorem that any two algebraically closed fields of the same characteristic and the same uncountable cardinality must be isomorphic. (The main point of the proof is that the cardinality of an uncountable field is equal to its transcendence degree over the prime subfield.) So $\overline{F} \cong \mathbb{C}$.

• Suppose that $F$ has cardinality $\lt 2^{\aleph_0}$. Then so does $\overline{F}$. For any two algebraically closed fields of the same characteristic, one embeds in the other. In this case, on cardinality grounds, it must be that $\overline{F}$ embeds in $\mathbb{C}$.

So in either case, $\overline{F}$ embeds in $\mathbb{C}$, as claimed. Choose a particular embedding. So we now have ring homomorphisms

$R \twoheadrightarrow F \hookrightarrow \overline{F} \hookrightarrow \mathbb{C}.$

Write $\phi: R \to \mathbb{C}$ for this composite. Since $F$ contains a square root of $-1$, we can choose some $r \in R$ such that $\phi(r) = \pm i$.

Finally, complex conjugation and the identity define two homomorphisms $\mathbb{C} \rightrightarrows \mathbb{C}$, which when composed with $\phi$ give two homomorphisms $\phi, \overline{\phi}: R \rightrightarrows \mathbb{C}$. But $\phi(r) = \pm i$, so $\phi(r) \neq \overline{\phi}(r)$. Hence $r$ is not absolute and $R$ is not solid.

Posted by: Tom Leinster on September 25, 2021 9:11 PM | Permalink | Reply to this

### Re: Solid Rings

Posted by: John Baez on September 26, 2021 6:58 PM | Permalink | Reply to this

### Re: Solid Rings

Posted by: Tom Leinster on September 26, 2021 10:31 PM | Permalink | Reply to this

### Re: Solid Rings

On second thoughts, I could take your question more seriously and try to describe the idea of the proof.

We’re trying to show that $R := \mathbb{Q} \times \prod \mathbb{Z}/p$ is not solid, which means showing that there exist a ring $S$ and distinct homomorphisms $\phi, \psi: R \rightrightarrows S$. And to say that $\phi$ and $\psi$ are distinct means, of course, that there exists $r \in R$ such that $\phi(r) \neq \psi(r)$. I had a hunch that it wasn’t going to be possible to give an actual construction of $S$, $\phi$, $\psi$ and $r$ — that we’d really need the axiom of choice.

So, I thought of replacing the product in the definition of $R$ with an ultraproduct. This gives a field, $F$, and a quotient map $R \twoheadrightarrow F$. The challenge now is to produce two distinct homomorphisms out of $F$.

We don’t have much control over what $F$ is. But we do know its characteristic ($0$) and we have an upper bound on its cardinality ($2^{\aleph_0}$). These facts alone imply that $F$ is a subfield of $\mathbb{C}$.

(Here I’m drawing a veil over my actual, messy, thought process and pretending that I already knew the following general result. Let $F$ and $K$ be fields of the same characteristic, with $K$ algebraically closed and uncountable. If $card(F) \leq card(K)$ then $F$ embeds in $K$. I gave the proof of this in the case $K = \mathbb{C}$, but the general proof is the same.)

So we have $R \twoheadrightarrow F \hookrightarrow \mathbb{C}$. But we want two homomorphisms $R \to \mathbb{C}$, not just one. We can get two by composing with an automorphism of $\mathbb{C}$, and the obvious choice is complex conjugation.

The final point is that conjugation might not have any effect on $F$; that is, $F$ might be $\subseteq \mathbb{R}$. That would spoil the plan. To avoid that, we make sure $F$ contains a square root of $-1$. And going back to the definition of $F$ as an ultraproduct, we can arrange this as long as there are infinitely many primes $p$ such that $-1$ has a square root mod $p$. Which, fortunately, there are.

As I said, this proof does seem very elaborate. Probably there’s a simpler way.

