Characteristic of a field is either $0$ or prime $p$.

Question

Things I dont understand from a definition I found:

Proof: If $F$ has characteristic $mn>0$ for integers, then $mn=0$ in $F$,

Why $mn$ must be $0$

so one of $m$ or $n$ is not invertible in $F^*$,

Why?

and thus either $m=0$ or $n=0$ as elements of $F$. If $m,n>1$, this contradicts the minimality property.

I am not sure how.

Can someone help me understand this?

Answer 1

Characteristic $mn$ means that the smallest positive $k$ such that $\underbrace{1+1+..+1}_{k \mbox{ times }}=0$ is $mn$.

This means that $$\underbrace{1+1+..+1}_{mn \mbox{ times }}=0$$ and no smaller number of times works.

This implies $$\left(\underbrace{1+1+..+1}_{m \mbox{ times }} \right)\cdot \left(\underbrace{1+1+..+1}_{n \mbox{ times }} \right)=0$$

Since $F$ is a Field, a product being zero implies that one of the factors is zero. But this contradicts the fact that the smallest positive $k$ such that $\underbrace{1+1+..+1}_{k \mbox{ times }}=0$ is $mn$.

Answer 2

The characteristic of a field is the smallest $k$ such that for any element $a\in F$, we have $a_1+\ldots+a_{k}=0$.

Since a field $F$ is an abelian group under addition, there exist $a\in F$ such that if $a_1+\ldots+a_{nm}=0$, then $a_1+\ldots+a_{n}=b\in F$ and $a\not=b\not=0$. Thus $b_1+\ldots+b_{m}=0$, but this contradicts the minimality of $mn$.