Morphisms in the category of rings 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 $B$?

Let $\phi:\Bbb R\to B$ be a ring morphism. Then $\ker\phi$ is an ideal of $\Bbb R$. But the only ideals of $\Bbb R$ are $\Bbb R$ and $\langle 0\rangle$. Thus $\ker\phi=\Bbb R$ or $\ker\phi=0$.
Suppose $\ker\phi=\Bbb R$. Then $B$ must be the trivial ring since $\phi(1)=1_B$ and the only ring satisfying $1_B=0_B$ is the trivial ring. In this case $\phi:\Bbb R\to B$ is unique.
Suppose $\ker\phi=0$. Then $\phi$ is injective. It follows that $\DeclareMathOperator{image}{image}\image(\phi)$ is a subring of $B$ isomorphic to $\Bbb R$. 
This proves that every nontrivial ring $B$ without a subring isomorphic to $\Bbb R$ admits no homomorphisms $\phi:\Bbb R\to B$.
Examples include $B=\Bbb Z$, $B=\Bbb Z_n$, $B=\Bbb Q$.
