Must an ideal generated by an irreducible element be a maximal ideal?

If you have an irreducible element say $b$ in a ring, is the ideal $\langle b\rangle$ a maximal ideal?

Solution 1:

Take the polynomial ring $\Bbb{Z}[x]$.

$2\in\Bbb{Z}[x]$ is irreducible. But $(2)$ is not maximal (what if $x$ is added to the ideal?)

Solution 2:

In general, an ideal $I=\langle b\rangle$ of $R$ generated by an iireducible element $b$ need not be maximal. For example, take $R=\mathbb{Z}[\sqrt{-5}]$ and $b=1+\sqrt{-5}$. Then $b$ is irreducible, but the ideal $(1+\sqrt{-5})$ is not maximal.On the other hand, for a PID we have: an ideal $I$ is maximal if and only if it is generated by an irreducible element (this has been shown multiple times here).

Solution 3:

Let's analyze what being irreducible says. By definition, $b$ is irreducible iff $b=xy$ implies at least one of $x$ or $y$ is a unit.

Suppose $a$ is another element such that $(b)\subsetneq (a)\subsetneq R$. Then $b=ar$ for some $r$. Since $(a)$ is proper, $a$ isn't a unit, and since $(a)\neq (b)$, it is impossible for $r$ to be a unit, so this would say that $b$ is reducible. If we ask for $b$ to be irreducible, then it implies that $(b)$ is maximal among proper principal ideals.

Conversely, suppose $(b)$ has this property of being maximal among proper principal ideals and suppose $b=xy$. Then $(b)\subseteq (x)\cap (y)$. If $x$ isn't a unit, then $(b)=(x)$. This implies that $x=br$ for some $r$, and then you get $b=bry$, hence $b(1-ry)=0$. If $b$ isn't a zero divisor (as the case would be if we were in a domain, say) then we could conclude that $1=ry$ and that $y$ is a unit. So if $b$ is regular (=isn't a zero divisor) then the converse is true: $(b)$ maximal among proper principal ideals implies $b$ is irreducible.

Of course, if you're in a principal ideal ring, being maximal among proper principal ideals is the same thing as being a maximal ideal. So in a PIR, $b$ irreducible implies $(b)$ maximal.

This leads you naturally to consider the example that freebird gave of $\Bbb Z[x]$, since we know the maximal ideals are all $2$-generated.