Why Does d and 1 Being Associates Imply 1 is a GCD?

[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 4.3.12 ... ...

Proposition 4.3.12 reads as follows:View attachment 8314In the above proof by Bland we read the following:

" ... ... Since \(\displaystyle x\) is primitive, \(\displaystyle d\) is a unit, so \(\displaystyle d\) and \(\displaystyle 1\) are associates. Thus \(\displaystyle 1\) is a greatest common denominator of \(\displaystyle \{ a_\alpha \ \mid \ a_\alpha \neq 0 \}\). ... ... "Can someone please explain exactly why \(\displaystyle d\) and \(\displaystyle 1\) being associates implies that \(\displaystyle 1\) is a greatest common denominator of \(\displaystyle \{ a_\alpha \ \mid \ a_\alpha \neq 0 \}\) ... ...Peter==============================================================================

It may help MHB readers of the above post to have access to Bland's definition of a primitive element ... so I am providing the same as follows:View attachment 8315Hope that helps ...

Peter

 

[GJA]

Re: Free Modules Over Prncipal Ideal Domains ... Bland, Proposition 4.3.12 ... ...

Hi Peter,

If $c$ is any common divisor of $\{a_{\alpha}: a_{\alpha}\neq 0\}$, then $c|d$ by definition of gcd. Since $d$ and $1$ are associates $d|1\Longrightarrow c|1,$ because $c|d$. Since $c$ is arbitrary, $1$ is a gcd of $\{a_{\alpha}: a_{\alpha}\neq 0\}$ by definition.

 

[Math Amateur]

Re: Free Modules Over Prncipal Ideal Domains ... Bland, Proposition 4.3.12 ... ...

Hi Peter,

If $c$ is any common divisor of $\{a_{\alpha}: a_{\alpha}\neq 0\}$, then $c|d$ by definition of gcd. Since $d$ and $1$ are associates $d|1\Longrightarrow c|1,$ because $c|d$. Since $c$ is arbitrary, $1$ is a gcd of $\{a_{\alpha}: a_{\alpha}\neq 0\}$ by definition.

Thanks for the help GJA ...

But ... just a clarification ...

I can verify that \(\displaystyle d \mid 1\) and that \(\displaystyle d|1\Longrightarrow c|1\) ... but I cannot follow your argument from there ... that because $c$ is arbitrary, $1$ is a gcd of $\{a_{\alpha}: a_{\alpha}\neq 0\}$ ... ? ... ... can you please explain more fully ...

... essentially ... how does \(\displaystyle d \mid 1\) and \(\displaystyle d|1\Longrightarrow c|1\) together with \(\displaystyle c\) being arbitrary imply that the gcd of $\{a_{\alpha}: a_{\alpha}\neq 0\}$ is \(\displaystyle 1\) ... ... ?

Thanks again ...

Peter

 

[GJA]

Re: Free Modules Over Prncipal Ideal Domains ... Bland, Proposition 4.3.12 ... ...

Hi Peter,

The definition of a gcd in a commutative ring (which $R$ is since it's a PID) is that any common divisor must also divide the gcd. By showing that the arbitrary common divisor $c$ divides $1$, we have shown $1$ is a gcd by definition. See the "GCD in Commutative Rings" section for a reference https://en.wikipedia.org/wiki/Greatest_common_divisor