When is the ring homomorphism $\mathbb{Z} \to R$ an epimorphism?

Question

For every ring $R$ there exists a unique ring homomorphism $f: \mathbb{Z} \to R$. For which rings is this an epimorphism?

The only examples I know are the subrings of $\mathbb{Q}$ and the quotient rings of $\mathbb{Z}$, namely the rings $\mathbb{Z}/\langle n \rangle$. Are these all, or are there more?

Three characterizations of epimorphisms of commutative rings are listed here:

• Epimorphisms of rings, The Stacks Project.

Maybe one will help.

(Why I'm interested: an object in a category is subterminal if its unique morphism to the terminal object is a monomorphism. Since $\mathbb{Z}$ is initial in $\mathrm{Ring}$, here I am asking what are the subterminal objects in $\mathrm{Ring}^{\rm op}$.)

Answer 1

Let me summarize what I've learned from the comments. A ring $R$ for which the unique homomorphism from $\mathbb{Z}$ to $R$ is an epimorphism is called a solid ring. Such rings are necessarily commutative, by Prop. 1.3 (b) of this paper:

• H. H. Storrer, Epimorphic extensions of non-commutative rings, Commentarii Mathematici Helvetici 48 (1973), 72–86.

By well-known results in commutative ring theory, a solid ring is thus the same as any of these:

• a commutative ring $R$ for which the multiplication map $m \colon R \otimes_{\mathbb{Z}} R \to R$ is an isomorphism;

• a commutative ring $R$ such that the forgetful functor $R\, \mathsf{Mod} \to \mathsf{AbGp}$ is full.

• a commutative ring whose core $ cR = \{r \in R: r \otimes 1 = 1 \otimes r \in R \otimes_\mathbb{Z} R \} $ is all of $R$.

Commutative solid rings, and thus all solid rings, were classified here:

• A. K. Bousfield and D. M. Kan, The core of a ring, Journal of Pure and Applied Algebra 2 (1972), 73–81.

The following rings are solid:

1. $\mathbb{Z}/n$ for any $n$.
2. any subring $R \subseteq \mathbb{Q}$. Such a subring is always of the form $\mathbb{Z}[P^{-1}]$, meaning the ring of fractions whose denominators (in lowest terms) are divisible only by the primes in some set $P$ of primes.
3. any ring of the form $\mathbb{Z}[P^{-1}] \times \mathbb{Z}/n$ where each prime factor of $n$ is in $P$.

Bousfeld and Kan show that in the category of commutative rings, every colimit of solid rings is solid. They also show that every solid ring is a colimit, in the category of commutative rings, of solid rings of the above three types.

Bousfeld and Kan also give a more explicit description of all the solid rings. They show that every solid ring is either of types 1-3 or of a fourth type:

4. $c(\mathbb{Z}[P^{-1}] \times \prod_{p \in Q} \mathbb{Z}/p^{e(p)}) $, where $P$ and $Q$ are infinite sets of primes with $Q \subseteq P$ and each $e(p)$ is a positive integer. (Here $c$ stands for the core, as defined above.)

Answer 2

I believe we can give a more conceptual characterization of both cores and solid rings using the concept of weak inverses. Here is the idea:

Definition: A weak inverse of an element $x$ of a monoid $M$ is an element $y$ satisfying $x = xyx$ and $y = yxy$.

This condition is part of the definition of an inverse semigroup. The motivating example for me is that in a product $\prod_i K_i$ of fields all weak inverses exist and are given by the operation which pointwise either computes the ordinary inverse (if the corresponding coefficient is nonzero) or zero (if the corresponding coefficient is zero). A commutative ring with all weak inverses is the same as a von Neumann regular or absolutely flat commutative ring, and is sometimes called a meadow, generalizing the terminology "field." Unlike fields, meadows are models of an algebraic theory, where we add a unary operator for weak inverse!

Proposition: If $M$ is commutative, then weak inverses are unique if they exist.

For a proof see the link; it is short and similar to the proof that ordinary inverses are unique when they exist. So, for example, an invertible element has weak inverse given by its ordinary inverse, and $0$ has weak inverse $0$.

Corollary: Let $R$ be a commutative ring and $S \subseteq R$ be any subset. Consider the weak localization $R[S^{\ast}]$ of $R$ at $S$, namely the ring obtained from $R$ by adjoining weak inverses $s^{\ast}$ to every $s \in S$. Then:

1. The map $R \to R[S^{\ast}]$ is an epimorphism of commutative rings.
2. If $S \subseteq \mathbb{Z}$ is any subset of $\mathbb{Z}$, then $\mathbb{Z}[S^{\ast}]$ is solid.

To connect this to the Bousfield-Kan classification, if $R$ is a commutative ring and $r \in R$ it's not hard to show that the weak localization $R[r^{\ast}]$ is given by

$$R[r^{\ast}] \cong R/r \times R[r^{-1}]$$

so it is the product of the quotient and the ordinary localization. This uses the fact that weak localization produces an idempotent $e_r = r^{\ast} r$ that splits the ring up. But we get something more complicated if we weak localize at infinitely many elements.

I believe the following should not be too hard to prove from here:

Claim:

1. Every solid ring is either a quotient $\mathbb{Z}/n$ of $\mathbb{Z}$, or a combination of a weak localization and a localization of $\mathbb{Z}$.
2. The core of a commutative ring $R$ consists of the image of $\mathbb{Z}$ together with all weak inverses of elements of $\mathbb{Z}$ that exist in $R$.

The weirdest and least familiar example is the solid ring produced by weakly inverting every element of $\mathbb{Z}$, or equivalently every prime; we might call this the meadowization

$$\widetilde{\mathbb{Z}} = \mathbb{Z}[p^{\ast}]/(p - p p^{\ast} p, p^{\ast} - p^{\ast} p p^{\ast})$$

of $\mathbb{Z}$. Here by "meadowization" I mean the left adjoint of the inclusion of meadows into commutative rings, so this is the initial meadow. In terms of the Bousfield-Kan classification it is the core of $\mathbb{Q} \times \prod_p \mathbb{Z}/p$, but in terms of Claim 2 we can describe this more explicitly: it consists of the image of $\mathbb{Z}$ inside this product together with the weak inverses $p^{\ast} \in \mathbb{Q} \times \prod_p \mathbb{Z}/p$, whose component at a prime $q$ is $p^{-1} \bmod q$ if $q \neq p$ and $0$ if $q = p$, and whose component at the generic point is $p^{-1} \in \mathbb{Q}$ (I suspect we can omit $\mathbb{Q}$ from this description but its presence emphasizes that $\widetilde{\mathbb{Z}}$ does in fact admit $\mathbb{Q}$ as a quotient).