# Topological proprieties of Specmax(A)

Hello,

We consider $A=C_{b}(X)$ the ring of continous bounded fonction on a completely regular space $X$, SpecMax(A) the set of maximal ideal of $A$ with the zariski topology.

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

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

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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). –  Buschi Sergio Nov 21 '12 at 19:57
Sergio, the definition of (quasi-)compact requires that every open covering contains a finite subcovering. –  Ralph Nov 22 '12 at 10:02
@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 $X$. –  Buschi Sergio Nov 23 '12 at 17:09
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## 3 Answers

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

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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"

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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. –  Ralph Nov 23 '12 at 20:10
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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.

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Dear Ali, could you please register a single account to avoid all this duplication? Then you can leave comments on your own answers –  Yemon Choi Nov 23 '12 at 20:43
Hi Yemon. Thank you very much of your hint. I will use Ali taherifar. –  Ali Nov 24 '12 at 5:08
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