# Why does the characteristic of a field always have to be prime? [duplicate]

I know little about group theory, and the concept of a field characteristic is new to me. Although there are some posts on this website that explain it very well.

"In layman's terms, it's the amount of times we can keep adding 1 to itself without looping back to 0. The characteristic being $p$ means that $\underbrace{1+1+\dots+1}_p=0$."

I think I can get my head around the idea, but there is still something I don't understand.

"We can see quickly using a zero divisor argument that the characteristic of any field with positive characteristic must be prime."

Why is it that it has to be prime? Why can't it be something else?

## marked as duplicate by rschwieb abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Sep 22 '17 at 13:00

Suppose the characteristic $n$ is composite with factors $1<p,q<n$. Then you could write $$\underbrace{(1+1+\dots+1)}_p\underbrace{(1+1+\dots+1)}_q=\underbrace{1+1+\dots+1}_{n}=0$$ but this means $pq=0$, so $p$ and $q$ are zero divisors, which contradicts the field axioms.
If the characteristic $n = jk \neq 0$ for some $j,k \in \mathbb{N}$ which we are interpreting as the sum of $1$s in the field then $j$ and $k$ are zero divisors and that's important because this means $j^{-1}$ and $k^{-1}$ don't exist, which violates the axiom of a field requiring multiplicative inverses.