Ring homomorphisms from $\Bbb Q$ into a ring 
Let $A$ be a ring. I'm trying to prove that there is only one ring homomorphism (different from the zero one) from $\Bbb Q$ into $A$ or there are no ring homomorphisms between $\Bbb Q$ and $A$. 

I have proven that if $A=\Bbb Z$ or $A=\Bbb R$ then there are no ring homomorphisms between $A$ and $\Bbb R$ (not taking into account the zero one) but I don't know how to prove it for any $A$.
 A: If you want to avoid invoking the universality of localization, you can show the result directly (although the demonstration here is pretty much the same argument as in the proof of that property). 
Let $f:\mathbb{Q} \to A$ be a ring homomorphism. Then $f$ restricts to a map $\mathbb{Q}^\times \to A^\times$, so every nonzero $n\in \mathbb{Z}$ must be invertible in $A$. Conversely, suppose every nonzero $n\in \mathbb{Z}$ lies in $A^\times.$ Then any $f:\mathbb{Q} \to A$ must have $f(n) = f(1) + \cdots + f(1) = n$ and thus $f(p/q) = f(p)/f(q) = p/q\in A$ for all $p, q$ with $q\not = 0$.
A: There is always a ring homomorphism $\Bbb Z\to A$ for any ring $A$. Now consider the universal property of localizations, that is: we will be able to pass to $K(\Bbb Z)=\Bbb Q$ iff... what?
A: Let $\phi$ be a ring homomorphism from $\mathbb{Q}$ into $A$, $\phi:\mathbb{Q}\rightarrow A$. Then ker $\phi= \{q\in \mathbb{Q},~\phi(q)=0_A\}$ is an ideal of $\mathbb{Q}$, and since $\mathbb{Q}$ is a field, the only ideals are $(0)$ and $\mathbb{Q}$ itself. So the only homomorphisms are either the zero homomorphism or one other which is injective. 
A: As you specifically exclude the zero homomorphism, you seem to accept that $x\mapsto 0$ is a ring homomorphism, i.e., you do not require a priori that $f(1)=1$ for all ring homomorphisms (or maybe you do not even require that all your rings are unital). Let $A=\mathbb Q\times \mathbb Q$. Then $x\mapsto(x,0)$ and $x\mapsto (0,x)$ and $x\mapsto (x,x)$ are three distinct nonzero homomorphisms (only the third one maps $1\in\mathbb Q$ to $1\in A$).
