Common divisors in a PID

Question

Suppose that R is an integral domain and that α, β, γ ∈ R. We say that γ is a common divisor of α and β if γ|α in R and γ|β in R. Suppose that R is a PID. Suppose that α, β ∈ R. Let I = (α) and let J = (β). Let K = I + J. Since R is a PID, we have K = (δ) where δ ∈ R. Prove that δ has the following properties:

δ is a common divisor of α and β
If γ ∈ R and γ is a common divisor of α and β, then γ|δ in R

So for 1, I noted that I is contained in K and J is contained in K, so therefore β = δx for some x in R and α=δy for some y in R. Then it follows that δ divides α and β.

Then for 2, I supposed that γ is a common divisor of α and β in R. So again, β=γm and α=γn for some m and n in R. And then from part 1, β=δx=γm and α=δy=γn. This is where I get tripped up, because I was thinking about multiplying both sides by the inverses of m and n and then I realized that m and n aren't necessarily invertible elements. So I could use some help. And also, does my answer for #1 work?

Any thoughts about the two answers that have been posted? – Gerry Myerson Mar 03 '14 at 08:58 — Gerry Myerson, Mar 03 '14 at 08:58

score 1 · Answer 1 · answered Mar 02 '14 at 02:26

1

$\gamma\mid\alpha$ implies $\alpha$ is in $(\gamma)$ so $I\subseteq(\gamma)$; similarly, $J\subseteq(\gamma)$; then $(\delta)=I+J\subseteq(\gamma)$, so $\gamma\mid\delta$.

answered Mar 02 '14 at 02:26

Gerry Myerson

179,216

score 0 · Answer 2 · answered Mar 02 '14 at 08:10

0

$\gamma\mid\alpha,\beta \!\iff\! (\gamma)\supseteq (\alpha),(\beta) \!\iff\! (\gamma)\supseteq(\alpha)+(\beta)=(\delta) \!\iff\! \gamma\mid\delta$

answered Mar 02 '14 at 08:10

Bill Dubuque

272,048

Remark. Above is the universal property of the gcd $,\delta = \gcd(\alpha,\beta).,$ Note that $(1)$ follows from the direction $,(\Leftarrow),$ with $,\gamma = \delta.\ \ $ – Bill Dubuque Mar 02 '14 at 08:30

Common divisors in a PID

2 Answers2