Thread: Irreducible and gcd.

1. 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 $\neq$ 1, in which case p|a, a contradiction. Hence if b = 1, g.c.d (a,p) = 1.

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

This look right?

2. 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 $\neq$ 1, in which case p|a, a contradiction. Hence if b = 1, g.c.d (a,p) = 1.

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

This look right?
let $d=\gcd(a,p).$ then $d \mid p.$ so $p=kd$ and thus either $k$ or $d$ is a unit, because $p$ is irreducible. if $d$ is a unit, then $\gcd(a,p)=1.$ if $k$ is a unit, then $ks=1,$ for some $s$ and hence $d=ps,$

i.e. $p \mid d.$ we also have $d \mid a.$ hence $p \mid a.$ contradiction.