We consider $A = C_{b}(X)$, the ring of continuous bounded functions on a completely regular space $X$. Let $\DeclareMathOperator{\SpecMax}{SpecMax} \SpecMax(A)$ be the set of maximal ideals of $A$ with the Zariski topology.

We know that there is an embedding of topological spaces: $$ \psi : X \longrightarrow \SpecMax(A) $$ defined by $\psi(x) = m_{x} := \lbrace f\in A \mid f(x)=0\rbrace$.

My question is: We can construct the Stone–Čech compactification of $X$ if we take $\bar{X}:=\overline{\psi(X)}$, but we must prove that $\overline{\psi(X)}$ is compact. We know that $\operatorname{Spec}(A)$ is quasi-compact, but I don't know if $\SpecMax(A)$ is also. Then how can we prove that $\overline{\psi(X)}$ is compact?

  • $\begingroup$ the base of Zarisky topology is maked by sets $V(I)$ of all promes $p$ containing $I$, for $I$ a subset (or a ideal is you want) of $A$. THen if $Max$-$Spec(A)\subset \cup_i U_i$ for a familiy of open sets $U_i\subset X$, by the definition of Zariski topology above follow that $X= \cup_i U_i$ (every prime ideals is contained in a maximal one). $\endgroup$ Nov 21, 2012 at 19:57
  • $\begingroup$ Sergio, the definition of (quasi-)compact requires that every open covering contains a finite subcovering. $\endgroup$
    – Ralph
    Nov 22, 2012 at 10:02
  • $\begingroup$ @Ralph, Rajkov ask "how prove that X is quasi-compact?" (then every close of X is quasi-compact too). Now from what I write above if $X$ has a open cover, i.e. $X= \cup_i (U_i\cap X)$ for open sets $U_i\subset Spec(A)$, then $X\subset \cup_i U_i$ and follow that $Spec(A)\subset \cup_i U_i$ then there are finite collection $U_1,\ldots U_n$ that cover $Spec(A)$, then cover $X$. $\endgroup$ Nov 23, 2012 at 17:09

1 Answer 1


You can prove that $\text{SpecMax}(A)$ is quasi-compact in the same way as you do for $\text{Spec}(A)$. Also note that $$\overline{\psi(X)} = \text{SpecMax}(A)=:\beta X.$$ I know of no (very) short proof that $\text{SpecMax}(A)$ is Hausdorff (in Zariski topology). However, a proof is presented in the proof of Theorem 9 (pdf-page 11) in the following well-written paper:

E. Hewitt: Rings of real-valued continuous Functions. Trans. Amer. Math. Soc. 64(1948),45-99

  • $\begingroup$ Hi Ralph. If $R$ is a comutative pm-ring( ring in which any prime ideal contained in unique maximal ideal), then the space of maximal ideals, Max(R), is a Hausdorff space. see "COMMUTATIVE RINGS IN WHICH EVERY PRIME IDEAL IS CONTAINED IN A UNIQUE MAXIMAL IDEAL, PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 30, No. 3, November 1971" $\endgroup$
    – user29350
    Nov 23, 2012 at 16:38
  • $\begingroup$ Yes, but I don't think the proof that each prime ideal of $A$ is contained in a unique max. ideal (can be found in Gillman-Jerison), is shorter than a direct proof that SpecMax(A) is Hausdorff. Anyway, the OP should now have enough references to see why the max spectrum is Hausdorff. $\endgroup$
    – Ralph
    Nov 23, 2012 at 20:10
  • $\begingroup$ Ok. You are right. But there is a question about the space of minimal prime ideals of a reduced ring $R$. We know that this space is Hausdorff and need not be compact. So it has a compactificaton. Now, what is a compactification of $\mathrm{Min}(R)$, the space of minimal prime ideals of $R$? Note that $\overline{\mathrm{Min}(R)}=\mathrm{Spec}(R)$, but $\mathrm{Spec}(R)$ is not a Hausdorff space. $\endgroup$
    – Ali
    Nov 23, 2012 at 20:32

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