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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 $$\{ x_1 \}$$ is a basis for $$F$$, then there is an $$a \in R$$ such that $$x = x_1 a$$. But $$x$$ is primitive, so $$a$$ is a unit in $$R$$. Hence $$x R = x_1 R$$ ... ... "
My question is as follows:
Why in the above quote, does it follow that $$x R = x_1 R$$ ... ... ?Is it because $$x = x_1 a$$ where $$a$$ is a unit ... ... ... ... ... (1)
Hence $$xR = x_1 a R$$ ... ... I presume this follows (1)
Therefore $$xR = x_1 (a R )$$
But $$aR = R$$ since a is a unit ... ''
So $$xR = x_1 R $$
Is that correct?
Peter
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 $$\{ x_1 \}$$ is a basis for $$F$$, then there is an $$a \in R$$ such that $$x = x_1 a$$. But $$x$$ is primitive, so $$a$$ is a unit in $$R$$. Hence $$x R = x_1 R$$ ... ... "
My question is as follows:
Why in the above quote, does it follow that $$x R = x_1 R$$ ... ... ?Is it because $$x = x_1 a$$ where $$a$$ is a unit ... ... ... ... ... (1)
Hence $$xR = x_1 a R$$ ... ... I presume this follows (1)
Therefore $$xR = x_1 (a R )$$
But $$aR = R$$ since a is a unit ... ''
So $$xR = x_1 R $$
Is that correct?
Peter