Is there a ring - homomorphism $\mathbb{F}_p \rightarrow \mathbb{F}_q $ (p,q prime , $p \not= q$ )?

Question

So we have two prime fields and seek a homomorphism between them. I assume that i have to find a homomorphism that is valid for all p,q prime , $p \not= q$, not just one for each choice. I would say that it is not possible.

• We can not use the zero map because we have the neutral element of multiplication (1) that needs to be mapped to 1.
• Any other mapping will fail the test as well, i mean if it does work i can just pick another q or p so i will get a different value, right?

As you see i am a bit unclear here, what i want to ask is

• Is my assumption correct?
• How could i start to formally prove my point?

Answer 1

$\mathbf F_q$ would have characteristic both $q$ and $p$, since it would contain a subfield isomorphic to $\mathbf F_p$, which is impossible if $p\neq q$.

Actually, the only homomomorphisms between finite fields are the canonical injections: $\,\mathbf F_{p^m}\hookrightarrow\mathbf F_{p^n}$, which exist if and only if $m\mid n$.

Answer 2

Hence, $F_p$ is a field. So every non_zero homomorphism define from $F_p$ to any ring $R$, is a injection homomorphism. And we know if $f:F_p\to R$ be a homomorphism when, $R$ is any ring, then: $\frac{F_p}{Ker(f)}\cong Im(f)$. But if $f\neq 0$, then $f$ should be on injection. That's mean, $F_p\cong Im(f)$. And when $p\neq q$, it is not possible.