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Let $A$ be a Boolean ring. One of my homework problems asks me to prove that the map of sets $\operatorname{Hom}_{Ring}(A,\mathbb{F}_2)\to \operatorname{Spec} A$ defined by $$ \phi\mapsto \ker(\phi) $$ is a bijection. I have no idea how to do it. My first question is that, how to show that there actually exists a homomorphism from $A$ to $\mathbb{F}_2$? I don't know how to assign elements of $A$ to elements of $\mathbb{F}_2$ so that it is a ring homomorphism.

Also any hints on this problem?

Thanks a lot!

KReiser
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Coco
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1 Answers1

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The kernel of a map from a ring to a field is a maximal ideal, and the points of $\mathrm{Spec} A$ are prime ideals, so your map is certainly an injection since maximal ideals are prime. Now, you know that the quotient by a prime ideal is an integral domain. What must a Boolean integral domain look like? Hopefully that's a sufficient hint.

As for existence (which actually isn't needed to do the exercise): that's as easy or difficult as showing that prime ideals (in a Boolean ring) exist in the first place. The usual approach is to use Zorn's lemma, or some other application of the axiom of choice. You don't need the full strength of AC, however. In fact the statement of existence is known as the Boolean Prime Ideal Theorem (BPIT), for which there is a huge literature concerning its set-theoretic significance.

user43208
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  • Thank you for your answer! Yes since a prime ideal of a Boolean ring exists (I do know this for a commutative nonzero ring), then $\phi: A\rightarrow A/\mathfrak{p}$ is a homomorphism, which induces a homomorphism $\phi':A\rightarrow \mathbb{F}_2$ since $\mathbb{F}_2\cong A/\mathfrak{p}$. Just curious, why existence isn't needed to do the exercise? I am always uncomfortable about functions between (possibly) empty sets so I always make sure that if $f:X\rightarrow Y$ is a function then both $X$ and $Y$ are nonempty.. – Coco Feb 08 '24 at 06:02
  • Coco, my advice is to let go of that habit, and get comfortable with it! Just to give an example: imagine a mathematician investigating properties that a hypothetical odd perfect number might have, even though there is every chance in the world that the set of these is empty. Similarly for a zero of the zeta function slightly to the right of the critical line. Anyway, existence isn't needed because it's just not. What you are trying to show is still true, even in universes of set theory where prime ideals don't exist! – user43208 Feb 08 '24 at 18:47