Prime and Maximal Ideals in PIDs .... Rotman, AMA Theorem 5.12

[Math Amateur]

I am reading Joseph J. Rotman's book: Advanced Modern Algebra (AMA) and I am currently focused on Section 5.1 Prime Ideals and Maximal Ideals ...

I need some help with understanding the proof of Theorem 5.12 ... ...Theorem 5.12 reads as follows:View attachment 5940
In the above text Rotman writes the following:" ... ... If \(\displaystyle (p) \subseteq J = (a)\), then \(\displaystyle a|p\). Hence either \(\displaystyle a\) and \(\displaystyle p\) are associates, in which case \(\displaystyle (a) = (p)\), or \(\displaystyle a\) is a unit, in which case \(\displaystyle J = (a) = R\). ... ... ... "My question is as follows:Rotman argues, (as I interpret his argument), that \(\displaystyle a|p\) implies that either \(\displaystyle a\) and \(\displaystyle p\) are associates ... or ... \(\displaystyle a\) is a unit ...Can someone please explain (slowly and clearly :) ) why this is the case ... ... ?Hope someone can help ... ...

Peter

 

[Math Amateur]

I am reading Joseph J. Rotman's book: Advanced Modern Algebra (AMA) and I am currently focused on Section 5.1 Prime Ideals and Maximal Ideals ...

I need some help with understanding the proof of Theorem 5.12 ... ...Theorem 5.12 reads as follows:
In the above text Rotman writes the following:" ... ... If \(\displaystyle (p) \subseteq J = (a)\), then \(\displaystyle a|p\). Hence either \(\displaystyle a\) and \(\displaystyle p\) are associates, in which case \(\displaystyle (a) = (p)\), or \(\displaystyle a\) is a unit, in which case \(\displaystyle J = (a) = R\). ... ... ... "My question is as follows:Rotman argues, (as I interpret his argument), that \(\displaystyle a|p\) implies that either \(\displaystyle a\) and \(\displaystyle p\) are associates ... or ... \(\displaystyle a\) is a unit ...Can someone please explain (slowly and clearly :) ) why this is the case ... ... ?Hope someone can help ... ...

Peter

I have been reflecting on my question and believe the answer is something like the following:Firstly ... we are given that \(\displaystyle p\) is irreducible ...

Now ... \(\displaystyle p\) irreducible

\(\displaystyle \Longrightarrow p\) is non-zero and \(\displaystyle p\) not a unit ... and ... where \(\displaystyle p\) equals a product,

say, \(\displaystyle p = ra\) ... then one of \(\displaystyle a\) and \(\displaystyle r\) is a unit ... Now, \(\displaystyle a|p \Longrightarrow p = ra\) for some \(\displaystyle r \in R\)

So then we have that:

\(\displaystyle p\) irreducible and \(\displaystyle p = ra \Longrightarrow\) one of \(\displaystyle a\) and \(\displaystyle r\) is a unit ...

If \(\displaystyle r\) is a unit then \(\displaystyle a\) and \(\displaystyle p\) are associates ... ...

... otherwise \(\displaystyle a\) is a unit ...
Can someone confirm that this is correct ... or alternatively point out shortcomings and errors in the analysis ...

Peter

 

[steenis]

I have this solution for you.

First, a notation:
$a\sim b$ for $a,b\in R$ iff $a$ and $b$ are associates iff there is a unit $u\in R$ such that $a=ub$.
$\sim$ is an equivalence relation.

R is a PID and commutative.
Given $a\mid p$ and $(p)$ is an prime ideal. And you want to prove that $a\sim p$ or $a\sim 1$ ($a$ is a unit).

$a\mid p$, so there is a $x\in R$ such that $p=xa$, this means that $xa\in (p)$.
$(p)$ is a prime ideal thus (1) $a\in (p)$ or (2) $x\in (p)$

Suppose (1) $a\in (p)$, then there is an $r\in R$ such that $a=rp$.
Then $p=xrp$, and because $R$ is a domain, $xr=1$, i.e., $x$ is a unit and $a\sim p$.

Suppose (2) $x\in (p)$, then there is an $s\in R$ such that $x=sp$.
Then $p=asp$, and because $R$ is a domain, $as=1$, i.e., $a$ is a unit and $a\sim 1$ $\Box$

You fill in the rest of the proof.

 

[Math Amateur]

I have this solution for you.

First, a notation:
$a\sim b$ for $a,b\in R$ iff $a$ and $b$ are associates iff there is a unit $u\in R$ such that $a=ub$.
$\sim$ is an equivalence relation.

R is a PID and commutative.
Given $a\mid p$ and $(p)$ is an prime ideal. And you want to prove that $a\sim p$ or $a\sim 1$ ($a$ is a unit).

$a\mid p$, so there is a $x\in R$ such that $p=xa$, this means that $xa\in (p)$.
$(p)$ is a prime ideal thus (1) $a\in (p)$ or (2) $x\in (p)$

Suppose (1) $a\in (p)$, then there is an $r\in R$ such that $a=rp$.
Then $p=xrp$, and because $R$ is a domain, $xr=1$, i.e., $x$ is a unit and $a\sim p$.

Suppose (2) $x\in (p)$, then there is an $s\in R$ such that $x=sp$.
Then $p=asp$, and because $R$ is a domain, $as=1$, i.e., $a$ is a unit and $a\sim 1$ $\Box$

You fill in the rest of the proof.

Thanks Steenis ... appreciate your help ...

Just working through your post now ...

Peter