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:

In the above text Rotman writes the following:" ... ... If ##(p) \subseteq J = (a)##, then ##a|p##. Hence either ##a## and ##p## are associates, in which case ##(a) = (p)##, or ##a## is a unit, in which case ##J = (a) = R##. ... ... ... "My question is as follows:Rotman argues, (as I interpret his argument), that ##a|p## implies that either ##a## and ##p## are associates ... or ... ##a## is a unit ...Can someone please explain (slowly and clearly

) why this is the case ... ... ?Hope someone can help ... ...

Peter

 

[Math Amateur]

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

Now ... ##p## irreducible

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

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

So then we have that:

##p## irreducible and ##p = ra \Longrightarrow## one of ##a## and ##r## is a unit ...

If ##r## is a unit then ##a## and ##p## are associates ... ...

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

Peter

 

[andrewkirk]

We have ##a|p## so ##p=ba## for some ##b##. Since ##p\in I=(p)## and ##I## is a prime ideal, it must be the case that either ##a## or ##b## is in ##I##.

If ##a\in I## we have ##p|a## which, together with the earlier-observed ##a|p## makes ##p,a## associates.

If ##a\not\in I## then ##b\in I## in which case ##b=pc## for some ##c##. So ##p=pca##. We write this as
$$0=p-pca=p(1-ca)$$
and since ##p\neq 0## and a PID is an Integral Domain, we must then have ##1-ca=0##, whence ##ca=1## so ##a|1##, making ##a## a unit.

 

[andrewkirk]

Can someone confirm that this is correct ... or alternatively point out shortcomings and errors in the analysis ...

That reasoning looks sound to me. It's a different route from mine, as yours explicitly uses the irreducibility of ##p## whereas mine uses the primeness of the ideal ##I##. I suspect the connection between the two lies in the proof of Proposition 5.6. But either one will suffice.

 

[Math Amateur]

Thanks Andrew ... your posts were REALLY helpful ...

Appreciate your help on this issue ...

Peter