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Proving that an ideal in a PID is maximal if and only if it is generated by an irreducible

I am trying to see whether the ideal generated by irreducible element in a principal ideal domain (PID) is maximal ideal.

Suppose r is irreducible in a PID say D

Let I be an ideal of D containing (r) the ideal generated by r

Since D is a principal ideal domain, there exist s in D such that I=(s), therefore (r) is a subset of (s).

So, r=st , for some t in D but r is irreducible, this implies that s or t is a unit.

If s is a unit then I= (s)= D .

If t is a unit then (r)= I=(s). But I am not sure this is true, because I do not have reason for saying (r)= I=(s), so that I can conclude and say (r) is maximal.

I need a little help for this. Thanks

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marked as duplicate by Bill Dubuque, Chris Eagle, Zev Chonoles Oct 4 '11 at 20:15

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

up vote 4 down vote accepted

In fact, we can generalize a bit.

Proposition: If $R$ is an integral domain and $x\in R\setminus \{0\}$, then $x$ is irreducible if and only if $xR$ is maximal amongst all principal proper ideals of $R$ (ie if $I=yR \subsetneq R$ and $xR\subseteq yR$ then $xR=yR$).

Proof: ($\Rightarrow$) Suppose $x$ is irreducible and choose $y\in R$ with $xR\subseteq yR\subsetneq R$. Then, for some $r\in R$, $x=yr$. Since $x$ is irreducible, either $y\in U(R)$ or $r\in U(R)$. However, the fact that $yR\neq R$ implies that $y\notin U(R)$, hence $r\in U(R)$, $y=r^{-1}x$, and it easily follows that $xR=yR$. Hence, $xR$ is maximal amongst proper principal ideals of $R$.

($\Leftarrow$) Suppose that $xR$ is maximal amongst all principal proper ideals of $R$, and assume that $x=yz$ for nonzero nonunits $y,z\in R$. Then, since neither $y$ nor $z$ are units, it is clear that $xR=yzR\subsetneq yR\subsetneq R$. This contradicts the maximality of $xR$ amongst proper principal ideals of $R$. Therefore $x$ is irreducible. $\blacksquare$

Now, if $R$ is a PID, then the above proposition implies that for all irredicuble $x\in R$, $xR$ is a maximal ideal.

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Of course, this proposition helps. My question here Jack is why do you conclude that the result hold in a PID, is it because of all the ideals in a PID are principal? – Hassan Muhammad Oct 5 '11 at 6:17
Yes, a PID is usually defined as an integral domain for which every ideal is principal. – user5137 Oct 5 '11 at 13:45

An element is irreducible iff the ideal it generates is maximal amongst the principal ideals.

If all ideals are principal, then an element is irreducible iff the ideal it generates is maximal.

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In a PID, to divide is to contain: $b \mid a$ iff $(a)\subseteq(b)$. Thus, $(a)$ is maximal iff $a$ has no non-trivial divisors iff $a$ is irreducible.

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HINT $\ $ For principal ideals: contains $\iff$ divides. Hence, having no proper containing ideal (maximal) is equivalent to having no proper divisor (irreducible).

More generally, in domains where ideals satisfy contains $\iff$ divides (e.g. Dedekind domains), prime ideals $\ne 0$ are maximal. This characterizes PIDs, i.e. PIDs are precisely the UFDs where every prime ideal $\ne 0$ is maximal (i.e. Krull dimension $\le 1$).

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