An exercise from Hungerford's Algebra.
Let $\mathbb Q$ be the field of rational numbers and $R$ any ring. If $f,g:\mathbb Q\rightarrow R$ are homomorphims of rings such that $f=g$ on $\mathbb Z$, then $f=g$ on $\mathbb Q$.
The hint given by author asks me to show $f(1/n)g(n)=g(1)$, which is easy to check, and hence $f(1/n)=g(1/n)$, where I get stuck because I think $nf(1/n)=ng(1/n)$ may not imply $f(1/n)=g(1/n)$ since $R$ can have character $\neq0$.
Can anyone help me to figure it out?
From $$f(1/n)g(n)=g(1)$$ multiply by $g(1/n)$ to get $$f(1/n)g(n)g(1/n)=g(1)g(1/n).$$ As $g$ is a homomorphism, this is $$f(1/n)g(1)=g(1/n)$$ and also $g(1)=1$.
Hint: fix a rational number $p/q$ with $p$ and $q$ integers. Using the relations of $f$ with respect to inverses and products, express $f(p/q)$ in terms of $f(p)$ and $f(q)$. Now use your hypothesis and proceed backwards but with the other morphism $g$.
Directly using that $f$ and $g$ are morphisms, for any fraction $p/q$ we have that $f(p/q) = f(pq^{-1}) = f(p)f(q)^{-1} = g(p)g(q)^{-1} = g(pq^{-1}) = g(p/q).$
