MHB Why is \{ x_1, x_2, \ldots, x_{n-1}, x \} a basis for F in Proposition 4.3.14?

[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 yet further help in order to fully understand the proof of Proposition 4.3.14 ... ...

Proposition 4.3.14 reads as follows:

View attachment 8323
View attachment 8324

In the above proof by Bland we read the following:

" ... ... If $$a_n \neq 0$$, let $$y = x_1 a_1 + x_2 a_2 + \ ... \ ... \ + x_{n-1} a_{n-1}$$, so that $$x = y + x_n a_n$$. If $$y = 0$$, then $$x = x_n a_n$$, so $$a_n$$ is a unit since $$x$$ is primitive. Thus $$\{ x_1, x_2, \ ... \ ... \ , x_{n-1}, x \}$$ is a basis for $$F$$. ... ..."Can someone please explain exactly why/how $$\{ x_1, x_2, \ ... \ ... \ , x_{n-1}, x \}$$ is a basis for $$F$$ ... ...
Help will be much appreciated ... ...

Peter==========================================================================================

It may help MHB
members reading this post to have access to Bland's definition of 'primitive element of a module' ... especially as it seems to me that the definition is a bit unusual ... so I am providing the same as follows:
https://www.physicsforums.com/attachments/8325
Hope that helps ...

Peter

 

[steenis]

Remember from linear algebra, that if
$$\{x_1, \cdots, x_i, \cdots, x_n \}$$
is a basis of F, then, if $c \neq 0$ then
$$\{x_1, \cdots, cx_i, \cdots, x_n \}$$
is also a basis of F.

Here we have that $x=x_n a_n$ and $a_n$ is a unit, thus $a_n \neq 0$ and if
$$\{x_1, \cdots, x_{n-1},x_n \}$$
is a basis of F, so
$$\{x_1, \cdots, x_{n-1}, x \}$$
is also a basis of F

 

[Math Amateur]

Remember from linear algebra, that if
$$\{x_1, \cdots, x_i, \cdots, x_n \}$$
is a basis of F, then, if $c \neq 0$ then
$$\{x_1, \cdots, cx_i, \cdots, x_n \}$$
is also a basis of F.

Here we have that $x=x_n a_n$ and $a_n$ is a unit, thus $a_n \neq 0$ and if
$$\{x_1, \cdots, x_{n-1},x_n \}$$
is a basis of F, so
$$\{x_1, \cdots, x_{n-1}, x \}$$
is also a basis of F

Thanks Steenis ...

Appreciate your help...

Peter

 

[steenis]

On second thoughts, my answer is nor correct, because R is not a field, but a commutative ring.

Correct answer, I hope:

$\{x_1,x_2, \cdots, x_n \}$ is a basis for F, thus

$F=x_1R \oplus x_2R \oplus \cdots \oplus x_nR$

we have: $x=x_n a_n$ and $a_n$ is a unit

then $R=a_nR$ and $x_nR=xR$

thus $F=x_1R \oplus x_2R \oplus \cdots \oplus xR$

thus $\{x_1,x_2, \cdots, x \}$ is a basis for F

 

[Math Amateur]

On second thoughts, my answer is nor correct, because R is not a field, but a commutative ring.

Correct answer, I hope:

$\{x_1,x_2, \cdots, x_n \}$ is a basis for F, thus

$F=x_1R \oplus x_2R \oplus \cdots \oplus x_nR$

we have: $x=x_n a_n$ and $a_n$ is a unit

then $R=a_nR$ and $x_nR=xR$

thus $F=x_1R \oplus x_2R \oplus \cdots \oplus xR$

thus $\{x_1,x_2, \cdots, x \}$ is a basis for F

Thanks for clarifying that Steenis ...

Peter