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

This is my try

Proof: ⇒: Suppose I is maximal and that I is generated by x. Write x=ab for some a,b∈R. Since a|x, I must be a subset of the ideal generated by a. Were this inclusion to be proper, by the maximality of 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.

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

Is my proof correct???

egreg
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Sara
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1 Answers1

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$\Rightarrow$ is basically correct, although instead of "$I$ must be a subset of the ideal generated by $a$, I would say $I\subseteq (a) \subseteq R$", and then say "Were the first inclusion to be proper" and so on.

$\Leftarrow$ seems a little confused, although clearly you understand what you are trying to do. You are given that for any $x$ generating $I$ that $x$ is irreducible. So your proof should go something like this: Suppose that $I = (x)$, and choose a maximal ideal $J$ with $I\subseteq J$. Let $J = (y)$. You can then continue as you did in your proof.

rogerl
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