knowing that the characteristic of an integral domain is $0$ or a prime number, i want to prove that the characteristic of a finite field $F_n$ is a prime number, that is $\operatorname{char}(F_n)\not = 0$. the proof i'm reading uses the fact that within $e,2e,3e,\ldots$ there are two that are necessarily equal $ie=(i+k)e$ so $ke=0$ for some positive $k$. But can i say the following argument:
$F_n=\{0,e,x_1,\ldots ,x_{n-2}\}$ ( $e$ is the multiplicative identity ) and since $e\not = 0$ then $e\not = 2e\not =\cdots ,\not = ne$ so we have $n$ distinct elements $ie, i=1,\ldots,n$ of $F_n$ hence one of them must equal $0$; $ie=0$ for some $i\in \{2,\ldots,n\}$, moreover the trivial field with one element $e=0$ has obviously characteristic $1$. Is there a non-trivial field where $e=0$?
[Math] characteristic of a finite field
field-theoryfinite-fields
Best Answer
Let $1$ denote the multiplicative identity of the field. Then, $1 \neq 0$ where $0$ is the additive identity. So every field must have at least two elements: your notion of a field with one element $0$ is not correct.
So, since the field contains $1$, it contains $1+1$, $1+1+1$, $1 + 1 + 1 + 1, \ldots, 1 + 1 + \cdots + 1$ (where the last sum has $n$ $1$s in it). All these $n$ sums cannot be distinct since the field has only $n-1$ nonzero elements. So, for some distinct $i$ and $j$, the sum of $i$ $1$s must equal the sum of $j$ $1$s, and by subtraction, the sum of $|i-j|$ $1$s must equal $0$. The smallest number of $1$s that sum to $0$ is called the characteristic of the finite field, and the characteristic must be a prime number. This is because if the characteristic were a composite number $ab$, then the sum of $a$ $1$s multiplied by the sum of $b$ $1$s, which equals the sum of $ab$ $1$s by the distributive law, would equal $0$, that is, the product of two nonzero elements would equal $0$, which cannot happen in a field.