Question

Let $\mathit{R}$ be an integral domain in which any two elements (not both zero) have a gcd. Let p be an irreducible element of $\mathit{R}$. Prove that whenever $\mathit{p}\mathbf{|}\mathit{c}\mathit{d}$, then $\mathit{p}\mathbf{|}\mathit{c}$ or$\mathit{p}\mathbf{|}\mathit{d}$ .

Short answer

If $p|cd$ then $p|c$ or $p|d$.

Step-by-step solution

Exercise 8 and Exercise 26

1.Let p be an irreducible element in an integral domain then $1_R$ is a gcd of p and a if and only if $p\cancel{|}a$.

2. Let $R$be an integral domain in which any two elements (not both$0_R$ ) have a gcd. If $(b,c)~1_R$ and $(b,d)~1_R$ then $(b,cd)~1_R$.

whenever p|cd then p|c   and p|d

Given that p is irreducible and $p|cd$ then by (1), $(p,cd)\ne 1$.

By contrapositive of (2), either $(p,c)\ne 1$ or $(p,d)\ne 1$. By statement (1), $(p,c)\ne 1$ implies$p|c$ and $(p,d)\ne 1$ implies $p|d$.

Therefore, if $p|cd$ then $p|c$ or $p|d$.