# $\langle r\rangle$ maximal $\iff r$ irreducible

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?

• I mean, "Easy" is not a proof of anything, so we're taking it on faith that you can indeed prove it. – Patrick Stevens Aug 10 '16 at 22:59
• 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... – Alex M. Aug 10 '16 at 23:00
• 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? – Chris Aug 10 '16 at 23:10
• Yes, if $r \in \langle a \rangle$ then $r = au$ with $u \in R$. But why do you ask this. – quid Aug 10 '16 at 23:19

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.

• You understood right. But why is this wrong? $J$ is maximal. Does it contain all the other ideals of $R$? – Chris Aug 10 '16 at 23:19
• 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. – quid Aug 10 '16 at 23:21
• 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 $I=R$? – Chris Aug 10 '16 at 23:26
• An ideal $J$ is maximal if and only if the only ideals that contain it are $J$ and the full ring, yes. – quid Aug 10 '16 at 23:28

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

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