If $F$ has characteristic $p$, then $pa$ =0 for all $a \in F$

Question

I have to prove the statement in the title, i.e

If $F$ has characteristic $p$, then $pa = 0$ for all $a \in F$, $p$ prime.

From the definition of a characteristic of a field, we have that

If F is a field of characteristic p then the prime field P of F is isomorphic to $\mathbb{Z}_p$.

i.e $\exists \phi :P ->\mathbb{Z}_p$, a bijective ring map.

Do I have to prove that (p) is an ideal in F so that $pa=0$ in $F/I$ ? ($I = (p)$)

Thanks

Answer 1

Since $F$ has a prime field $\Bbb{F}_p$ so $p1=0$ (where $1$ is a multiplicative identity of $F$.) So by distributive law we get $$ pa=\underbrace{ a+a+\cdots+a }_{p \text{ times}}=a(\underbrace{ 1+1+\cdots+1 }_{p \text{ times}})=a0=0. $$