Show that if you add the identity element ($1$) a finite amount of times will result to the neutral element ($0$).
I started with saying that in every field there's an element $a \in F$ so that $a + 1 =0$. But I'm not even sure if this is true because I'm not sure about the example in $\mathbb{R}$. I thought that if $F = \mathbb{R}$ so $a = -1$.
True or not true, I don't really know how to continue further here.
On
HINT: Describe the subring generated by 1. Rather, describe the subring generated by 1 in any ring in terms of the free rings.
On
if $1,2,3,...$ were all distinct, the field would infinite, hence $n=m$ for some $m<n$ and $n-m=0$
On
Proof
Let $|F| = n$, then $\{1,2\cdot 1,3\cdot 1,\cdots, (n+1)\cdot 1\}$ cannot be all distinct. Thus suppose WLOG $a>b$ and $a\cdot 1 = b\cdot 1\Longrightarrow (a-b)\cdot 1 = 0$.
This says that Char $F>0$.
Show that Char $F = p$ where $p$ is a prime.
Suppose not, then let char $F= n = c\cdot d$ where $1<c<n$ and $1<d<n$ are natural numbers. Then
$$1+1+\cdots + 1 = (1+1+\cdots +1)(1+1+\cdots +1) = 0$$
LHS is $n$ times the sum, right hand side is the product of $c$ and $d$ times the sum. Then since $c<n$ and $d<n$, both factor are non-zero, but their product is zero. Contradiction since in a field there is no zero divisor.
Assuming basically no background, consider the elements $$ 1, \quad 1+ 1, \quad 1+ 1 + 1, \quad \dots \quad, \underbrace{1+ \dots + 1}_{n}, \quad \dots $$ Since $F$ is finite, they cannot all be distinct. So there exist $n < m$ such that $$ \underbrace{1+ \dots + 1}_{n} = \underbrace{1+ \dots + 1}_{m} = \underbrace{1+ \dots + 1}_{n} + \underbrace{1+ \dots + 1}_{m-n}, $$ so adding to both sides the opposite of $\underbrace{1+ \dots + 1}_{n}$ you get $$ 0 = \underbrace{1+ \dots + 1}_{m-n}, $$ with $m - n > 0$.