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Let $A,B$ be complete Boolean algebras and $\varphi,\psi:A\rightarrow B$ be maps preserving $0,1$, and arbitrary joins and meets. Let $C$ be the equalizer of these two; so $C=\left\{a\in A:\varphi(a)=\psi(a)\right\}$. It can be easily checked that $C$ itself is a complete Boolean algebra.

Question: Let $a\in A$ be given so that for any prime ideal $\mathfrak{r}$ in $B$, we have $$\varphi(a)\notin\mathfrak{r}\quad\Leftrightarrow\quad \psi(a)\notin\mathfrak{r}.$$ Then for any prime ideal $\mathfrak{p}$ in $A$ with $a\notin\mathfrak{p}$, is $\mathfrak{p}\cap C=\mathfrak{q}\cap C$ implies $a\notin\mathfrak{q}$?

A motivation of the question was to check continuity of a certain functor. Let us denote the Stone space of prime (or equivalently, maximal) ideals as $\mathrm{Spec}$, so that $A$ and $B$ can be identified with the clopen algebras of $\mathrm{Spec}A$ and $\mathrm{Spec}B$. Then $\mathrm{Spec}C$ becomes a quotient space of $\mathrm{Spec}A$, by defining an equivalence relation $\sim$ on $\mathrm{Spec}A$ as $$\mathfrak{p}\sim\mathfrak{q}\quad \textrm{if and only if}\quad\mathfrak{p}\cap C=\mathfrak{q}\cap C.$$ By the well-known duality, certainly $\mathrm{Spec}C$ should be the coequalizer of the induced maps $\varphi^{*},\psi^{*}:\mathrm{Spec}B\rightarrow\mathrm{Spec}A$ in the category of Stone spaces (with continuous maps as morphisms). Completeness conditions in fact says that this is the coequalizer in the category of Stonean spaces. But it is not clear to me about the relation between the "usual" quotient space of $\mathrm{Spec}A$ by the relation $\varphi^{*}(\mathfrak{r})\sim'\psi^{*}(\mathfrak{r})$ and the space $\mathrm{Spec}C$. The question asks that if we have a clopen set $a$ in $\mathrm{Spec}A$ which is closed under the relation $\sim'$, then it is also closed under $\sim$.

I'm also not sure about that completeness really matters or not.

Thank you.

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The condition $\varphi(a)\notin\mathfrak{r}\iff \psi(a)\notin\mathfrak{r}$ is equivalent to $\varphi(a)=\psi(a)$, since an element of a Boolean algebra is determined by the prime ideals that contain it (if $a\neq b$, then there is a prime ideal containing $1+a+b$, which then must contain exactly one of $a$ and $b$). So in fact $a\in C$, so trivially if $\mathfrak{p}\cap C=\mathfrak{q}\cap C$ then $a\not\in\mathfrak{p}$ implies $a\not\in\mathfrak{q}$.

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  • $\begingroup$ Thank you very much. But unfortunately, I think I misunderstood myself about the real problem.. The question I should ask was the equivalence of $\sim$ and $\sim'$, instead of the original question I wrote. Do you have any idea about that? Again, thanks. $\endgroup$
    – Junekey Jeon
    Commented Feb 5, 2016 at 9:11
  • $\begingroup$ No, they aren't the same in general. For instance, let $I$ be any connected compact Hausdorff space, and let $\pi:S\to I$ be any continuous surjection from a Stone space to $I$. Taking the kernel pair of $\pi$ gives a pair of maps $\varphi^*,\psi^*:T\to S$ from some Stone space $T$ whose coequalizer is $\pi$. Thus the quotient by the equivalence relation $\sim'$ in this case is $I$. But the quotient by $\sim$ will be just a point, since $I$ is connected. $\endgroup$
    – Eric Wofsey
    Commented Feb 5, 2016 at 9:25

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