Let $K$ be a field of characteristic $0$. Show that there is a unique ring homomorphism $\mathbb{Q}\to K$. 
Let $K$ be a field of characteristic $0$. Show that there is a unique ring homomorphism $\mathbb{Q}\to K$.

I know that, considering $f$ as a homomorphism, $f$ must be the zero-homomorphism or an injective one, but I wasn't able to prove that an injective homomorphism is unique!
 A: Presumably, a ring homomorphism is expected to take $1$ to $1$, and as such it cannot be the zero map with a nonzero target ring. That means that you know what $1$ has to map to. That should allow you to deduce what $\mathbb{Z}$ must be mapped to, and from there, what $\frac{1}{n}$ needs to be mapped to for each $n$, and from there what each rational needs to be mapped to.
If you do not know that a ring homomorphism must take $1$ to $1$, you can show that $f(1)$ here must be either $0$ or $1_K$. Because $1^2=1$, so $f(1)$ must be a root of $x^2-x\in K[x]$. 
A: Since $\Bbb Q$ happens to be a field itself, its only ideals are $0$ and itself.  Thus, any nontrivial ring homomorphism from $\Bbb Q$ is injective (as you did mention).  Thus $\varphi $ is indeed an embedding of $\Bbb Q$ into $\Bbb K$.  
But there can only be one such embedding.  For there's only one automorphism of $\Bbb Q$, and any other embedding would differ from $\varphi$ by an automorphism.
A: Hope this helps you.
Define $\Psi:\mathbb{Z}\rightarrow K$ by $\Psi(n)=n\cdot 1,~\forall n\in\mathbb{Z}$, then $\Psi$ is a ring homomorphism, and since the characteristic of $K$ is $0$ consequently is one-to-one. Hence $\mathbb{Z}\hookrightarrow K$.
Then there exists a unique embbeding $f:\mathbb{Q}\hookrightarrow K$ such that $f\circ\varphi=\Psi$ where $\varphi$ is the natural embedding from $\mathbb{Z}$ to $\mathbb{Q}$.
Generally speaking the field of quotients of an integral domain is the smallest field in which it can be embbeded. So if an integral domain $D$ is embbeded in a field $K$ there must be an intermediate embbeding $$D\hookrightarrow Quot(D)\hookrightarrow K$$ 
