How would I find all the ring morphisms from $\mathbb{Z}_7$ to $\mathbb{Z}_4$? I know that if the map function is $f(x)$, we must have that $f(1) = 1$ and that $f(a \cdot b) = f(a) \cdot f(b)$, as well as $f(a + b) = f(a) + f(b)$.
In other such questions, I understand that if it's from say $\mathbb{Z}_{15}$ to $\mathbb{Z}_3$ then it's much easier since 15's a multiple of 3.
However, I'm not sure how to find the morphisms if their relatively prime numbers.
Considering ring homomorphisms adds unnecessary complications. Let's start looking at homomorphisms $\phi:\mathbb{Z}_m\rightarrow\mathbb{Z}_n$ of the additive groups.
We know that
Putting all these things together the number of elements in $\phi(\mathbb{Z}_m)$ must divide both $m$ and $n$.
Hence, if $m$ and $n$ are coprime the only possibility is that $\phi(\mathbb{Z}_m)$ has only one element, which must be $0$ and $\phi=0$ identically.
Now observe that a ring homomorphism must be a homomorphism of the additive groups.
Let $f \colon \mathbb{Z}_7 \to \mathbb{Z}_4$ be a homomorphism. Then $$ f(4) = f(1+1+1+1) = f(1) + f(1) + f(1) + f(1) = 0 $$ But then $$ f(1) = f(4+4) = f(4) + f(4) = 0 $$ So $f(x) = 0$ for all $x \in \mathbb{Z}_7$.
In general, suppose that $m$ and $n$ are relatively prime, and consider $f \colon \mathbb{Z}_m \to \mathbb{Z}_n$. Then $f(n) = n \cdot f(1) = 0$ in $\mathbb{Z}_n$. There exists an integer $a$ such that $a n \equiv 1 \pmod{m}$. But then also, $f(1) = f(an) = a f(n) = 0$. So $f(x) = 0$ for all $x \in \mathbb{Z}_m$.
