Example of quasi-compact, non-quasi seperated scheme where qcqs fails?

Question

The qcqs lemma (in Ravi Vakil's notes) says that if $X$ is a quasi-compact (qc) and quasi-separated (qs) scheme, for any global section $f$, the natural map from $\Gamma(X, O_X)_f \to \Gamma(X_f, O_X)$ is an isomorphism. (We are defining $X_f$ to be the complement of the set where $f(p) = 0$, i.e. where $f = 0$ in $O_{X,x} / m_x$. So the natural map is the one guaranteed by the universal property of localization as $f$ becomes a unit over $X_f$.)

The proof of this lemma uses in a crucial way these finiteness assumptions in order to take advantage of the willingness of localization to commute with finite products. (One builds the exact sequence describing the global sections via the sheaf axioms, plugs a finite affine cover with a finite affine cover of the overlaps into it. Then localizing this sequence expresses $\Gamma(X, O_X)_s$ as the kernel of the exact sequence describing $\Gamma(X_s, O_X)$, bootstrapping off of the case of this theorem in the affine case.)

I want counter examples in the situation when this hypothesis has been dropped.

Here is what I have been thinking so far:

I tried the example of the only qc but not qs scheme that I know: the infinite affine space with a doubled origin. The functions on this space and on any open subscheme are the same as functions on the corresponding open $\mathbb{A}^{\infty}$, since the gluing identifies the variables on a dense open set. So I don't think this will provide an example (but I am not sure - now after thinking about this I am more convinced that this example would provide a counter example, but I don't see how). Here one would (obviously) need to use that infinitely many affines are required to cover the intersection of the two $\mathbb{A}^n$ in order to produce something...

I think I found an example of a quasi-seperated non-quasi compact scheme that provides a counter example. Essentially the point is that localization doesn't commute with infinite products in general, so one can take the infinite disjoint union of $Spec Z / (2^i)$, for all $i \geq 0$, and try to invert the element $2$. $X_2 = \emptyset$, but $(\Pi Z / 2^i)[2^{-1}] \not = 0$, since for instance $(1,1,1,1, \ldots)$ is not 2-torsion.

Answer 1

Let me give a two-part answer: I will give a counterexample when $X$ is not quasi-compact, and I will prove that for $X$ integral and quasi-compact the result holds (without quasi-separatedness).

To ease notation, since the only sheaf we use is $\mathcal O_X$, I will just write $\Gamma(X)_f \to \Gamma(X_f)$. (You'll thank me later.)

Example. Let $X = \coprod \mathbb A^1_k = \coprod \operatorname{Spec} k[t]$, where the disjoint union is indexed over $\mathbb N$. Let $f = (t, \ldots) \in \Gamma(X)$. Then $$\Gamma(X)_f = \left(\prod k[t]\right)_{(t,\ldots)} \subsetneq \prod k[t]_t = \Gamma(X_f)$$ is the subset of sequences $(P_i)$ with bounded denominator, i.e. there exists $n \in \mathbb N$ such that $t^n P_i \in k[t]$ for all $i$.

Lemma. For $X$ integral and $f \in \Gamma(X)$, the map $\Gamma(X)_f \to \Gamma(X_f)$ is injective.

Proof. Let $\{U_i \subseteq X\}$ be an open cover of $X$ by affines. We know that $X_f \cap U_i = (U_i)_f$, and the map $$\Gamma(U_i)_f \to \Gamma((U_i)_f)$$ is an isomorphism. The sheaf condition and exactness of localisation gives a commutative diagram $$\begin{array}{ccccc} 0 & \rightarrow &\Gamma(X)_f & \rightarrow & \left(\prod \Gamma(U_i)\right)_f \\ & & \downarrow & & \downarrow \\ 0 & \rightarrow & \Gamma(X_f) & \rightarrow & \prod \Gamma(U_i)_f \end{array}$$ with exact rows. I claim that the second vertical map is injective. For $f = 0$ both sides are $0$, so assume $f \neq 0$. Since $X$ is connected, $f$ does not vanish identically on any nonempty open. Hence, $f|_{U_i} \neq 0$. This implies that the second vertical map is injective: if $(x_i)_i/f^j$ maps to $0$, then each $x_i/f^j$ is zero, so $x_i = 0$ (here we use that $X$ is integral). Hence, the first vertical map is injective as well. $\square$

Remark. The lemma also holds for $X$ quasi-compact without the integrality assumption: in this case we can choose a finite such covering, and the second vertical map is actually an isomorphism.

Remark. The lemma fails for general $X$, cf. OPs example $\coprod \operatorname{Spec} \mathbb Z/2^i$ (with $f = 2$).

Corollary. If $X$ is integral and quasi-compact, then the map $\Gamma(X)_f \to \Gamma(X_f)$ is an isomorphism.

Proof. Cover $X$ by finitely many open affines $U_i$. The sheaf condition gives a commutative diagram with exact rows $$\begin{array}{ccccccc} 0 & \rightarrow &\Gamma(X)_f & \rightarrow & \left(\prod \Gamma(U_i)\right)_f & \rightarrow & \left(\prod \Gamma(U_i \cap U_j)\right)_f \\ & & \downarrow & & \downarrow & & \downarrow\\ 0 & \rightarrow & \Gamma(X_f) & \rightarrow & \prod \Gamma((U_i)_f) & \rightarrow & \prod \Gamma((U_i \cap U_j)_f). \end{array}$$ By the affine case and the finiteness of the cover, the second vertical map is an isomorphism. The third vertical map is injective by the lemma (and finiteness of the cover). Thus, the result follows from the five lemma. $\square$

This should give a bit of a hint of in what direction to look for a counterexample: we need affine opens $U, V \subseteq X$ whose intersection is 'something like' $\coprod \operatorname{Spec} \mathbb Z/2^i$. As long as we don't have a conclusive answer, it would also be interesting to come up with more examples where the map is not injective.

There will be zero-divisors.