We have a canonical homomorphism from a ring $R$ to its field of fractions $Q(R)$ given by $f : r\mapsto r/1$.
Is this homomorphism unique?
So letting $g: R\rightarrow Q(R)$ be a homomorphism, by the universal property of the field of fractions, there exists a unique homomorphism $h:Q(R)\rightarrow Q(R)$ such that $g = h\circ f$. By this universal property it follows that it should be enough to prove that $h$ is the identity map, but I am not sure if this has to be the case.
Added part: As follows from the answers/discussion below, in general the above is not true. Now I wonder, is it true when $R$ is a discrete valuation ring?
Best Answer
No, it need not be unique in general. To see why, consider the chain of maps
$$R\xrightarrow{g} R\xrightarrow{f} Q(R)\xrightarrow{h}Q(R).$$
We can “toggle” what $g$ and $h$ are to get different maps from $R$ to $Q(R)$. Since $R$ and $Q(R)$ might have nontrivial endomorphisms, your canonical morphism won’t always be unique.
I wonder when it is unique, though. Certainly $\mathbb{Z}\hookrightarrow\mathbb{Q}$ is unique, but I’m not sure about other rings...