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What is the set $S$ of rings $R$ such that, for all rings $R'$, there is at most one nonzero homomorphism $R \to R'$?

We're dealing only with commutative rings with unity, and our definition of ring homomorphisms require $1_R \mapsto 1_{R'}$, and we're excluding considering the zero homomorphism. We know $\mathbb Z \in S$ and also $\mathbb Z/n\mathbb Z \in S$, which leads us to posit that any element of $S$ will have some special property $\mathbb Z$ and $\mathbb Z/n\mathbb Z$ have that makes them have a unique homomorphism, if it exists. After some thinking, I've come up with the possible hypothesis that the special property is the following: $R \in S$ if and only if any $r \in R$ is a finite sum of $1_R$, which will imply membership of $S$ by virtue of the property of ring homomorphisms forcing $1_R \mapsto 1_{R'}$.

It turns out that all such rings are indeed members of $S$, but I'm uncertain whether the converse holds.

Any tips, hints, or thoughts?

shintuku
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  • Thoughts: 1) is it a problem you are allowing the zero homomorphism into $R'$ from every ring? 2) Most people read "given an arbitrary ring R' " as "for all rings $R'$". Is that what you meant? Or did you just mean to consider when $Hom(R,R')$ is very simple for two particular fixed rings $R,R'$ in the category of rings? Otherwise question 1) becomes important again when you hit $R'=R$. If you allow the zero homomorphism and are quantifying over all rings, then nothing except the zero ring itself works. – rschwieb Jun 07 '23 at 18:09
  • $R = \mathbb{Q}$ is an example of a ring satisfying this property, but which does not satisfy the hypothesis that every element is a finite sum of $1_R$. (Also, how do you count $\mathbb{Z}$ as satisfying your alternate property?) – Daniel Schepler Jun 07 '23 at 18:14
  • @rschwieb thanks a lot for commenting; unfortunately I haven't done category theory yet so I can't exactly parse your phrasing, but (1) yes this turns out to be a problem, I'll modify the question accordingly, and (2) yes I understand "given an arbitrary ring R'" as you say. Thanks a lot for your comment! – shintuku Jun 07 '23 at 18:16
  • @DanielSchepler thanks for your comment! I'll check $\mathbb Q$, the counterexample is much appreciated. For $\mathbb Z$, if I understood your question correctly, we have $z \in \mathbb Z$ has $z = z\cdot 1 = 1 + \cdots + 1$ a total of $z$ times – shintuku Jun 07 '23 at 18:18
  • But what if $z$ is negative? – Daniel Schepler Jun 07 '23 at 18:20
  • @DanielSchepler hm... you're right. We have $-1 - \cdots - 1$ a total of $z$ times but this is not the same as the general property I've stated. I will rethink this a bit, your thoughts are much appreciated – shintuku Jun 07 '23 at 18:21
  • A trivial remark is that $R$ has this property if and only if every subring of $R$ has this property. So by Daniel Schepler's comment any subring of $\mathbb{Q}$ has this property – math54321 Jun 08 '23 at 02:07
  • How do we know $\Bbb Q$ has this property? – Karl Jun 08 '23 at 04:10
  • The rings in this set have only one ring automorphism, the identity. – lhf Jun 08 '23 at 13:40

1 Answers1

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Let $l,r:R\to R\otimes_{\mathbb{Z}}R$ be the ring homomorphisms given by $l(x)=x\otimes 1$ and $r(x)=1\otimes x$ for $x\in R$.

Suppose that there are two homomorphisms $\alpha,\beta:R\to R'$ for some $R'$. Then defining $\gamma:R\otimes_{\mathbb{Z}}R\to R'$ by $\gamma(x\otimes y)=\alpha(x)\beta(y)$, we get $\gamma\circ l=\alpha$ and $\gamma\circ r=\beta$, so if $\alpha$ and $\beta$ are distinct, then so are $l$ and $r$.

So the property asked for is equivalent to $l$ being equal to $r$.

This is one of many well-known properties equivalent to the unique ring homomorphism $\mathbb{Z}\to R$ being a ring epimorphism (an epimorphism in the category of rings, which is not the same thing as a surjective ring homomorphism).

Rings $R$ for which $\mathbb{Z}\to R$ is a ring epimorphism have been called solid rings, and were classified by Bousfield and Kan. See this thread for more details and references, but some examples of solid rings include

  • $\mathbb{Z}/n\mathbb{Z}$ for any $n$,
  • any subring of $\mathbb{Q}$, such subrings being of the form $\mathbb{Z}[P^{-1}]$, the rationals whose denominator is only divisible by primes in $P$, for $P$ any set of primes,
  • $\mathbb{Z}[P^{-1}]\times \mathbb{Z}/n\mathbb{Z}$, where all the prime factors of $n$ are in the set $P$.
Jeremy Rickard
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