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?
Best Answer
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