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

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 ${a_n:ninmathbb{N}}$ 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.

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

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

Now $mcdot1=ncdot1$ implies $(m-n)cdot1=0$, a contradiction. Can you complete the argument?

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 $1in F$ is finite. That order is, by definition, the characteristic of the field.