Deduce Stone's theorem in commutative algebra

Question

I am working on Ex.1.25 of Atiyah's "Commutative Algebra", which asks to deduce Stone's Theorem: every Boolean lattice is isomorphic to the lattice of open-and-closed subsets of some compact Hausdorff topological space.

By the previous exercises, I can prove that any Boolean lattice is isomorphic to the lattice associated with a Boolean ring, and it suffices to prove that for a Boolean ring $A$, its associated Boolean lattice, $L$, is isomorphic to the lattice of clopen subsets of $\mathrm{Spec}(A)$, since $\mathrm{Spec}(A)$ is compact Hausdorff. Then I considered a map sending $a\in L$ to $X_{a}$, where $X_a := \mathrm{Spec}(A)-V(a)=V(1-a)$. But I cannot prove this map is injective. Any idea is welcome!

score 1 · Accepted Answer · answered Jun 06 '23 at 21:21

1

If you know that two ideals define the same variety iff their radicals are equal, then you can prove injectivity as follows. Suppose $X_a=X_b$. Then the ideals generated by $a$ and by $b$ have the same radical.

In particular, some power of $a$ is divisible by $b$. But in a Boolean ring, every element satisfies $a^2=a$, so the only power of $a$ is $a$ itself. Thus, $a$ is divisible by $b$. That means $a\leq b$ in the Boolean lattice you started with.

Now repeat the last paragraph with the roles of $a$ and $b$ interchanged, to get $b\leq a$. So we have both $a\leq b$ and $b\leq a$, and therefore $a=b$.

answered Jun 06 '23 at 21:21

Andreas Blass

71,833

From "$a$ is divisible by $b$" I can get $a=bc$ for some $c\in A$, but in the Boolean lattice associated with $A$ the partial order is defined by $a\leq b \Longleftrightarrow a=ab$, so I am still confused. – Yuheng Shi Jun 06 '23 at 21:32
@YuhengShi If $a=bc$ then, by commutativity and idempotence, $ab=bcb=b^2c=bc=a$, and so $a\leq b$. – Andreas Blass Jun 06 '23 at 21:36
Oh, that's right! Thanks! – Yuheng Shi Jun 06 '23 at 21:37

Deduce Stone's theorem in commutative algebra

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