1
$\begingroup$

Let $R$ be a PID, and $r\in R- \{0\}$. Prove that $\langle r\rangle$ maximal $\iff r$ irreducible.

"$\Leftarrow$"Easy.

"$\Rightarrow$"If $J=\langle r \rangle$ then we will prove that $r$ is irreducible. If $r=ab$, we want to prove that $a\in U(R)$ or $b \in U(R)$.

If we take the ideal which is generated by $\langle a\rangle$ then (because $J$ is maximal)$$\langle a\rangle \subseteq\langle r \rangle \iff r\mid a \iff a=kr, k\in R \Longrightarrow r=krb\iff r(1-kb)=0_R \iff kb=1_R$$ so $b\in U(R)$. Same way if we work with $\langle b \rangle$.

Is this proof right?

$\endgroup$
  • $\begingroup$ I mean, "Easy" is not a proof of anything, so we're taking it on faith that you can indeed prove it. $\endgroup$ – Patrick Stevens Aug 10 '16 at 22:59
  • 1
    $\begingroup$ If $r=ab$, then $\langle r \rangle \subseteq \langle a \rangle$, not the opposite. On the other hand, since $\langle r \rangle$ is maximal, then $\langle r \rangle = \langle a \rangle$, therefore... $\endgroup$ – Alex M. Aug 10 '16 at 23:00
  • $\begingroup$ Patrick: My friend,i did this proof and i stack in the opposite direction. Alex: If $r \in \langle r \rangle \subseteq \langle a \rangle $ then $r \in \langle a \rangle$ and then $r=au, u\in R$. Is this wrong? $\endgroup$ – Chris Aug 10 '16 at 23:10
  • $\begingroup$ Yes, if $r \in \langle a \rangle$ then $r = au$ with $u \in R$. But why do you ask this. $\endgroup$ – quid Aug 10 '16 at 23:19
2
$\begingroup$

The argument is not completely correct.

If I understand correctly you start with "(because $J$ is maximal) $\langle a\rangle \subseteq\langle r \rangle $", but it is not true that if you chose some maximal ideal then every other ideal is contained in it.

Instead argue like this if $r= ab$ then $r \in \langle a \rangle$ and thus $\langle r \rangle \subset \langle a \rangle$. Since $\langle r \rangle$ is maximal it follows that $\langle r \rangle = \langle a \rangle$ or that $ \langle a \rangle =R$.

Then continue from there.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ You understood right. But why is this wrong? $J$ is maximal. Does it contain all the other ideals of $R$? $\endgroup$ – Chris Aug 10 '16 at 23:19
  • 1
    $\begingroup$ No. A maximal ideal does not contain all the other ideals. For example in the integers, $2\mathbb{Z}$ is a maximal ideal but it does not contain the ideal $9\mathbb{Z}$. Or just note that in general there are several maximal ideals. Not each of them can contain all the other ones. $\endgroup$ – quid Aug 10 '16 at 23:21
  • $\begingroup$ Nice example. So by definition the only thing that we can conclude is that: If for every $I$ ideal of $R$ such that $J \subseteq I$, $J$ is maximal iff $I=J$ or $I=R$? $\endgroup$ – Chris Aug 10 '16 at 23:26
  • 1
    $\begingroup$ An ideal $J$ is maximal if and only if the only ideals that contain it are $J$ and the full ring, yes. $\endgroup$ – quid Aug 10 '16 at 23:28
4
$\begingroup$

Hint $ $ Note that for principal ideals: $\ \rm\color{#0a0}{contains} = \color{#c00}{divides}$, $ $ i.e. $(a)\supset (b)\iff a\mid b,\,$ thus

$\qquad\quad\begin{eqnarray} (r)\,\text{ is maximal} &\iff&\!\!\ (r)\, \text{ has no proper } \,{\rm\color{#0a0}{container}}\,\ (a)\\ &\iff&\ r\ \ \text{ has no proper}\,\ {\rm\color{#c00}{divisor}}\,\ a\\ &\iff&\ r\ \ \text{ is irreducible}\\ \end{eqnarray}$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.