Prove irreducibility implies prime

Question

Let R be an integral domain such that any pair b,c has a gcd. Prove if a is irreducible, then a is prime. I think we have to prove gcd(a,bc) | gcd(a,b)gcd(a,c). But I couldn't. Any help?

Answer 1

Let $p\in R$ be irreducible, and suppose $p\,|\,ab$ but $p \nmid a$. We also know that $p\,|\,pb$, so $p\,|\,\gcd(ab,pb)$. But $\gcd(pb,ab) = \gcd(a,p)\,b$. Since $p$ is irreducible and $p \nmid a$, $\gcd(a,p) = 1$. Indeed, $\gcd(a,p)\,|\,p$, so as $p$ irreducible we must have $\gcd(a,p)= 1$ or $p$. But $\gcd(a,p)\,|\,a$, so it cannot be $p$. So $p\,|\,\gcd(ab,pb) = \gcd(a,p)\,b = b$ and we're done.