Properties of $Hom_{\mathbb{Z}}(\mathbb{Q}, \mathbb{Q} )$

Question

I have to prove that :

1. $\text{Hom}_{\mathbb{Z}}(\mathbb{Q}, \mathbb{Q}) \cong \mathbb{Q}$ as abelian groups

2. $\text{End}_{\mathbb{Z}}(\mathbb{Q}) \cong \mathbb{Q}$ as rings

What I have done:

1. We can define $$\lambda : \text{Hom}_{\mathbb{Z}}(\mathbb{Q}, \mathbb{Q}) \to \mathbb{Q}$$ $$\lambda(f) = f(1)$$

2. Any hint?

Answer 1

Let $\lambda(f)=f(1)$. then if $n$ is an integer, $$ f(nx)=nf(x)$$ for all $x\in \Bbb{Q}$. If $m$ is integer which is not $0$, $$ mf\left(\frac{1}{m}\right)=f\left(\frac{1}{m}+\cdots+\frac{1}{m}\right)=f(1)=\lambda(f)$$ Therefore if $f(1)=g(1)$ for $f,g\in \mathrm{Hom}_{\Bbb{Z}}(\Bbb{Q},\Bbb{Q})$, $$ f(x)=g(x)$$ for all $x\in \Bbb{Q}$. Hence $f(1)$ derermines $f$.

If $f\in \mathrm{Hom}_{\Bbb{Z}}(\Bbb{Q},\Bbb{Q})$, then $f\in \mathrm{End}_{\Bbb{Z}}(\Bbb{Q})$ because of $$ \begin{aligned} nf(x)&=f(nx) \\ \frac{1}{m}f(x)&=f\left(\frac{x}{m}\right) \end{aligned} $$ for all $x\in \Bbb{Q}$ and $(m,n)\in (\Bbb{Z}\setminus\{0\})\times \Bbb{Z}$, and $$ \lambda(fg)=(fg)(1)=f(1)g(1)=\lambda(f)\lambda(g)$$ therefore $\lambda:\mathrm{End}_{\Bbb{Z}}(\Bbb{Q})\to \Bbb{Q}$ is ring homomorphism.