Let $X$ be a scheme finite type over a field $k$ such that each of its irreducible components $X_i$ is quasi-affine. Is it true that $X$ is quasi-affine?

  • 3
    $\begingroup$ No, that is not true. Let $X_1$ and $X_2$ be isomorphic to the deleted affine plane $\mathbb{A}^2\setminus\{(0,0)\}$. For $i=1,2,$ let $L_i$ be a deleted affine line through the origin in $X_i$. Thus, both $L_1$ and $L_2$ are isomorphic to $\text{Spec}\ k[t,t^{-1}]$. Let $f:L_1\xrightarrow{\cong} L_2$ be an isomorphism that maps the origin to $\infty$. Glue $X_1$ and $X_2$ according to this isomorphism. Then every global section of the structure sheaf of $X$ restricts on $L_1=L_2$ as a constant. $\endgroup$ May 1, 2018 at 13:32

1 Answer 1


I am just writing my comment as an answer. Probably this example has appeared before on MO. Let $\mathbb{P}^3_k$ denote the projective space $\text{Proj}\ k[s_1,s_2,t_1,t_2]$. Let $\overline{X}$ denote the closed, reduced subscheme $$\overline{X} =\text{Zero}(s_1s_2) = \overline{X}_1 \cup \overline{X}_2, \ \ \overline{X}_1 = \text{Zero}(s_2), \ \ \overline{X}_2 = \text{Zero}(s_1).$$ This is a union of two hyperplanes, $\overline{X}_1$ and $\overline{X}_2,$ whose intersection is a line, $$\overline{L} = \overline{X}_1\cap \overline{X}_2 = \text{Zero}(s_1,s_2) = \text{Proj} \ k[t_1,t_2].$$ Inside $\overline{X}_1$, define $C_1$ to be the closed subscheme that is a line, $$C_1 = \text{Zero}(s_2,t_1).$$ Inside $\overline{X}_2$, define $C_2$ to be the closed subscheme that is a line, $$C_2=\text{Zero}(s_1,t_2).$$ Note that $C_1\cap \overline{X}_2$ is the singleton set of $p_2=[0,0,0,1]$, and $C_2\cap \overline{X}_1$ is the singleton set of $p_1=[0,0,1,0].$ Define $C$ to be the closed union of $C_1$ and $C_2$ in $\overline{X}$.

Define $X$ to be the open complement in $\overline{X}$ of the closed subset $C$. By construction, $X$ is quasi-projective. Moreover, $X$ is the union of the two irreducible components, $$X_1 = \overline{X}_1\setminus(C_1\cup \{p_1\}), \ \ X_2 = \overline{X}_2 \setminus (C_2\cup \{p_2\}).$$
The intersection of the two irreducible components is $$L=\overline{L}\setminus\{p_1,p_2\}.$$

Each of the two irreducible components is isomorphic to a punctured affine plane. In particular, $\mathcal{O}_{X_1}(X_1)$ equals $k[s_1/t_1,t_2/t_1]$. Similarly, $\mathcal{O}_{X_2}(X_2)$ equals $k[s_2/t_2,t_1/t_2]$. Finally, the intersection $L$ is a twice-punctured projective line with coordinate ring $\mathcal{O}_L(L)$ equal to $k[t_2/t_1,t_1/t_2]$. The restriction map $\rho_1$, resp. $\rho_2$, to $\mathcal{O}_L(L)$ from $\mathcal{O}_{X_1}(X_1)$, resp. from $\mathcal{O}_{X_2}(X_2)$, is the $k$-algebra map that sends $s_1/t_1$ to $0$, resp. that sends $s_2/t_2$ to $0$, and that sends $t_2/t_1$ to itself, resp. that sends $t_1/t_2$ to itself. The fibre product of these restriction maps is the $k$-algebra $$\mathcal{O}_X(X) = \{(f_1,f_2)\in \mathcal{O}_{X_1}(X_1)\times \mathcal{O}_{X_2}(X_2) : \rho_1(f_1) = \rho_2(f_2)\}.$$ This is infinitely generated as a $k$-algebra by the following monomials, $$\frac{s_1}{t_1}, \ \frac{s_2}{t_2}, \ \left(\frac{s_1^mt_2}{t_1^{m+1}}\right)_{m\geq 1}, \left(\frac{s_2^nt_1}{t_2^{n+1}}\right)_{n\geq 1}. $$ In particular, the image in $\mathcal{O}_L(L)$ of $\mathcal{O}_X(X)$ equals the constant subfield $k$. Thus, $X$ is not quasi-affine.

  • $\begingroup$ I discussed this example years ago with Ravi Vakil. I very much believe this example has appeared before on MathOverflow (maybe as an example of a quasi-projective $k$-scheme whose global sections $k$-algebra is not finitely generated). $\endgroup$ May 1, 2018 at 16:02
  • $\begingroup$ A very similar example (but slightly different) appears in the following MathOverflow answer: mathoverflow.net/questions/209443/… $\endgroup$ May 1, 2018 at 16:42
  • $\begingroup$ The example in the answer gives a quasi-projective scheme with quasi-affine components. Are there non-quasi-projective examples? Or is such a scheme necessarily quasi-projective? $\endgroup$ May 2, 2018 at 6:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.