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For a field K with multiplicative identity $1$, consider the numbers $2=1+1$, $3=1+1+1$, $4=1+1+1+1$, etc. Either these numbers are all different, in which case we say that $K$ has characteristic $0$, or two of them will be equal. In the latter case, it is straightforward to show that, for some number $p$, then $p$ will be a prime, and we say that $K$ has characteristic $p$. I will show that $p$ is prime.

Proof. Assume that. Thus, $p=p_{1}p_{2}$. Thus, $$p=p_{1}(1+1+1+...+1)+p_{2}(1+1+1...+1)=0.$$

Can you hint me for this proof?

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    $\begingroup$ It is product of $1+1+\cdots +1$ ($p_1$ times) and $1+1+\cdots+1$ ($p_2$ times). $\endgroup$ – André Nicolas Feb 20 '16 at 17:28
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Remember that characteristic is the least positive integer such that $m\cdot 1 = 0.$

If you have $p$ to be a characteristic of a field and $p = p_1p_2$ with $1<p_1, p_2< p.$ Then

$$ p\cdot 1 = (p_1p_2)\cdot 1 = (p_1\cdot 1)(p_1\cdot 1) = 0 $$

Now you are in a field and every field is an integral domain so either $p_1\cdot 1 = 0$ or $p_2\cdot 1 = 0.$ Which contradicts the minimality of $p.$ So $p$ must be a prime.

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Hint: Show that if the characteristic of a field is composite number then field must have zero divisor.

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Hint: if $p=p_1p_2$, then $1<p_1<p$ and $1<p_2<p$. The characteristic of $K$ is is the smallest number $p$ such that $1+...+1=0$ $p$ times.

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