For a locally ringed space $(X,\mathcal{O})$, is there a sheaf of ideals whose stalks are the max ideals of the stalks of $\mathcal{O}$?

Question

For $X$ a locally ringed space with structure sheaf $\mathcal{O}$, for each $x \in X$ let $\mathcal{M}_x$ denote the max ideal of the stalk $\mathcal{O}_x$ of $\mathcal{O}$ at $x$.

In general, does there exist an $\mathcal{O}$-ideal sheaf $\mathcal{I} \subset \mathcal{O}$ on $X$, whose stalk at each $x \in X$ is given by $\mathcal{I}_x = \mathcal{M}_x$?

Answer 1

Edit: this answer originally erroneously said the answer was “yes”. In fact, the answer is “no”. Consider the sheaf of real-valued continuous functions on $X = \mathbb{R}$. We see that $\mathcal{O}_x$ consists of the germs of functions which vanish at $x$.

Consider $f(z) = z - x$. The germ of $f$ is in $M_x$.

If we had such an $I$, we could find some neighbourhood $U$ of $x$ such that $f|_U \in I(U)$. But we could find $y \neq x$, $y \in U$. Then we should have that the germ of $f$ is in $M_y$, but $f$ doesn’t vanish at $y$.

However, there is a “co-ideal” $C \subseteq \mathcal{O}$, defined by $C(U) = \{x \in \mathcal{O}(U) \mid x$ is a unit$\}$. The stalks $C_x$ are exactly $\mathcal{O}_x \setminus M_x$.