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Let $X$ be a compact Hausdorff space and $\mathcal P$ be a prime ideal of the ring of all real valued continuous functions on $X$. Then $\mathcal P$ must contained in one maximal ideal i.e. there is a point $y\in X$ such that each element of $\mathcal P$ vanishes at the point $y$. Does this imply that, $\{ y\}$ is $G_{\delta}$ in $X$?

Sumanta
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No. Why would it imply $\{y\}$ is $G_\delta$ in $X$? For a simple example where it is not, let $X$ be any product of uncountably many compact Hausdorff spaces with more than one point. Then since any nonempty open set in the product topology is restricted on only finitely many coordinates, any nonempty $G_\delta$ set is restricted on only countably many coordinates, so no singleton is $G_\delta$.

Eric Wofsey
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  • In this case , i.e. whenever X is compact hausdorff and no point of X is of type G-delta, can we show a prime ideal of C(X) which is not maximal. – Sumanta Sep 27 '18 at 16:04
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    There are prime ideals that are not maximal in $C(X)$ for any infinite compact Hausdorff space $X$. See https://math.stackexchange.com/questions/527658/prime-ideals-in-c0-1. See also https://math.stackexchange.com/questions/2323548/what-is-the-krull-dimension-of-cx-for-x-infinite-compact-and-hausdorff for a stronger result. – Eric Wofsey Sep 27 '18 at 16:08
  • $C(\beta \omega)$ is also a good candidate space for such a ring. – Henno Brandsma Sep 28 '18 at 20:56