Is it true that $End_{\mathbb{Z}}(\mathbb{Q}) \simeq \mathbb{Q}$ as a ring? It is clear that for every ring $R$ with multiplicative identity, then $End_{R}(R) \simeq R$ as a ring. I read a paper and they said that $End_{\mathbb{Z}}(\mathbb{Q}) \simeq \mathbb{Q}$ as a ring structure. But, I don't know how to prove it. Please help me to find the isomorphism between the ring $End_{\mathbb{Z}}(\mathbb{Q})$ and $\mathbb{Q}$
 A: An endomorphism
$\phi \in End_{\Bbb Z}(\Bbb Q) \tag 1$
is determined by its value on $1 \in \Bbb Z$, to wit: if
$\phi(1) = r \in \Bbb Q, \tag 2$
then
$\phi(p) = \phi(p \cdot 1) = p\phi(1) = pr, \; \forall p \in \Bbb Z; \tag 3$
also,
$q \phi \left ( \dfrac{1}{q} \right ) = \phi \left ( \dfrac{q}{q} \right ) = \phi(1) = r, \; \text{for} \; 0 \ne q \in \Bbb Z; \tag 4$
whence
$\phi \left ( \dfrac{1}{q} \right ) = \dfrac{r}{q}; \tag 5$
then for
$\dfrac{p}{q} \in \Bbb Q, \; p, q \in \Bbb Z, \tag 6$
$\phi \left ( \dfrac{p}{q} \right ) = p \phi \left ( \dfrac{1}{q} \right ) =  \dfrac{p}{q}r =\dfrac{p}{q}\phi(1). \tag 7$
Voila!!! Is it not a miracle!!! That a mere $\Bbb Z$-endomorphism $\phi$ of $\Bbb Q$ is in fact witnessed to be $\Bbb Q$-linear as well?  For lucky (7) reveals that, for $s, t \in \Bbb Q$, $\phi(st) = st\phi(1) = s \phi(t \cdot 1) = s \phi(t)$!!!
We now define the map
$\Theta: End_{\Bbb Z}(\Bbb Q) \to \Bbb Q \tag 8$
via
$\Theta(\phi) = \phi(1); \tag 9$
then
$\Theta(\phi + \psi) = (\phi + \psi)(1) = \phi(1) + \psi(1) = \Theta(\phi) + \Theta(\psi); \tag{10}$
that is, $\Theta$ is a homomorphism 'twixt the additive groups of $End_{\Bbb Z}(\Bbb Q)$ and $\Bbb Q$; 
we examine the multiplicative structure of $\Theta$:
$\Theta(\psi \phi) = (\psi \phi)(1) = \psi(\phi(1)) = \psi(\phi(1) \cdot 1) = \phi(1) \psi(1) = \psi(1) \phi(1), \tag{11}$
where $\psi(\phi(1) \cdot 1) = \phi(1) \psi(1)$ has in fact been established in (7), since $\phi(1) \in \Bbb Q$. (10) and (11) together show that in fact $\Theta$ is a ring homomorphism; it is clearly surjective since we may in fact take (2), (7) as defining a $\phi \in End_{\Bbb Z}(\Bbb Q)$ with $\phi(1) = r \in \Bbb Q$; finally,
$\ker \Theta = \{0\}, \tag{12}$
for if
$\phi(1) = \Theta(\phi) = 0 \tag{13}$
then
$\phi \equiv 0, \tag{14}$
again by (2), (7). Since $\Theta$ is thus seen to be a surjective ring homomorphism with kernel $\{0\}$, it is an isomorphism and we have achieved our desired goal:
$\Theta: End_{\Bbb Z}(\Bbb Q) \simeq \Bbb Q. \tag{15}$
.
A: Hint: $f \in End_{\mathbb{Z}}(\mathbb{Q})$ is determined by $f(1)$.
More generally, if $D$ is a domain and $K=Q(D)$, then $End_{D}(K) \simeq K$ as $D$-algebras, by the same argument.
