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

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

-
How do you define characteristic? – Prahlad Vaidyanathan Oct 6 '13 at 13:00
The definition is given in the fourth line of the post : If $F$ is a field of characteristic p then the prime field $P$ of $F$ is isomorphic to $\mathbb{Z}_p$. – Alexis Marchand Oct 6 '13 at 13:02
Does that mean I can use straight away that $p=0$ in this case( i.e $p=0$ mod $p$ ?) – Alexis Marchand Oct 6 '13 at 13:05
Wow... how did the top comment get (at least) one upvote after the OPs answer? And although I can't be 100% sure, I think the OP had the definition on the question from the start. – Git Gud Oct 6 '13 at 13:36

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 $$pa=\underbrace{ a+a+\cdots+a }_{p \text{ times}}=a(\underbrace{ 1+1+\cdots+1 }_{p \text{ times}})=a0=0.$$
You mean $1$ is the multiplicative identity element right? – Pratyush Sarkar Oct 6 '13 at 14:26