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

1. Sep 1, 2016

### 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

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2. Sep 1, 2016

### 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

3. Sep 1, 2016

### 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.

4. Sep 1, 2016

### andrewkirk

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.

5. Sep 1, 2016

### Math Amateur

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

Appreciate your help on this issue ...

Peter