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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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    Can you prove one direction? What exactly is you problem? It is almost always better to identify your conceptual difficulties with the concepts involved and ask about those, than to ask us solve your homework.2011-05-17
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    This is on our practice final exam. Our exam is this Friday. There is no one collecting our work.2011-05-17
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    I am not at all worried that you are trying to get some kind of unfair advantage. My point was that somebody solving this particular question for you will not help you in the exam. Understanding the concepts well enough to be able to solve it yourself will.2011-05-17
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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

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