5

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

rschwieb
  • 160,592
Jam
  • 2,812

3 Answers3

7

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?)

freebird
  • 130
  • i think <2> is maximal on Z .So you are saying <2> < <2x> (subset) – Jam Jun 13 '14 at 13:41
  • Well you can check by Z[x]/<2> if it is a field then <2>is maximal? But Z[x]/<2> seems like it is isomorphic with Z[x] which is just an Integral domain am i right? I cant think of an isomorphism at the moment though – Jam Jun 13 '14 at 13:50
  • @ManolisLyviakis: so you have your answers... – Dietrich Burde Jun 13 '14 at 13:51
  • Im just not so sure is Z[x]/<2> isomorphic to Z[x]? it just change by multiples of 2 ... – Jam Jun 13 '14 at 13:53
  • I think what freebird is saying is $\langle2\rangle\subsetneq\langle2,x\rangle\subsetneq\mathbb{Z}[x]$ – robjohn Jun 13 '14 at 14:22
  • ohh <2,x> not 2*x xD thanks – Jam Jun 13 '14 at 14:26
4

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).

Dietrich Burde
  • 140,055
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.

rschwieb
  • 160,592