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In my course of spectral theory and operator algebras I came across the following exercise:

Let $\mathcal{A}=C_0(X)$ where $X$ is a locally compact Hausdorff space. Describe the multiplier algebra $M(\mathcal{A})$. Moreover, since the multiplier algebra is commutative, it is of the form $C(\beta X)$ where $\beta X$ is a compact Hausdorff space. Show that there exists an injection $j:X\rightarrow \beta X$ such that $j(X)$ is open in $\beta X$.

I was able to show that $M(\mathcal{A})\cong C_b(X)$, the bounded continuous functions on $X$, however the second part is annoying me. It seems very intuitive though.

I guess the more general statement is that if $\mathcal{A}\cong C_0(X)$ and $\mathcal{B}\cong C(Y)$ such that $\mathcal{A}$ is an essential two-sided ideal of $\mathcal{B}$, then $X$ injects into $Y$ such that $X$ is open in $Y$.

I'm aware that $\beta X$ is the Stone-Cech compactification, however I do not want to use that knowledge. How can I construct this map $j$ more or less constructively? Hints and tips are welcome, I tried certain things but I failed to prove the essential claims that I made.

Thank you in advance :)

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You are almost there. The ideals of $C (Y) $ are given by its closed subsets, and the essential ideals are those that correspond to closed nowhere dense subsets. In other words, the essential ideals of $C (Y) $ are precisely $C _0 (T) $, where $T\subset Y $ is open and dense.


Edit: proof.

Let $Y_0=\{t\in Y:\ f(t)=0,\ \text{ for all }f\in J\}$. Note that $$ Y_0=\bigcap_{f\in J} f^{-1}(\{0\}), $$ so $Y_0$ is closed and thus compact. Write $J_0=\{f\in C_0(T):\ f|_{Y_0}=0\}$. It is clear that $J\subset J_0$.

Let $V\subset Y$ be open with $Y_0\subset V$. For any $g\in J$, we have $|g|^2=\overline g\,g\in J$. Given $t\in Y\setminus V$, there exists $g_1\in J$ with $g_1(t)\ne0$; by replacing $g_1$ with $|g_1|^2$, we may assume that $g_1(t)>0$. As $Y\setminus V$ is compact, there exists $c>0$ with $g_1(t) \geq c$ for all $t\in Y\setminus V$. Define $g_2:Y\setminus V\to\mathbb C$ by $g_2(t)=1/g_1(t)$. Using again that $Y\setminus V$ is compact, we get from Tietze's Extension an extension, that we still call $g_2$, to all of $Y$. Let $g_V=g_1g_2\in J$ (since $g_1\in J$). We have $g_V|_{Y\setminus V}=1$, $g_V|_{Y_0}=0$, and $\|g_V\|\leq1$.

Now let $f\in J_0$, and fix $\varepsilon>0$. For each $t\in Y_0$ we have $f(t)=0$, so there exists an open set $V_t$ with $t\in V_t$ and $|f|<\varepsilon$ on $V_t$. As $Y_0$ is compact, it is covered by finitely many $V_{t_1},\ldots,V_{t_n}$. Let $V=\bigcup_{j=1}^nV_{t_j}$, open. We have $Y_0\subset V$ and $|f|<\varepsilon$ on $V$. Using again that $J$ is an ideal, $fg_V\in J$.

For any $t\in Y\setminus V$, since $g_V(t)=1$ we get $f(t)-f(t)g_V(t)=0$. For $t\in V$ we have $|f(t)-f(t)g_V(t)|=|f(t)|\,|1-g_V(t)|\leq 2\varepsilon$. Thus $$ \|f-fg_V\|\leq2\varepsilon. $$ As $\varepsilon$ was arbitrary, we have shown that $f\in \overline J=J$. So $J_0\subset J$, and the equality follows.

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  • $\begingroup$ I don't think so. $\endgroup$ – Martin Argerami Nov 16 '18 at 12:55
  • $\begingroup$ Would you mind showing me the proof of the above conclusion? $\endgroup$ – math112358 Jun 10 at 14:00
  • $\begingroup$ Edited. $ \ \ \ \ $ $\endgroup$ – Martin Argerami Jun 10 at 21:36
  • $\begingroup$ in where the condition "$T\subset Y$ is open and dense" be used? $\endgroup$ – math112358 Jun 12 at 17:17
  • $\begingroup$ The argument is a characterization of all the ideals. I'll let you deal with the essential ones. $\endgroup$ – Martin Argerami Jun 12 at 17:18

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