$I_p$ is a commutative ring of residue classes ie. $I_p = \{\bar{0}, \bar{1}, \ldots, \overline{p-1}\}$. I have a doubt in proving the necessary condition that when p is prime => $I_p$ is a field.
We basically try to prove that there are no zero divisors in $I_p$. So the proof starts with let $\bar{a}, \bar{b} \in I_p$ then
$\bar{a}\bar{b} = 0$ => $\bar{ab} = 0$
=> p is divisor of ab i.e. p | ab
=> p | a or p | b (This is the step that is not clear)
=> $\bar{a}$ = 0 or $\bar{b}$ = 0
=> there are no zero divisor in $I_p$
And since $I_p$ is a finite commutative ring with no zero divisor it is a field.
If p|ab how can we say p|a or p|b. According to Euclid's lemma isnt it that p divides atleast one of a or b. That mean that there is a possibility where p | a and p |b.
In that case the proof is still valid. since both $\bar{a} = 0$ and $\bar{b} = 0$ and there are no zero divisors.
But isn't the step incorrect?
I also found this definition of Prime element in Ring. The definition says the same what was used in the proof. In the article it states
With this definition, Euclid's lemma is the assertion that prime numbers are prime elements in the ring of integers.
but that is not the definition of prime numbers in integer, again there is a case where p|a and p|b.