Let $X$ be a compact topological space and $I\subseteq C^0(X)$ a proper ideal. How to show that $$V(I):=\{x\in X: f(x)=0,\ \forall f\in I\}\neq \varnothing?$$ Thanks.
How to show $V(I)=\{x\in X: f(x)=0, \forall f\in I\}\neq \emptyset$?
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general-topology
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0This isn't true. Take $X= [0,1]$. Then, $x \mapsto x+1$ is never zero, so $V(C^0(X)) = \varnothing$. Perhaps there's some missing hypothesis? Or am I misinterpreting something? – 2017-01-11
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0Can you provide a bit more context? That is, what are your thoughts? What have you tried so far? – 2017-01-11
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0@OpenBall presumably $I$ is a proper ideal in this problem – 2017-01-11
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0What if $I=C^0(X)$? – 2017-01-11
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0Indeed, $I$ is a proper ideal and $C^0(X)$ stands for the ring of continuous functions. – 2017-01-11
1 Answers
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Take $M$ be a maximal ideal such that $I\subset M$ and show that there is $a\in X$ such that $M=\{f\in C^0(X)\ \ / f(a)=0\}$. Hence $a\in I$.
Let $M$ be a maximal ideal and $I\subset M$: suppose that for all $x\in X$ there is $f_x\in M$ such that $f_x(x)\neq 0$. As $f_x$ is continuous there is an open $U_x$ such that $f_x$ not vanish on $U_x$. Now as $X$ is compact it can be covered by a finie open, say $U_{x_1},\ldots,U_{x_n}$. Let $f=f_{x_1}^2+\ldots+f_{x_n}^2$. Observe tha $f$ is invertible and $f\in M$.
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0Nice proof, but here is a question: Is there any reason you are considering a maximal ideal $M$ containing $I$? The proof seems to work equally with $I$ instead of $M$. – 2017-01-11
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1we can replace $M$ by $I$, and the resultat is that $I\subset \{f\in C^(X)\ / \ f(a)=0\}$. It work. And if $I$ is maximal ideal we have equality. – 2017-01-11
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0Aha, it sounds like you not only wanted to show that $V(I) \neq \varnothing$ but also tried to characterize maximal ideals in $C(X)$. That makes sense. – 2017-01-11