Help Understanding Bland's Proposition 4.3.14 in Rings and Their Modules

[Math Amateur]

I am reading Paul E. Bland's book, "Rings and Their Modules".

I am focused on Section 4.3: Modules Over Principal Ideal Domains ... and I need some further help in order to fully understand the proof of Proposition 4.3.14 ... ...

Proposition 4.3.14 reads as follows:View attachment 8320
View attachment 8321In the above proof by Bland we read the following:

" ... ... If \(\displaystyle \{ x_1 \}\) is a basis for \(\displaystyle F\), then there is an \(\displaystyle a \in R\) such that \(\displaystyle x = x_1 a\). But \(\displaystyle x\) is primitive, so \(\displaystyle a\) is a unit in \(\displaystyle R\). Hence \(\displaystyle x R = x_1 R\) ... ... "

My question is as follows:

Why in the above quote, does it follow that \(\displaystyle x R = x_1 R\) ... ... ?Is it because \(\displaystyle x = x_1 a\) where \(\displaystyle a\) is a unit ... ... ... ... ... (1)

Hence \(\displaystyle xR = x_1 a R\) ... ... I presume this follows (1)

Therefore \(\displaystyle xR = x_1 (a R )\)

But \(\displaystyle aR = R\) since a is a unit ... ''

So \(\displaystyle xR = x_1 R \)

Is that correct?

Peter

 

[steenis]

Yes, you did it correct.

You can do it also this way:

Let $y \in xR$ then $y=xr=x_1ar \in x_1R$
and
Let $y \in x_1R$ then $y=x_1s=xa^{-1}s \in xR$

 

[Math Amateur]

Yes, you did it correct.

You can do it also this way:

Let $y \in xR$ then $y=xr=x_1ar \in x_1R$
and
Let $y \in x_1R$ then $y=x_1s=xa^{-1}s \in xR$

Thanks for the help, Steenis ...

Peter