Is the ideal generated by irreducible element in principal ideal domain maximal? [duplicate]

Possible Duplicate:
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

Solution 1:

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.

Solution 2:

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.