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How can one prove that if a field $K$ has characteristic zero, then $K$ is infinite?

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    $\begingroup$ If $K$ is characteristic zero, then there is an injective map from $\mathbb{Q}$. $\endgroup$ – Michael Burr Jan 5 '19 at 17:56
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Consider the following sequence of elements

$a_1 = 1$, $a_2=1+1$,..., $a_n=1+1+...+1$ (we sum $n$ times) and so on.

we prove that $\{a_n:n\in\mathbb{N}\}$ is infinite. If by contradiction it's finite then $a_n=a_m$ for some $n<m$ in this case we have $a_m-a_n = a_{m-n}=0$ but then the characteristic is $m-n$ or smaller.

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According to definition, a field $k$ is characteristic zero if there does not exist a number $n\in \mathbb{Z}$ so that $$n\cdot 1= \underbrace{1+\cdots+1}_{n\:\text{times}}=0.$$ It follows that the map $\mathbb{Z}\to k$ given by $n\mapsto n\cdot 1$ is an injection. So, $\lvert \mathbb{Z}\rvert\le \lvert k\rvert$. In particular $k$ is infinite.

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If a field is of characteristic $0$,there is no $n\in \Bbb N $ such that $n\cdot 1=\underbrace{1+1+\cdots+1}_{n\; \text{times}}=0$.

Now $m\cdot1=n\cdot1$ implies $(m-n)\cdot1=0$, a contradiction. Can you complete the argument?

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If $F$ is a finite field, then viewed as a group with the additive structure, it is a finite group whose identity element is $0$. In any finite group, all elements have finite order. Thus the order of $1\in F$ is finite. That order is, by definition, the characteristic of the field.

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