# Weak Nullstellensatz in locally ringed spaced?

Is there a version of the weak Nullstellensatz valid in general locally ringed spaces? The statement I imagine is something like this:

Let $$(X, \mathcal{O}_X)$$ be a locally ringed space (possibly with some assumptions). Let $$I$$ be an ideal in $$\Gamma(X, \mathcal{O}_X)$$, and suppose for all $$p \in X$$, there is some $$f \in I$$ such that $$f|_p$$, i.e. in the image of $$f$$ in $$\mathcal{O}_{X, p}$$, is a unit. Then $$I = \Gamma(X, \mathcal{O}_X)$$.

One can show that this is true if $$I$$ is a principal ideal (if $$f$$ is invertible in all stalks, then we can find an inverse around each point and glue them together).

• With those assumptions, wouldn't it follow that $1 \in I_p$ for all $p$, and therefore $1 \in \Gamma(X, I)$? – Daniel Schepler Oct 16 '18 at 23:18
• @DanielSchepler Can you elaborate? This would give a very easy proof of the weak Nullstellensatz. – vukov Oct 17 '18 at 1:51
• Where do you need elaboration? (Though I think to prove the weak Nullstellensatz from this, you would need to show that all the closed points of $\operatorname{Spec}(k[x_1, \ldots, x_n])$ are of the form $\langle x_1 - a_1, \ldots, x_n - a_n \rangle$ for $k$ alg. complete anyway, which is just one of the alternate formulations of the Nullstellensatz.) – Daniel Schepler Oct 17 '18 at 17:11
• Think of the example of the ideal $\langle t^2 + 1 \rangle$ in $\operatorname{Spec} \mathbb{R}[t]$: the space has points $\langle t - a \rangle$ where the residue field is $\mathbb{R}$, but it also has points $\langle t^2 + at + b \rangle$ for $a^2 - 4b < 0$ where the residue field is $\mathbb{C}$. Now $t^2+1$ is invertible at every real point, but not invertible at the complex point $\langle t^2+1 \rangle$. So, just being invertible at the real points is not sufficient for $t^2+1$ to be invertible globally. – Daniel Schepler Oct 17 '18 at 17:16
• @DanielSchepler I think the result is false, at least without assuming that X is compact. For example, consider the sheaf of continuous function to $\mathbb{R}$ on $\mathbb{R}$. Consider the ideal $I$ of compactly supported functions. This is a proper ideal, but at every point in $\mathbb{R}$, there is a function that does not vanish at that point. – vukov Oct 27 '18 at 16:23