Question

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 ?

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

Answer 2

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"

Answer 3

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.