MHB Primitive Elements and Free 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 help in order to fully understand the proof of Proposition 4.3.14 ... ...

Proposition 4.3.14 reads as follows:View attachment 8318
View attachment 8319
In the above proof by Bland we read the following:

" ... ... The induction hypothesis gives a basis $$\{ x, x'_2, \ ... \ ... \ x'_{n -1} \}$$ of $$M$$ and it follows that $$\{ x, x'_2, \ ... \ ... \ x'_{n - 1}, x'_n \}$$ is a basis of $$F$$ that contains $$x$$. ... ... "My question is as follows:

Why/how exactly does it follow that $$\{ x, x'_2, \ ... \ ... \ x'_{n - 1}, x'_n \}$$ is a basis of $$F$$ that contains $$x$$. ... ... Help will be 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:


View attachment 8322

Hope that helps ...

Peter

 

[steenis]

$F$ is a free module, $rank(F) = n$, $x$ is a primitive element of $F$, and suppose the induction hypothesis is true for free modules with $rank < n$

If $\{ x_1, x_2, \cdots, x_n \}$ is a basis for $F$ ($rank(F) = n$)

then $x = x_1 a_1 + x_2 a_2 + \cdots + x_n a_n$ for some $a_i \in R$, $i=1, 2, \cdots, n$

Suppose $a_n = 0$

Then $x \in M = x_1R \oplus x_2R \oplus \cdots \oplus x_{n-1}R$ (definition)

Notice that $F = M \oplus x_nR$

$M$ is a free module and $rank(M) = n-1 < n$

By the induction hypothesis, $M$ has a basis $\{ x, x’_2, \cdots, x’_{n-1} \}$, containing $x$

Thus $M = xR \oplus x’_2R \oplus \cdots \oplus x’_{n-1}R$

and $F = xR \oplus x’_2R \oplus \cdots \oplus x’_{n-1}R \oplus x_nR$

Thus $\{ x, x’_2, \cdots, x’_{n-1}, x_n \}$ is a basis of $F$ that contains $x$

 

[Math Amateur]

$F$ is a free module, $rank(F) = n$, $x$ is a primitive element of $F$, and suppose the induction hypothesis is true for free modules with $rank < n$

If $\{ x_1, x_2, \cdots, x_n \}$ is a basis for $F$ ($rank(F) = n$)

then $x = x_1 a_1 + x_2 a_2 + \cdots + x_n a_n$ for some $a_i \in R$, $i=1, 2, \cdots, n$

Suppose $a_n = 0$

Then $x \in M = x_1R \oplus x_2R \oplus \cdots \oplus x_{n-1}R$ (definition)

Notice that $F = M \oplus x_nR$

$M$ is a free module and $rank(M) = n-1 < n$

By the induction hypothesis, $M$ has a basis $\{ x, x’_2, \cdots, x’_{n-1} \}$, containing $x$

Thus $M = xR \oplus x’_2R \oplus \cdots \oplus x’_{n-1}R$

and $F = xR \oplus x’_2R \oplus \cdots \oplus x’_{n-1}R \oplus x_nR$

Thus $\{ x, x’_2, \cdots, x’_{n-1}, x_n \}$ is a basis of $F$ that contains $x$

Thanks Steenis ...

Appreciate your help ...

Peter