Suppose $\phi: \mathbb{Q} \to \mathbb{Z}$ is a homomorphism, where both $\mathbb{Q}$ and $\mathbb{Z}$ are under addition. Prove $\phi$ is the zero map 
Suppose $\phi: \mathbb{Q} \to \mathbb{Z}$ is a homomorphism, where
both $\mathbb{Q}$ and $\mathbb{Z}$ are under addition. Prove $\phi$ is
the zero map.

Suppose $\phi$ is a non zero map. Meaning there is at least one $x \in \mathbb{Q}$ such that $\phi(x) = a$ for some non zero integer a. Since the definition of homomorphism involves two elements from the domain, I tried taking another element $y \in \mathbb{Q}$ such that $\phi(y) = b$ for some integer b. Since $\phi$ is homomorphism, $\phi(x+y) =\phi(x) + \phi(y)= a + b$.
My intention is to arrive at a contradiction that $\phi$ is not a homomorphism. But I am stuck.
 A: Hint: for every $q \in \Bbb{Q}$ and $n \in \Bbb{N}_{>0}$, there is an element $p \in \Bbb{Q}$ such that $np=q$ (where $np$ means $\sum_1^np$). In the usual jargon, this says that $\Bbb{Q}$ is a divisible group - any element of $\Bbb{Q}$ can be divided by any positive integer. Now if $np = q$, then $n\phi(p) = \phi(q)$, so a homomorphic image of a divisible group is divisible. What subgroups of $\Bbb{Z}$ are divisible?
A: Let me explain it in two ways.
Method 1. If $\phi \colon \mathbf Q \to \mathbf Z$ is an additive group homomorphism then for each $r \in \mathbf Q$ the integer $\phi(r)$ is divisible by arbitrarily high powers of $2$, and that forces $\phi(r)$ to be $0$.  After all, a nonzero integer can be divisible by $2$ only finitely many times.
Step 1. Since $\phi$ is additive, $\phi(ax) = a\phi(x)$ for $a \in \mathbf Z^+$ and $x \in \mathbf Q$.
Step 2. For $r \in \mathbf Q$ and $k \geq 1$ we can write $r = 2^ks$ where $s$ is the fraction $r/2^k$.  Applying $\phi$ to both sides of the equation,
$\phi(r) = \phi(2^ks) = 2^k\phi(s)$ by Step 1 (with $a = 2^k$). Thus the integer $\phi(r)$ is divisible by $2^k$ no matter how large $k$ we choose. That forces $\phi(r)$ to be $0$.
Method 2. If $\phi \colon \mathbf Q \to \mathbf Z$ is an additive group homomorphism then for each $r \in \mathbf Q$ the integer $\phi(r)$ is divisible by infinitely many primes, so it must be $0$.
Step 1. Since $\phi$ is additive, $\phi(ax) = a\phi(x)$ for $a \in \mathbf Z^+$ and $x \in \mathbf Q$. (Same as previous Step 1.)
Step 2. For $r \in \mathbf Q$ and a prime $p$, we can write $r = ps$ where $s$ is the fraction $r/p$.  Applying $\phi$ to both sides of the equation,
$\phi(r) = \phi(ps) = p\phi(s)$ by Step 1 (with $a = p$). Thus the integer $\phi(r)$ is divisible by every prime number, and that forces $\phi(r)$ to be $0$.
This idea of proving something is zero because it's divisible by arbitrarily high powers of a prime (like $2$) or because it's divisible by infinitely many primes (in fact by all primes) really has been used to prove some famous theorems in the past. For example, in the first proof of the Coates-Wiles theorem (a special case of the Birch and Swinnerton-Dyer conjecture), an algebraic number was shown to be $0$ by proving it is divisible by infinitely many prime ideals.
A: Let $k$ be the smallest/largest integer $>0$/$<0$ such that for some $x\in\mathbb{Q}$, $\phi(x)=k$. Then $\phi(\frac{x}{2})+\phi(\frac{x}{2})=k$, so $\phi(\frac{x}{2}) = \frac{k}{2}$ which is smaller/greater than $k$ and greater/smaller than $0$.
