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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. –  Alex B. May 17 '11 at 16:20
    
This is on our practice final exam. Our exam is this Friday. There is no one collecting our work. –  Jessica May 17 '11 at 16:27
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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. –  Alex B. May 17 '11 at 16:33

4 Answers 4

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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The ideal containing <x> would have to be <y> or <z> –  Jessica May 17 '11 at 16:28
    
=> Suppose that <x> is a maximal ideal so x = yz where neither y nor z is a unit. Then <x> is a subset of <y> but <y> is not a subset of the ring R. This contradicts the definition of maximality. (I hope this part is legit) –  Jessica May 17 '11 at 16:38
    
$\langle y \rangle$ is certainly still a subset of the ring. –  Hans Parshall May 17 '11 at 17:12
    
This is less proof by contradiction and more proof by contrapositive. This proof might be easier attempted directly; see my answer. –  Hans Parshall May 17 '11 at 17:12
    
@Hans: I don't think there's much of a difference between a proof by contradiction and a proof by contrapositive. –  Zhen Lin May 17 '11 at 19:14

HINT $\ $ For principal ideals: contains $\iff$ divides. Thus having no proper containing ideal (maximal) is the same as having no proper divisor (irreducible).

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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.

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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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