{wiki=Line_with_two_origins#Related_examples}

The affine plane with doubled origin is formed by gluing two copies of $\mathbb A_k^2$ by the identity away from the origin. The overlap is the <punctured affine plane>, so this scheme is nonseparated and the two-open affine cover is not acyclic for the structure sheaf.

Affine plane with doubled origin

For the cover $\mathcal U=\{X_1,X_2\}$ of the <affine plane with doubled origin>, the overlap is the <punctured affine plane> $U$. Since $H^0(U,\mathcal O_U)=k[x,y]$, the Čech complex begins
$$
k[x,y]\oplus k[x,y]\longrightarrow k[x,y],
\qquad (a,b)\longmapsto b-a.
$$
This map is surjective, and the normalized complex has no terms in degrees at least two. Consequently
$$
\check H^p(\mathcal U,\mathcal O_Y)=0\qquad(p>0).
$$

The <Mayer-Vietoris sequence for sheaf cohomology> also gives $H^1(Y,\mathcal O_Y)=0$, but its next part gives
$$
H^2(Y,\mathcal O_Y)\cong H^1(U,\mathcal O_U).
$$
Part b with $f=1$ shows that the group on the right is infinite-dimensional. Thus
$$
\check H^2(\mathcal U,\mathcal O_Y)=0
\ne H^2(Y,\mathcal O_Y).
$$

This does not contradict the <acyclic cover theorem>. Although $X_1$ and $X_2$ are affine, their intersection $U$ is not acyclic: it has nonzero first structure-sheaf cohomology. Equivalently, this affine cover does not satisfy the theorem's hypotheses; the doubled-origin plane is not a <semi-separated scheme>.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 113 / 3 / c / Solution

{title2=$\mathbb A_k^2\setminus\{0\}$}

The punctured affine plane is obtained by deleting the closed point $(x,y)$ from $\operatorname{Spec}k[x,y]$. Its regular functions still form $k[x,y]$, while
$$
H^1(\mathbb A_k^2\setminus\{0\},\mathcal O)\cong k[x^{\pm1},y^{\pm1}]/\bigl(k[x^{\pm1},y]+k[x,y^{\pm1}]\bigr),
$$
which has basis represented by $x^{-a}y^{-b}$ for $a,b\geq1$.


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 Incoming links: Punctured affine plane