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I want to prove the following:

Let $A$ be a Boolean ring and let $\mathfrak{a}\neq A$ be an irreducible ideal. Then $\mathfrak{a}$ is maximal.

I already know that the prime ideals of $A$ are maximal and that $\mathfrak{a}$ is the intersection of the maximal ideals that contain $\mathfrak{a}$.

Suppose $\mathfrak{a}$ is not maximal and let $M$ be the set of maximal ideals containing $\mathfrak{a}$. I would like to find two subsets $R,S\subset M$ such that $\bigcap_{\mathfrak{m}\in R}\mathfrak{m}\neq\mathfrak{a}$, $\bigcap_{\mathfrak{m}\in S}\mathfrak{m}\neq\mathfrak{a}$ and $\bigcap_{\mathfrak{m}\in R\cup S}\mathfrak{m}=\mathfrak{a}$. But I'm not sure if this is possible, especially if $M$ is very large.

Can someone give me a hint?

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2 Answers 2

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Taking by quotient, we may assume $\mathfrak{a}=0$ is irreducible. In a Boolean ring $A$, for any element $x$, $0=xA\cap (1-x)A$, thus $xA=0$ or $(1-x)A=0$. Thus $x=0$ or $x=1$, that is to say $A$ has only two elements, $A$ must be $\mathbb{F}_2$. We are done.

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  • $\begingroup$ This looks good, but at the moment I don't see why $xA\cap(1-x)A=0$. $\endgroup$
    – Stefan
    May 26, 2012 at 18:04
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    $\begingroup$ @StefanWalter In general if $e$ is an idempotent in a ring $R$, $eR\cap(1-e)R=0$. To see this, suppose $er=(1-e)s$ is in the intersection. Then multiplying on the left with $e$, $er=e^2r=0s=0$. $\endgroup$
    – rschwieb
    May 26, 2012 at 18:38
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    $\begingroup$ @Stefan $ $ Hint $\rm\ \ I + J = 1\:\Rightarrow\: I\cap J = I\:\!J\:$ since $\rm\:(I+J)(I\cap J)\subseteq I\:\!J\subseteq I\cap J.\ $ **QED** $\ $ Now specialize $\rm\: I = xA,\ J = (1-x)A\:\Rightarrow\:I\:\!J = 0.$ $\ \ $ $\endgroup$
    – Bill Dubuque
    May 26, 2012 at 18:46
  • $\begingroup$ @Stefan Above is an ideal generalization of $\rm\:gcd(i,j) = 1\:\Rightarrow\:lcm(i,j) = i\:\!j.$ Indeed, this is simply the special principal ideal case $\rm\:I = (i), J = (j)\:$ in $\:\mathbb Z\:$ (or any PID/Bezout). $\endgroup$
    – Bill Dubuque
    May 26, 2012 at 19:09
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It is well-known (and easy to prove) that $\mathrm{Spec}(A)$ is Hausdorff (in fact, a Stone space). Besides, $\mathfrak{a}$ is irreducible iff $V(\mathfrak{a})$ is irreducible (see Ted's comment; here we use that every ideal in a boolean ring is radical). The only irreducible Hausdorff space is a point. Points correspond to maximal ideals.

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    $\begingroup$ Dear @Martin Brandenburg, It seems that it is not true for the statement "Besides, $\mathfrak{a}$ is irreducible iff $V(\mathfrak{a})$ is irreducible." in general, see mathoverflow.net/questions/87870/… $\endgroup$
    – wxu
    May 26, 2012 at 17:21
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    $\begingroup$ In the ring $k[x,y]/(x^2,xy, y^2)=k[\xi, \eta]$, the ideal $(0)=(\xi)\cap (\eta)$ is reducible although $V(0)$ is irreducible.The statement "$\mathfrak{a}$ is irreducible iff $V(\mathfrak{a})$ is irreducible" in the answer is thus indeed false in general. $\endgroup$
    – Georges Elencwajg
    May 26, 2012 at 17:57
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    $\begingroup$ But it is true in a Boolean ring, since every ideal is its own radical. We have $V(a) = V(b)$ iff $\sqrt{a} = \sqrt{b}$ (in every ring) iff $a=b$ (in a Boolean ring). So $V$ is injective on ideals. So $a = b \cap c$ with $a \ne b,c$ iff $V(a) = V(b \cap c) = V(b) \cup V(c)$ with $V(a) \ne V(b), V(c)$. $\endgroup$
    – Ted
    May 26, 2012 at 19:02

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