Question

Give an alternative proof of Lemma 10.11 as follows. If $\mathit{p}\mathbf{|}\mathit{b}$, there is nothing to prove. If $\mathit{p}\cancel{\mathbf{|}}\mathit{b}$ then ${\mathbf{1}}_{\mathbf{R}}$ is a gcd of p and b by Exercise 8. Now show that $\mathit{p}\mathbf{|}\mathit{c}$ by copying the proof of Theorem 1.4 with p in place of a and Exercise 20 in place of Theorem 1.2.

Short answer

If $R$ is a PID, p is irreducible in $R$ and $p|bc$ then $p|b$ or $p|c$.

Step-by-step solution

Exercise 8 and 20

If p is an irreducible element in an integral domain then $1_R$ is the gcd of p and a if and only if role="math" localid="1654681191281" $p\cancel{|}a$.

The elements a, b have a greatest common divisor that can be written as a linear combination of a and b.

Proof of p|c

Given that if $p|b$ then there is nothing to prove. If $p\cancel{|}b$then by exercise 8, $1_R$ is the gcd of p and b.

Now by exercise 20, the linear combination of p and b can be written as $1_R=up+vb$.

Multiply both the sides by c then $c=upc+vbc$.

Assume that there is $q\in R$such that $bc=pq$then,

$$\begin{gathered} c=upc+vbc \\ =upc+vpq \\ =p(uc+vq) \end{gathered}$$

Thus, $c=p(uc+vq)$ implies $p|c$.

Therefore, if $R$ is a PID, p is irreducible in $R$ and $p|bc$ then$p|b$ or $p|c$.