i am starting to study abstract algebra and at this question i missed some points of the proof i wish you can explain:

prove that-->for all integer a>1 there can be found at least 1 prime divisor.

proof:let S={d (element) Z: d>1 and d|a} be this set.we know that S is not empty since a|a and a>1 so a (element) S and Sis not empty set.

positive integer set is well-ordered set so there exist the minimum element of S and say it p.if we show that p is prime number then the proof is over.

from here;

assume contrary that p is not prime then p=q.r and there can be found at least one q,r (element) Z such that 1<q<p.

and q|p and p|a -->q|a and q (element) S and this contradicts witht the assumption of p be the smallest element of S since p is prime.

my question is why do we try to prove that p is smallest element of S.we are searching at least one prime divisor why that must be the smallest one,what happened if the smallest one is not prime?may there is another elemnt is prime of S.