# On a *ringed space*, show that the non vanishing set of $f$ is open, and that it is invertible there

This is an exercise of Ravi Vakil that I solved by a very trivial argument without using the hint. For this reason I'm worried that I might have missed something.

If $f$ is a function on a locally ringed space $X$, show that the subset of $X$ where $f$ doesn't vanish is open. (Hint: show that if $f$ is a function on a ringed space $X$, show that (sic) the subset of $X$ where the germ of $f$ is invertible is open)

Solution: Assuming that the first sentence means "$f$ is an element of $\mathcal O_X$", and that $\mathcal O_X$ is a ring of continuous functions on $X$, we just pull back the closed set $\{0\}$ to $X$ via $f^{-1}$. Under a continuous function the pull back of a closed set is closed, and therefore the complement of the locus where $f$ doesn't vanish is closed.

Questions:

1) is my solution correct?

2) I know it is true for the special example of schemes, but for a ringed space in general why is the locus of where a function $f$ doesn't vanish the same of where the germ of $f$ is invertible? The most I can say is that this is true for locally ringed spaces (and not for just ringed spaces as the hint suggests), because if the germ of a function is non-zero at a point $p$ in a locally ringed space $X$, $f_p$ can't belong to the maximal ideal $\mathfrak m_p$. For if it did, then under the homomorphism $ev_p: \mathcal O_{X,p} \rightarrow k$ the maximal ideal of $k$, which is $(0)$, would not pull back to $\mathfrak m_p$ (it would miss $f_p$).

There is a subtle and interesting point in this context that one really has to understand:
let me try to be as clear as possible rather than concise.

a) Let $$(X,\mathscr O_X)$$ be a ringed space that is not necessarily locally ringed.
Then given $$f\in \mathscr O_X(X)$$, we can ask whether the germ $$f_x\in \mathscr O_{X,x}$$ at $$x\in X$$ is invertible i.e. if $$f_x\in \mathscr O_{X,x}^\star$$.
Note carefully that the ring $$\mathscr O_{X,x}$$ is absolutely arbitrary and should not be assumed local.

In this set-up the set of points $$x\in X$$ with $$f_x\in \mathscr O_{X,y}^\star$$ is open:
Indeed if $$f_x\in \mathscr O_{X,x}^\star$$ there exists a germ $$g_x$$ with $$f_xg_x=1\in \mathscr O_{X,x}$$ and thus on a suitable open neighbourhood $$U$$ of $$x$$ there exists a representative $$g\in \mathscr O_X(U)$$ of $$g_x$$ satisfying $$f|U\cdot g=1\in \mathscr O_X(U)$$.
It is then clear that for all $$y\in U$$ we also have $$f_y\in \mathscr O_{X,y}^\star$$ since $$f_y\cdot g_y=1\in \mathscr O_{X,y}$$ : the promised openness has been proved .

As a digression, let me add that the reasoning above easily implies (by covering $$X$$ by suitable open $$U$$'s and using the unicity of inverses in a ring) that a global section $$f\in \mathscr O_X(X)$$ is invertible if and only if all its germs are invertible: $$f\in \mathscr O_X(X)^\star \iff (\forall x\in X) \; f_x\in \mathscr O_{X,x}^\star \quad (\bigstar)$$

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b) Suppose now that $$(X,\mathscr O_X)$$ is a locally ringed space.
Each point $$x\in X$$ then acquires a residue field $$\kappa(x)= \mathscr O_{X,x}/\mathfrak m_x$$ and a function $$f\in \mathscr O_X(X)$$ now receives a value $$f(x)=\text {class} (f_x) \in \mathscr O_{X,x}/\mathfrak m_x=\kappa(x)$$.
This value is close to the classical interpretation of the value of a function but it is crucial to understand the enormous difference between $$f_x$$ and $$f(x)$$.
The charm of local rings is that invertibility of $$f_x$$ is now exactly equivalent to the rather concrete requirement of having non-zero value: $$f_x\in \mathscr O_{X,x}^\star \iff f(x)\neq 0\in \kappa(x) \quad \quad (\bigstar \bigstar)$$ In particular, taking $$(\bigstar)$$ into account, we get: $$f\in \mathscr O_X(X)^\star \iff (\forall x\in X) \; f(x)\neq 0\in \kappa(x)$$

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c) So the set of points where $$f(x)\neq 0$$ is open thanks to the above equivalence $$(\bigstar \bigstar)$$ and the openness result in a).
As an aside, I find it a little intriguing that the trivial-looking assertion of openness of $$f(x)\neq 0$$ requires such a roundabout proof.

• Thank you Georges! It's always great to receive a detailed explanation :-) Jan 21, 2014 at 21:33
• Dear Rodrigo, you are welcome. Actually this is a subject that took me some time to really understand and organize. Mar 16, 2014 at 7:39
• @GeorgesElencwajg How do we show that a global section $f\in \mathscr O_X(X)$ is invertible if and only if all of its germs are invertible? Jan 13, 2021 at 3:08

When he says function, he refers to any $f \in \mathcal{O}_X(U)$

1) No, $\mathcal O_X$ is just a sheaf and $f$ need not be a function at all.

2) To vanish at $p$ means that the germ $[f]_x$ is in the maximal ideal of the local ring $\mathcal O_{X,p}$

3) To be invertible at $p$ means that the stalk $[f]_x$ is in $\mathcal O_{X,p}^*$, a unit.

Now:

If $f$ is invertible at $p$ means that $[f]_p [g]_p=_p$. By the definition this holds on a open nhood $f|_V g|_V = 1|_V$ so that the invertible locus is open.

Now if $\mathcal O_{X,p}$is a local ring and $[f]_x$ does not vanish, so $[f]_x \notin \mathfrak{m}_x$ is invertible

• Your (2) is incorrect: $f$ vanishes at $p$ if it is zero in the residue field of $\mathscr{O}_{X, p}$. Jan 21, 2014 at 14:09
• Thank you Blah, but tapping in to ZhenLin's remark it should also be "Now if your stalk is a local ring $[f]_x \neq 0 \Rightarrow [f]_x \notin \mathfrak m \Rightarrow [f]_x$ is a unit. Also, at 2) and 3) I gather you meant to say germ. Jan 21, 2014 at 14:13
• @Zhen, Rodrigo: sorry, made the necessary edits.
– Blah
Jan 21, 2014 at 17:42
• When he says function on $X$, he refers to any $f \in \mathcal O_X(X)$. Jan 13, 2021 at 3:07