MHB Why Does \( a_1 \mid b_1 b_2 \cdots b_n \) in Theorem 7.2.20?

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I am reading The Basics of Abstract Algebra by Paul E. Bland ...

I am focused on Section 7.2 Euclidean, Principal Ideal, Unique Factorization Domains ... ...

I need help with the proof of Theorem 7.2.20 ... ... Theorem 7.2.20 and its proof reads as follows:https://www.physicsforums.com/attachments/8280
View attachment 8281
In the last paragraph of the above proof by Bland we read the following:

" ... ... If $$a = a_1 a_2 \ ... \ ... \ a_m = b_1 b_2 \ ... \ ... \ b_n$$ where each $$a_i$$ and $$b_i$$ is irreducible, then $$a_1 \mid b_1 b_2 \ ... \ ... \ b_n$$ ... ... "
Can someone please explain exactly and in detail why/how $$a_1 \mid b_1 b_2 \ ... \ ... \ b_n$$ ... ... Peter
 
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Notate $x=b_1 b_2 \cdots b_n$
What does it mean if $a_1$ divides $x$ ?
It means that there is a $y \in D$ such that $a_1y=x$
Can you tell now what is $y$ ?
 
steenis said:
Notate $x=b_1 b_2 \cdots b_n$
What does it mean if $a_1$ divides $x$ ?
It means that there is a $y \in D$ such that $a_1y=x$
Can you tell now what is $y$ ?
Thanks for the help Steenis ...

Basically you have pointed out that:

$$a_1 ( a_2 a_3 \ ... \ ... \ a_m ) = b_1 b_2 \ ... \ ... \ b_n $$

$$\Longrightarrow a_1 \mid b_1 b_2 \ ... \ ... \ b_n$$The above implies that in what you have written we have $$y = a_2 a_3 \ ... \ ... \ a_m$$ ... ...Is that correct?

Peter
 
Correct, and do you understand that therefore $a_1|b_1 \cdots b_n$ ?
 

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