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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

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    $\begingroup$ How do you define characteristic? $\endgroup$ – Prahlad Vaidyanathan Oct 6 '13 at 13:00
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    $\begingroup$ 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 $\mathbb{Z}_p$. $\endgroup$ – ALM Oct 6 '13 at 13:02
  • $\begingroup$ Does that mean I can use straight away that $p=0$ in this case( i.e $p=0$ mod $p$ ?) $\endgroup$ – ALM Oct 6 '13 at 13:05
  • $\begingroup$ 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. $\endgroup$ – Git Gud Oct 6 '13 at 13:36
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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 $$ pa=\underbrace{ a+a+\cdots+a }_{p \text{ times}}=a(\underbrace{ 1+1+\cdots+1 }_{p \text{ times}})=a0=0. $$

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  • $\begingroup$ You mean $1$ is the multiplicative identity element right? $\endgroup$ – Pratyush Sarkar Oct 6 '13 at 14:26
  • $\begingroup$ @PratyushSarkar Yes. It is my mistake and I fixed it. $\endgroup$ – Hanul Jeon Oct 6 '13 at 14:27

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