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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.

Bill Dubuque
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Jessica
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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
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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. – Santropedro Jun 06 '17 at 22:00

4 Answers4

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

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

Bill Dubuque
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    But what if $p=0$? For instance, let $K$ be a field. Then, $(0_K)$ is a maximal ideal of $K$ but $0_K$ is not irreducible. So, wouldn't it be more accurate to say that if $p$ is non-unit and non-zero, then $(p)$ is maximal if and only if $p$ is irreducible? – Chris May 14 '21 at 23:19
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    @Chris That depends on definitions. This (possible) exception was already mentioned in the comments on the question many years ago. Generally my hints concentrate on the essence of the matter - leaving trivial exceptional cases to the reader. – Bill Dubuque May 14 '21 at 23:32
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    Thank you for your comment. Indeed this has been already mentioned, you're right. But I was wondering what happens if we want to see what is the set $\operatorname{Mspec}(D)$ of all the maximal ideals of a PID, say $D$. – Chris May 14 '21 at 23:40
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    Following up on @Chris's comment, the last implication is the one that uses that $p \ne 0$: if we have $p = dd'$ then, since $p$ has no proper divisors $(d) = (1)$ or $(d) = (p)$, and similarly for $(d')$. If $(d) = (1)$ or $(d') = (1)$ we are done, else $p = dd'$ with $(d) = (p)$ and $(d') = (p)$, and in this case $p = p^2 u$ for some unit $u$, thus $p(1 - pu) = 0$, since $p$ is not a zero divisor $1 - pu = 0$ hence $p$ is a unit, contradiction. – Caleb Stanford Feb 18 '24 at 19:30
  • Counterexamples are likely in a principle ideal ring rather than a PID where we can take $p$ to be some idempotent element. For instance $(2) = (4)$ is idempotent and maximal in $\mathbb{Z}/6\mathbb{Z}$ but not irreducible since $4 = 4^2$. – Caleb Stanford Feb 18 '24 at 19:32
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    @Caleb OP's context is a domain. Generally divisibility theory is more complicated when zero-divisors are present, e.g. basic notions such as "irreducible" bifurcate into a few inequivalent notions. – Bill Dubuque Feb 18 '24 at 19:44
  • Thanks for the links, very interesting! Yes, and as to the relevance to OP, I like to think of $p = 0$ as a zero divisor even in a domain. – Caleb Stanford Feb 18 '24 at 19:54
  • @BillDubuque You might be interested in answering my new question here. – Caleb Stanford Feb 18 '24 at 20:46
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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.

ncmathsadist
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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.

Hans Parshall
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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 $x$?

Zhen Lin
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