Characteristic of a field is $0$ or prime I'm trying to prove that the characteristic of any field $F$ is either $0$ or a prime number, but I have no idea what to do.  Help?
 A: Hint: Show that if the characteristic of $F$ were a composite number, say $n=ab$, then $F$ would have zero-divisors (since $n=0$...). Then show that a zero-divisor cannot be a unit unless $0=1$, which is not the case for fields.
Alternate Hint: Look at the unique homomorphism $\phi:\mathbb{Z}\to F$, defined by $\phi(0)=0_F$, $\phi(1)=1_F$, $\phi(2)=1_F+1_F$, etc. and note that its image, being a subring of a field, must be an integral domain.
A: Suppose $char(F) = m \neq 0$.
By definition, $m$ is the smallest integer so that $\underbrace{1 + ... + 1}_{m-times} = 0_F$
Suppose m is not prime and say $m = \alpha.\beta$, where $0 < \alpha,\beta < m$.
Therefore we can write:
$$
\eqalign{
0_F &= \underbrace{1 + ... + 1}_{m-times}\\
&= (\underbrace{\underbrace{1 + ... + 1}_{\alpha-times}) \ + \ ... \ + \ (\underbrace{1 + ... + 1}_{\alpha-times})}_{\beta-times} \tag{1}\label{1}
}
$$
By closure property, $\underbrace{1 + ... + 1}_{\alpha-times} \in F$. Let $\underbrace{1 + ... + 1}_{\alpha-times} = \zeta \tag{2}\label{2}$ [It's important to not call it $\alpha$. See Note]  
There are 2 cases:  
Case $\zeta = 0_F$:
Since $\alpha < m$ and $\underbrace{1 + ... + 1}_{\alpha-times} = 0_F, \ \ \ \therefore char(F) = \alpha$.
We arrive at a contradiction.  
Case $\zeta \neq 0_F$:
$\exists \zeta^{-1} \in F$
$$
\eqalign{
\therefore 0_F &= 0_F.\zeta^{-1}\\
&= (\underbrace{\zeta + ... + \zeta}_{\beta-times}).\zeta^{-1} \ [by \ (\ref{1}) \ and \ (\ref{2})]\\
&= \underbrace{1 + ... + 1}_{\beta-times}
}
$$
Since $\beta < m, char(F) = \beta$ and we again arrive at a contradiction.   
Hence $m$ is a prime.    QED  
Note: We should not write $\underbrace{1 + ... + 1}_{\gamma-times} = \gamma$, as $\gamma$ is just an integer and not necessarily a member of the field $F$. Also this may lead to an erroneous conclusion, e.g. say since $char(F) = m$, if we write $\underbrace{1 + ... + 1}_{m-times} = m = 0$, we can say $m = m.1 = 0$ and since $1 \neq 0, \therefore m = 0$.
