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

Question

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?

Answer 1

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?

Answer 2

The only additive group homomorphism $\mathbb{Z}/3\mathbb{Z} \to \mathbb{Z}/5\mathbb{Z}$ is the zero map because the size of the image has to be a common divisor of $3$ and $5$. (*)

Therefore, there are no ring homomorphisms $\mathbb{Z}/3\mathbb{Z} \to \mathbb{Z}/5\mathbb{Z}$ that send $1 \mapsto 1$, and so no field homomorphisms.

(*) More elementary, and along your line, $0=f(0)=f(3)=3f(1)$ and $0=0f(1)=5f(1)$ imply that $f(1)=0$ and so $f=0$.

Answer 3

$\text{char}\left(\mathbb{Z}/3\mathbb{Z}\right) = 3 \neq 5 = \text{char}\left( \mathbb{Z}/5\mathbb{Z}\right)$

However, a field homomorphism would require equal characteristic.

$\Rightarrow$ There is no field homomorphism $\mathbb{Z}/3\mathbb{Z} \to \mathbb{Z}/5\mathbb{Z}$