Thread: Irreducible and gcd.

R is an integral domain.
Assume that greatest common divisors exist in R. Let p be an irreducible element of R and let a be an element of R that is not divisible by p. Show that 1 is a greatest common divisor of a and p.

Thoughts...
p is irreducible, hence if p|ab then either a=1 or b=1.
If b = 1 then either a = 1, in which case g.c.d (a,p) = 1, or a 1, in which case p|a, a contradiction. Hence if b = 1, g.c.d (a,p) = 1.

So assume b 1, hence a = 1. Hence g.c.d (a,p) = 1.

This look right?

Originally Posted by Deadstar

R is an integral domain.
Assume that greatest common divisors exist in R. Let p be an irreducible element of R and let a be an element of R that is not divisible by p. Show that 1 is a greatest common divisor of a and p.

Thoughts...
p is irreducible, hence if p|ab then either a=1 or b=1. incorrect! p is irreducible not prime!
If b = 1 then either a = 1, in which case g.c.d (a,p) = 1, or a 1, in which case p|a, a contradiction. Hence if b = 1, g.c.d (a,p) = 1.

So assume b 1, hence a = 1. Hence g.c.d (a,p) = 1.

This look right?

let then so and thus either or is a unit, because is irreducible. if is a unit, then if is a unit, then for some and hence

i.e. we also have hence contradiction.