Find all endomorphisms of $\mathbb{Q}$. ($\mathbb{Q}$ is the field.)
When finding isomorphism of $2\mathbb{Z}$ and $3\mathbb{Z}$, we define the mapping like $φ: 2\mathbb{Z} → 3\mathbb{Z}$ and then we say $φ(2) = 3k, k∈Z$.
How to start in this case of $\mathbb{Q}$? Can you just tell me how does the mapping look like? Thanks for your time!
This is different if we are looking for ring endomorphisms or group endomorphisms.
Let's start with ring endomorphisms, as they are easier. Suppose $\phi : \mathbb{Q} \rightarrow \mathbb{Q}$ is a ring homomorphism. Then $$\phi(q) = q\phi(1) = q$$ for any $q\in\mathbb{Q}$ and hence $\phi$ is just the identity. Therefore $\{\textrm{ring endomorphisms of } \mathbb{Q}\} \cong \{\textrm{id}\}$.
Now suppose $\phi : \mathbb{Q} \rightarrow \mathbb{Q}$ is a group homorphism. For any $n \in \mathbb{Z}$ and $q\in\mathbb{Q}$, we have $$\phi(nq) = \phi(q + \cdots + q) = \phi(q) + \cdots +\phi(q) = n\phi(q)$$ Therefore if $q = \frac{m}{n}$ is a fraction, $n\phi(q) = \phi(m)$ so $$\phi(q) = \frac{\phi(m)}{n} = \frac{m}{n}\phi(1)$$ Hence, $\phi$ is determined by $\phi(1)$. Since $\phi(1)$ can take any value in $\mathbb{Q}$, and each homomorphism is distinct, we obtain a (ring) isomorphism:
$$ \begin{align} \{\textrm{group endomorphisms of } \mathbb{Q}\} & \rightarrow \mathbb{Q} \\ \phi & \mapsto \phi(1)\end{align}$$
and so $\{\textrm{group endomorphisms of } \mathbb{Q}\}\cong \mathbb{Q}$.
This example shows it is always important to be precise which category we are working over!