Problem of the Week #225 - Sep 20, 2016

[Euge]

Here is this week's POTW:

-----
Suppose $X$ is a compact Hausdorff space. Show that there is homeomorphism between $X$ and the collection $Y$ of maximal ideals in $C(X,\Bbb R)$, the space of continuous real-valued functions on $X$ (here, $Y$ is topologized with the Zariski topology).

-----

Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Euge]

No one answered this week's problem. You can read my solution below.

Spoiler

For each $x\in X$, let $\mathbf m(x) := \{g\in C(X) : g(x) = 0\}$. I claim that the mapping $X \to Y$ sending $x$ to $\mathbf{m}(x)$ is a homeomorphism. Given an element $\mathbf{m}\in Y$, let $$X_{\mathbf{m}} = \bigcap_{g\in \mathbf{m}} \{x\in X : g(x) = 0\}$$ Then $X_\mathbf{m}$ is nonempty. Otherwise, to each $x\in X$ there corresponds a $g_x\in \mathbf{m}(x)$ for which $g(x)$ is nonzero. By continuity of $g_x$, there is an open neighborhood $U_x\ni x$ on which $g_x$ is nowhere zero. By compactness of $X$, finitely many $U_x$ cover $X$, say $U_{x_1},\ldots U_{x_n}$. Then $g := \sum\limits_{i = 1}^n g_{x_i}^2$ is non-vanishing on $X$; this means $g$ is invertible in $C(X)$. This contradicts the fact that $g$ belongs to the maximal ideal $\mathbf{m}$. So $X_\mathbf{m} \neq \emptyset$.

Since $X$ is compact Hausdorff, Urysohn's lemma shows that if $x \neq y$, there is a $g\in C(X)$ so that $g(x) \neq g(y)$. This implies $\mathbf{m}(x) \neq \mathbf{m}(y)$, proving injectivity of the mapping $X \to Y$.

To prove surjectivity, take $\mathbf{m}\in Y$ and $x\in X_{\mathbf{m}}$. For all $g\in \mathbf{m}$, $g(x) = 0$, so then $g\in \mathbf{m}(x)$. Consequently, $\mathbf{m}\subset \mathbf{m}_x$. Maximality of $\mathbf{m}$ forces $\mathbf{m} = \mathbf{m}(x)$, proving surjectivity.

Given $g\in C(X)$, the sets $U(g) := \{x \in X : g(x) \neq 0\}$ and $\mathbf{U}(g) := \{\mathbf{m}\in Y : g\notin \mathbf{m}\}$ are basic open sets in $X$ and $Y$, respectively. So it suffices to show that $U(g)$ maps onto $\mathbf{U}(g)$ under the map $X\to Y$. To each $\mathbf{m}\in \mathbf{U}(g)$, there corresponds an $x\in X$ for which $\mathbf{m}(x) = \mathbf{m}$. Then $g(x) \neq 0$ (as $g\notin \mathbf{m}$), whence $x\in U(g)$ and $U(g)$ maps onto $\mathbf{U}(g)$.