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Let $R$ be a PID (Principal Ideal Domain) and $x$ is an element R. Prove that the ideal $\langle x\rangle$ is maximal if and only if $x$ is irreducible.

Ok, so I know what an irreducible is. I'm thinking that this problem is asking us to set up a proof by contradiction but I can't see how. No one in my study group has any clue.

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    Important: It is needed as additional hypothesis that $a \neq 0$. For example, $\mathbb{Z}_p$ with $p$ prime is a ring, whose only subgroups are the trivial ones, and considering they ARE ideals (that's trivial), then they are the 2 ONLY ideals. It's easy to check that then $\mathbb{Z}_p$ it's a PID: $0$ it's generated by the element 0, and $\mathbb{Z}_p$ is generated by $1$. Then, ${0}$ is a maximal ideal, however , $0$ is not irreducible [by definition.](https://en.wikipedia.org/wiki/Irreducible_element)2017-06-06

4 Answers 4

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Hint $\ $ For principal ideals: $\ \rm\color{#0a0}{contains} = \color{#c00}{divides}$, $ $ i.e. $(a)\supseteq (b)\iff a\mid b,\,$ thus having no proper containing ideal (maximal) is the same as having no proper divisor (irreducible), $ $ i.e.

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

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Suppose that $R$ is a PID and that $\mathcal{I}$ is an an ideal. Then $\mathcal{I}$ is maximal iff for any $x$ generating $\mathcal{I}$, $x$ is irreducible.

Proof: $\Rightarrow$: Suppose $\mathcal{I}$ is maximal and that $\mathcal{I}$ is generated by $x$. Write $x = ab$ for some $a,b\in R$. Since $a|x$, $\mathcal{I}$ must be a subset of the ideal generated by $a$. Were this inclusion to be proper, by the maximality of $\mathcal{I}$, we would have $R$ being generated by $a$. This make $a$ a unit. By symmetry, $a$ or $b$ is a unit. We conclude that $x$ is irreducible.

$\Leftarrow$: Suppose that $x$ is irreducible. If $x$ is not a unit, there is a maximal ideal $\mathcal{I}$ of $R$ with $x\in\mathcal{I}$. Since $R$ is a PID, we can choose $y\in R$ so that $\mathcal{I}$ is generated by $y$. Since $x\in \mathcal{I}$, $y|x$. Write $x = ay$ for some $y\in R$. Since $x$ is irreducible $a$ is a unit. Hence $x$ and $y$ generate the same ideal; this ideal is maximal.

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    What about the case $R$ is not a PID. For example $\mathbb{R}[x,y]$?2015-02-10
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Proof by contradiction is a perfectly good idea.

First, suppose $x = y z$, and neither $y$ nor $z$ are units. Can you find an ideal containing $\left< x \right>$?

Second, suppose $\left< x \right>$ is not maximal, so there is an ideal containing it. Can you find a factor of $x$?

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    @Zhen: The d$i$sti$n$ction is s$u$btle and probably not worth arguing about.2011-05-17
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Suppose $\langle x \rangle$ is maximal and $x = yz$. Then $\langle x \rangle \subseteq \langle y \rangle$. From here, you should be able to show that $y$ is either a unit or an associate of $x$, showing $x$ is irreducible.

Suppose instead $x$ is irreducible. Then $\langle x \rangle \subseteq \langle y \rangle$ would imply $x = yz$ for some $z$. From here, you should be able to show that $y$ is either a unit or an associate of $x$, showing $\langle x \rangle$ is maximal.