If $p$ is a prime $p\mid m\cdot n \implies p\mid m$ or $p\mid n$ by Euclid's lemma. I want to prove that if $p$ is a prime number, and $p = a\cdot b$, then $a = 1$ or $b = 1$.
I know that a prime number does not have any other positive divisors than $1$ and $p$, but I can't seem to get started from here. Is it possible to use gcd?
I know that a prime number does not have any other positive divisors than $1$ and $p$
So let's use that. If $p = a\cdot b$, then $a$ and $b$ are both divisors of $p$, so they are both either $1$ or $p$. And they cannot both be equal to $p$. Thus we conclude that one of them must be $1$. Assuming $a$ and $b$ are positive, of course.
If $p$ is a prime $p\mid m\cdot n \implies p\mid m$ or $p\mid n$ by Euclid's lemma. I want to prove that if $p$ is a prime number, and $p = a\cdot b$, then $a = 1$ or $b = 1$.
EL does it: $\ p = ab\,\Rightarrow\, p\mid ab\,\Rightarrow\ p\mid a\,$ or $\,p\mid b.\,$
If $\ p\mid a\ $ cancelling $\,p\,$ in $\,p=ab\,$ $\,\Rightarrow\,1 = (a/p)b\,$ so $\,b=1\,$ (by $b>0).\,$ Similarly $\,p\mid b\,\Rightarrow\,a = 1$.
Remark $ $ In general domains this shows that primes are irreducible. Generally primes are defined to be noninvertibles satisfying Euclid's Lemma and irreducibles are nonzero noninvertibles having only trivial factorizations. Generally the converse fails, i.e. irreducibles need not be prime, i.e. they needn't satisfy Euclid's Lemma, and this means that factorizations into irreducibles needn't be unique. But when such factorizations are unique (as in $\Bbb Z)$ we have $\,p\,$ is prime $\!\iff p$ is irreducible.
