Q:

Field homomorphism $\mathbb{Z}/3\mathbb{Z} \to \mathbb{Z}/5\mathbb{Z}$

I'm supposed to look for a field homomorphism $f: \mathbb{Z}/3\mathbb{Z} \to \mathbb{Z}/5\mathbb{Z}$.
I started at $f(1) = 1$ which has to be true, due to the definition of a field homomorphism, right?
Then I can follow $f(0) = f(3) = f(1+1+1) = f(1) + f(1)+f(1) = 1 + 1 + 1 = 3$.
But also $f(1) = f(1+0) = f(1) + f(0) = 1 + 3 = 4 \neq 1 = f(1)$.
What's wrong here?

A:

Hint: Think about a group homomorphism first, since obviously a field homomorphism is a group one. Every non-identity element of $\mathbb Z/3\mathbb Z$ has order $3$. What elements of $\mathbb Z/5\mathbb Z$ have order $3$? So, what does every element of $\mathbb Z/3\mathbb Z$ have to map to?
Field homomorphisms have to be injective (since the kernel of the morphism is an ideal). What do you conclude?

