Let $X$ be an infinite compact Hausdorff space and let $C(X)$ be the ring of real-valued continuous functions on $X$. Does $(0)$ have a primary decomposition in this ring?
I feel like the answer is no. My reasoning is as follows: suppose $(0)$ has a minimal primary decomposition $(0) = \bigcap^n_{i = 1} \mathfrak{q}_i$ with $r(\mathfrak{q}_i) = \mathfrak{p}_i$. Now every ideal in $C(X)$ is contained in a maximal ideal of the form $\mathfrak{m}_x$ consisting of the functions which vanish at $x$ (and all the maximal ideals have this form). It must also be the case that each $\mathfrak{q}_i$ is contained in a unique $\mathfrak{m}_i$ by the minimality of the primary decomposition (i.e. $\mathfrak{q}_i \not\supseteq \bigcap_{j \neq i} \mathfrak{q}_j$). Then all the $\mathfrak{q}_i$ intersect at $0$ only; therefore the $\mathfrak{q}_i$ are not unique and our decomposition is not minimal.
I am unsure of the step where I claim that each $\mathfrak{q}_i$ is contained in a unique $\mathfrak{m}_x$. I was thinking if that wasn't the case then we would have $\mathfrak{q}_i \supset \mathfrak{a}$ for some $\mathfrak{q}_i$ and thus $\mathfrak{q}_i \supseteq \bigcap_{i \neq j} \mathfrak{q}_j$, which contradicts the minimality of the prime decomposition.
