TITLE: Morphisms in the category of rings
QUESTION [9 upvotes]: We know that in the category of (unitary) rings, $\mathbb{Z}$ is the itinial object, i.e. it is the only ring such that for each ring $A,$ there exists a unique ring homomorphism $f:\mathbb{Z} \to A$.
This means, in particular, that $\mathbb{R}$ does not satisfy this property, so for a certain ring $B$, we can construct $g:\mathbb{R} \to B$ and $h:\mathbb{R} \to B$ such that $g \ne h$.
So far, I have proven that $B$ cannot be an ordered ring and cannot be $\mathbb{R}^n \ (n\in\mathbb{N})$. Can you help me finding such a ring $B$?

REPLY [7 votes]: Many useful and (sometimes surprisingly) true things have been said in other answers; let me just make some simple remarks.
To begin, you could have consulted the proof of what you implicitly used, namely that if $\def\Z{\Bbb Z}\Z$ is an initial object, then it is automatically unique, up to canonical isomorphism (indeed, for once, up to unique isomorphism). The (very standard) proof goes: suppose $R$ is another initial object in the category, then by assumption there are unique morphisms $f:\Z\to R$ and $g:R\to\Z$. Then $g\circ f:\Z\to\Z$ is a morphism and $\Z$ being initial it is unique and therefore equal to the identity of$~\Z$; similarly $f\circ g:R\to R$ is a morphism and $R$ being initial it is unique and therefore equal to the identity of$~R$, from which it follows that $f$ and $g$ are inverse morphisms, so $R$ is isomorphic to $Z$.
Now taking $\def\R{\Bbb R}R=\R$ it is clear that $f:\Z\to\R$ fails to be surjective so cannot have an inverse; by (the contrapositive of) the above argument this shows that $\R$ cannot be initial, and it does not admit a morphism $g:\R\to\Z$ at all. [That is a small lie. The argument uses the hypothesis that $R$ is initial twice: for the existence of $g$, and for equating $f\circ g$ to the identity of$~R$; one or the other must fail. But it is fairly easy to see that for $R=\R$ the former fails, and indeed one does have the property that any morphism $\R\to\R$ equals the identity, though it takes a bit of works to show that.]

Next, and rather unrelated, some words about the tensor product proof that $\R$ does not even have, as the above might suggest, the weaker property that if a morphism $\R\to A$ exists, then it is unique. You can think of the $\otimes$ in the construction of the ring $S\otimes S$ as similar to the fraction bar used to construct $\Bbb Q$, with the following difference: in fractions we decree that one can cancel (or introduce) equal factors ($an/nb=a/b$), but for the tensor product one decrees that one can move a factor across the $\otimes$ symbol in either direction ($an\otimes b=a\otimes nb$). The difference has as consequence that while sums of fractions can always be combined to a single fraction, sum of tensors cannot always.
The factors allowed to be moved are integers (the bare minimum), or possibly some larger class $R$ of scalars, which must then be attached as $\otimes_R$ to indicate this attribute of the construction; it serves as a gatekeeper that allows elements of (the ring) $R$ to cross the $\otimes_R$, but not others. Nonetheless, even for a bare $\otimes_\Z$, one can effectively pass rational factors across: to move $p/q$ from left to right, it suffices to move a factor $p$ from left to right and a factor $q$ from right to left (therefore $\otimes_\Bbb Q$ is not really different from $\otimes_\Z$). However one cannot go beyond rationals: $\sqrt2\otimes_\Z1$ and $1\otimes_\Z\sqrt2$ are different, for if they could be shown equal using a finite number of tensor-equivalences, one would basically have produced a fraction $p/q$ equal to $\sqrt2$, which is impossible. (A formal proof requires work.)
As a concluding remark: the ring $\Bbb Q$ and the finite fields $\Z/p\Z$ with $p$ prime do have the property that morphisms from them are unique if they exist. These are exactly the prime fields: fields without any proper sub-fields.