# Definition of prime numbers and equivalence

A prime number $$p$$ is a positive integer that has exactly two different dividers ($$1$$ and itself). This is a very clear and universal definition.

My notes say that this definition is equivalent to the previous one: $$p$$ is a prime number if from $$p \mid ab$$ follows that $$p \mid a$$ or $$p \mid b$$ for $$a,b \in \mathbb{Z}$$. I was wondering why this is an equivalent definition for a prime number.

• Formally the first definition states that $p$ is irreducible and the second that $p$ is prime. Every prime $p$ is irreducible and in the special case of the ring of integers the concepts irreducible and prime coincide. But in a more general setting an irreducible element of a ring is not necessarily prime. So actually prime is stronger than irreducible. – drhab Jan 26 at 11:16

Let $$p$$ be prime according to the first definition and let $$p\nmid a$$ and $$p\nmid b$$ where $$a$$ and $$b$$ are positive integers.

It is well known that there are unique expressions $$a=r_1^{u_1}\cdots r_n^{u_n}$$ and $$b=s_1^{v_1}\cdots s_m^{v_m}$$ where the $$s_i$$ and $$r_j$$ are primes according to the first definition and the $$u_i$$ and $$v_j$$ are positive integers.

Then from $$p\nmid a$$ and $$p\nmid b$$ it follows that $$p\notin\{r_1,\dots, r_n,s_1,\dots, s_n\}$$ which is evidently the set of prime factors of $$ab$$. So we conclude that $$p\nmid ab$$.

Proved is now that a prime according to the first definition is also a prime according to the second definition.

For the converse see the answer of greedoid.

Formally the first definition states that $$p$$ is irreducible and the second that $$p$$ is prime. Every prime $$p$$ is irreducible and in the special case of the ring of integers the concepts irreducible and prime coincide. But in a more general setting an irreducible element of a ring is not necessarily prime. So actually prime is stronger than irreducible.

Suppose there is positive not prime $$n$$ such that $$n \mid ab\implies n \mid a \;\;\;{\rm or}\;\;\; n \mid b$$ for $$a,b \in \mathbb{Z}$$.

Then $$n= x\cdot y$$ and $$xy\ne 1$$, so $$n\mid xy$$ and thus $$n\mid x$$ or $$n\mid y$$.

But $$x,y so we have a cotnradiction.

The other way is well known fact in number theory.