How to Show that a Field of Characteristic $0$ is Infinite How can one prove that if a field $K$ has characteristic zero, then $K$ is infinite?
 A: 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.
A: 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.
A: 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? 
A: 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. 
