Field theory, how was the multiplication associated with addition? I was reviewing the definition of the characteristic of a field where it was defined to be:
"The characteristic of a field $F$, denoted ch$(F)$, is defined to be the smallest positive integer $p$ such that $p\cdot 1_F = 0$ if such a p exists and is defined to be $0$ otherwise."
The further proposition said:" If ch$(F) = p$ then for any $\alpha \in F. 
p\cdot \alpha=\alpha+\alpha+...(p\, times)+\alpha=0$.
My question was that:


*

*how was the addition associated multiplication? 

*Further, how did the prime elements associated with the prime number? 
Where I found a sentence at the beginning of the chapter:
"For $n$ a positive integer, let $n\cdot 1_F = 1_F+...+ 1_F$ ($n$ times).
"
So, question:


*if we didn't have this relation, does that mean the above proposition would be incorrect and we can no longer associate the addition with multiplication?

 A: It's important to distinguish between the ring multiplication, which is the operation in the ring, and 
the shorthand notation $kr := r+r+\dots +r \ $ ($k$ times)
$\\ $
I like to use $ \ k \cdot  r$ for ring multiplication and $kr$ for the shorthand notation.
This basically defines the characteristic of a field to be the order of $1_F$ under addition, if it is finite,  
and $0$ if the order of $1_F$ infinite.
You then have $$p\alpha = \alpha + \alpha + \cdots \alpha \ (p \text{ times}) = \alpha \cdot 1_F \ + \alpha \cdot 1_F \ + \cdots + \ \alpha \cdot 1_F $$
$ \\ $
$$ = \alpha \ \cdot \left( 1_F + 1_F + \cdots + 1_F\right) = \alpha \ \cdot ( \ p 1_F \ ) = \alpha  \ \cdot 0 = 0$$
$ \\ $
For integers, a number $p$  is prime if and only if $$p | ab \implies p|a \text { or } p|b$$
That was the property that was generalized in defining prime elements. The "elementary school" definition of a prime, that its only divisors are itself and 1, was generalized to the notion of an irreducible element. They are equivalent in PID's, and hence in the integers.
A: This is not directly linked to field multiplication. By the properties of the integers, given any element $\alpha\in F$, there exists a unique group homomorphism $(\mathbb{Z},+)\to (F,+)$ mapping $1$ to $\alpha$.
This homomorphism maps any positive integer $k$ to $\underbrace{\alpha+\dots+\alpha}_{k\text{ times}}$ and a negative integer $k$ to $-(\underbrace{\alpha+\dots+\alpha}_{-k\text{ times}})$. This is general group theory. It is customary to write $k\alpha$ for the element so described.
Let $\chi_F$ be the group homomorphism we get for $\alpha=1_F$. Like every homomorphism with domain $\mathbb{Z}$, the kernel of $\chi_F$ is a subgroup, so it is of the form $p\mathbb{Z}$, for a unique $p\ge0$. In particular, if $p>0$, we have $p1_F=\chi_F(p)=\underbrace{1_F+\dots+1_F}_{p\text{ times}}=0$ and $p$ is the least positive integer having this property.
If the kernel of $\chi_F$ is $\{0\}=0\mathbb{Z}$, no such integer exists.
What happens in the case of fields (more generally of rings) is that $\chi_F$ is also a ring homomorphism: $\chi_F(mn)=\chi_F(m)\chi_F(n)$, which can be proved by induction on $n$ when $n\ge0$ and by the rule of signs for $n<0$.
This is the key for proving that when the characteristic of $F$ is $p>0$, then $p$ is prime. Indeed, if $p=mn$ with $m,n>0$, then
$$
0=\chi_F(p)=\chi_F(mn)=\chi_F(m)\chi_F(n)
$$
so either $\chi_F(m)=0$ or $\chi_F(n)=0$, which implies $m\in p\mathbb{Z}$ or $n\in p\mathbb{Z}$, that is, $m=p$ or $n=p$.
