# How to Show that a Field of Characteristic $0$ is Infinite [closed]

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

• If $K$ is characteristic zero, then there is an injective map from $\mathbb{Q}$. – Michael Burr Jan 5 '19 at 17:56

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 $$n in this case we have $$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 $$n\in \mathbb{Z}$$ 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.

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?

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.