Uniformly closed ideals of smooth/real analytic functions

Question

Consider $U\subseteq \mathbb{R}^n$ an open subset and denote by $R$ either the algebra of real-valued smooth or real analytic functions on $U$. In either case suppose that $R$ is equipped with the subspace topology from $C^0(U,\mathbb{R})$, which is equipped with the topology of uniform convergence on compact sets.

I am interested which ideals of $R$ are closed with respect to this topology. In particular, are ideals generated by finitely many real analytic functions closed in this topology?

I know that if one provides the ring of smooth functions with the Frechet topology of uniform convergence of derivatives on compact sets, then the closed ideals are well understood. In particular, any ideal generated by finitely many analytic functions is then closed in that topology. I also know that the topology I consider on the smooth/real analytic functions here is not particularly well-behaved. For example, it is not complete.

It should be noted that if $X\subseteq U$ is a closed subset then the ideal $J_X\subseteq R$ of all functions in $R$ that vanish on $X$ is closed in the given topology, as uniform convergence implies point-wise convergence.

I am also aware that if one considers the ring of continuous functions on $U$ with the topology of uniform convergence, then the only closed ideals are the ideals of functions vanishing on a given closed subset.

I tried tracking down discussions of this in the literature, but my search has not yielded any fruitful insights.

Answer 1

Suppose that $g \in I \subset C^\infty(U)$ belongs to a $C^0$-closed ideal, while $f$ has the same zero-set as $g$. Choose a family of smooth functions $\varphi_\varepsilon \colon \mathbb{R} \to \mathbb{R}$ that smoothly and monotonously interpolates between $\varphi_\varepsilon(x) = 0$ for $|x| \le \varepsilon/2$ and $\varphi_\varepsilon(x) = x$ for $|x| \ge \varepsilon$. Then the ratio $\varphi_\varepsilon(f)/g$ is smooth $$ |f - \frac{\varphi_\varepsilon(f)}{g} g| = |f - \varphi_\varepsilon(f)| < 2\varepsilon $$ uniformly over $U$. So $f$ is arbitrarily close to elements of $I$, hence $f\in I$. This shows that the only $C^0$-closed ideals in $C^\infty(U)$ are of the form $J_X$, which is essentially an adaptation of the argument from the $C^0(U)$ case.

In the analytic case, one could choose $\varphi_\varepsilon(x)$ to vanish at high but non-infinite order at $x=0$. This may or may not work depending on how $g$ vanishes at its zero set. I'm not sure about the general conditions. Division by analytic functions is known to be delicate (see some references from this question for instance).