Posted by: Tom Leinster on September 26, 2021 11:52 PM | Permalink | Reply to this

### Re: Solid Rings

An anonymous MathOverflow contributor found a much simpler, choice-free, proof that the ring

$\mathbb{Q} \times \prod_p \mathbb{Z}/p$

(where the product runs over prime numbers $p$) is not solid. Their proof is in the comments to this answer. I’ll give an expanded version of it here.

Recall that to say $\mathbb{Q} \times \prod_p \mathbb{Z}/p$ is not solid is to say that there exist a ring $S$ and two different homomorphisms

$\mathbb{Q} \times \prod_p \mathbb{Z}/p \rightrightarrows S.$

Here’s how the simpler proof goes.

• The ring $\prod_p \mathbb{Z}/p$ has an ideal $\bigoplus_p \mathbb{Z}/p$ consisting of the elements $(x_p) \in \prod_p \mathbb{Z}/p$ such that $x_p = 0$ for all but finitely many primes $p$. It’s a proper ideal, since there are infinitely many primes.

• Let $S$ be the quotient ring $\bigl( \prod \mathbb{Z}/p\bigr) / \bigl( \bigoplus \mathbb{Z}/p \bigr)$. Then the ring $S$ is nontrivial, since the ideal by which we’re quotienting is proper.

• In the ring $S$, each nonzero integer $n$ is invertible, since $n$ is invertible in $\mathbb{Z}/p$ for all but finitely many primes $p$ (its prime factors).

• It follows that the unique homomorphism $\mathbb{Z} \to S$ extends uniquely to a homomorphism $\mathbb{Q} \to S$. (You can view this as an instance of the universal property of fields of fractions.)

• We now have two homomorphisms $\mathbb{Q} \times \prod_p \mathbb{Z}/p \rightrightarrows S$. The first projects onto the first factor and then applies the map $\mathbb{Q} \to S$ just mentioned. The second projects onto the second factor then applies the quotient map $\prod_p \mathbb{Z}/p \to S$.

• These two homomorphisms are distinct. Indeed, let’s evaluate each of them at the element $(1, (0, 0, \ldots))$ of $\mathbb{Q} \times \prod_p \mathbb{Z}/p$. The first homomorphism outputs $1_S \in S$, and the second outputs $0_S \in S$. And since the ring $S$ is nontrivial, $1_S \neq 0_S$.

Posted by: Tom Leinster on July 25, 2023 1:33 PM | Permalink | Reply to this

### Re: Solid Rings

Nice! The mills of math grind slowly, yet they grind exceedingly small.

Posted by: John Baez on August 1, 2023 2:45 PM | Permalink | Reply to this

### Re: Solid Rings

Not only do mathematicians use words in funny ways, but they can’t agree with each other, or even with themselves, about how to use words.

Momentarily caught by surprise by your statement

Epimorphisms of rings aren’t always onto.

I checked the first abstract algebra book I grabbed off my shelf (Hungerford’s Algebra) and found that, as I thought, it defines a ring epimorphism to be an onto ring homomorphism. But in a later chapter it defines an epimorphism in a category to be an epic morphism, and uses the same example you did to illustrate that with these definitions, an epimorphism in the category of rings is not necessarily a ring epimorphism. Yuck.

This feels to me like an algebraist’s version of the stupid terminological knots we analysts get tied up in with the word “positive”.

Posted by: Mark Meckes on September 17, 2021 2:59 AM | Permalink | Reply to this

### Re: Solid 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.

Posted by: John Baez on September 17, 2021 3:42 AM | Permalink | Reply to this

### Re: Solid Rings

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.

Posted by: Tom Leinster on September 21, 2021 10:29 AM | Permalink | Reply to this

### Re: Solid Rings

That’s all very satisfactory, and it makes me wonder whether the same techniques or results can be used to solve a problem I came across years ago.

