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For a manifold $X$ with a point $x \in X$ define the ring of germs of the smooth functions at $x$ to be $C^{\infty}_x(X)=C^{\infty}(X)/\sim$ where $f_1 \sim f_2 \iff \exists U \in \tau_X.f_1|U=f_2|U$ where $\tau_X$ is a topology on $X$. I assume the operations to be the pointwise addition and multiplication.

Show that the subset $m=C^{\infty}_x(X)\setminus C^{\infty}_x(X)^\times$ of non-units is an ideal.

My problem is to prove that the subset of non-units is closed under the additivity. Consider $f \in C^{\infty}_x(X)$ such that $\forall U \in \tau_X. x \in U \implies \exists y \in U. f(y)=0$. Clearly, $f$ is a non-unit. However, one could show that there exists an analogous $g$ to this $f$ such that both satisfy $\forall x \in X. \neg (f(x)=0 \wedge g(x)=0)$, hence the subset of non-units $m$ would not be closed under the addition as $f+g \notin m$ and $m$ cannot be an ideal.

Dávid Natingga
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You are thinking too complicated. The set of non-units in $C_x^\infty(X)$ is precisely the set (let's call it $V_x$) of germs of (smooth) functions vanishing in $x$. It is easily seen that that is an ideal.

To see that the set of non-units is $V_x$, first, it is clear that $V_x$ contains only non-units. On the other hand, if we have a germ $f_x \in C_x^\infty(X)\setminus V_x$, let $f \in C^\infty(X)$ be any representant of $f_x$. Let $c = f(x)$. There exists a neighbourhood $U$ of $x$ such that $y\in U \Rightarrow \lvert f(y) - c\rvert < \frac{\lvert c\rvert}{2}$. Choose a $\varphi \in C^\infty(X)$ with $\varphi(x) = 1$ in a neighbourhood $V$ of $x$ with $\overline{V} \subset U$, $0 \leqslant \varphi \leqslant 1$, and $\operatorname{supp}\varphi \subset U$. Let

$$g(y) = \varphi(y)\cdot f(y) + (1-\varphi(y))\cdot c.$$

Then $g \equiv f$ on $V$, so $g\sim f$, and $g$ has no zero on $X$, since $\lvert g(y)\rvert \geqslant \lvert c\rvert - \varphi(y)\lvert f(y)-c\rvert \geqslant \lvert c\rvert/2$, so $g$ is invertible, hence $f_x$ is a unit.

Daniel Fischer
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