Say that an element $r$ of a ring $R$ is absolute if $f(r) = g(r)$ whenever $f$ and $g$ are homomorphisms from $R$ to another ring $S$. So absoluteness means that whenever $S$ is a ring to which $R$ admits a homomorphism, there’s a well-defined image of $r$ in $S$, independent of the homomorphism chosen.

Can you classify or describe the absolute elements of an arbitrary ring?

Obviously $0$ and $1$ are absolute in any ring $R$. Hence $n \in R$ is absolute for all $n \in \mathbb{Z}$. In fact, any ring element $r$ is absolute if it’s “rational”, meaning that $n \cdot r = m$ for some coprime integers $m, n$.

But not every absolute element is rational. A friend found a counterexample which I don’t remember just now.

Absoluteness seems like it should be related to solidity, surely…?

Posted by: Tom Leinster on September 17, 2021 10:18 AM | Permalink | Reply to this

### Re: Solid Rings

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

Posted by: John Baez on September 17, 2021 5:54 PM | Permalink | Reply to this

### Re: Solid Rings

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!

Posted by: Tom Leinster on September 17, 2021 7:21 PM | Permalink | Reply to this

### Re: Solid Rings

Nice! My “argument” was not only longer, it didn’t reach the desired conclusion: it was just some stuff that made me suspect the conclusion was true.

Posted by: John Baez on September 18, 2021 10:23 PM | Permalink | Reply to this

### Re: Solid Rings

If someone could check whether there’s anything to add to the nLab page core of a ring that would be helpful.

You’ll also see there a hint to an anlogue in higher algebra and a reference to a generalisation to monoids in monoidal categories.

Posted by: David Corfield on September 17, 2021 2:17 PM | Permalink | Reply to this

### Re: Solid Rings

There are a few related questions I’d like to settle, that would make nice additions to the nLab page:

• Does the category CommRing have a (epi, regular mono) factorization system? How about Ring? If so, material in the nLab implies that applying this to $\mathbb{Z} \to R$ gives the core of $R$.

• Given a subring inclusion $i \colon S \to R$, does the ‘dominion’ of $S$ in $R$ provide the (epi, regular mono) factorization of $i$? I see now that Anton Geraschenko writes:

George Bergman gave me a reference (Isbell’s “[Epimorphisms and dominions, IV]”) and a very pretty counterexample. In particular, he says that the characterization of epimorphisms Andrew gave us works for non-commutative rings as well:

Recall that an inclusion A in B is an epimorphism if and only if the “dominion” of A in B is all of B, where this dominion is defined as the subring of elements b of B which behave the same under all pairs of homomorphisms on B that agree on elements of A.

Now the Silver–Mazet–Isbell Zigzag Lemma for rings says that the dominion of A in B consists of those elements of B which can be written XYZ, where X is a row, Y a matrix, and Z a column over B, such that XY and YZ have entries in A. (It is easy to verify that such a product is in the dominion of A – a generalization of the proof that if Y is in A and has an inverse in B, then this inverse is in the dominion of A.)

• Does the core of $R$ consist exactly of the ‘absolute’ elements of $R$, as defined by Tom? Note that his absolute elements of $R$ are, by definition, the dominion of the image of $\mathbb{Z} \to R$.

If the answers to all these are yes, the concepts of ‘solid ring’, ‘core’, ‘dominion’, ‘absolute element’ and ‘Silver–Mazet–Isbell Zigzag Lemma’ should all be thought of as aspects of an (epi, regular mono) factorization system for CommRing — and maybe also Ring.

If so, spelling this out would make everything a bit clearer, a bit less of a mess. (Maybe Isbell already did it!)

Posted by: John Baez on September 17, 2021 6:58 PM | Permalink | Reply to this

### Re: Solid Rings

Is there any relationship between solid rings as described in this blog post and Peter Scholze’s solid objects (abelian groups, modules, etc) in his condensed mathematics?