The Stacks project

109.22 Non-quasi-affine variety with quasi-affine normalization

The existence of an example of this kind is mentioned in [II Remark 6.6.13, EGA]. They refer to the fifth volume of EGA for such an example, but the fifth volume did not appear.

Let $k$ be a field. Let $Y = \mathbf{A}^2_ k \setminus \{ (0, 0)\} $. We are going to construct a finite surjective birational morphism $\pi : Y \longrightarrow X$ with $X$ a variety over $k$ such that $X$ is not quasi-affine. Namely, consider the following curves in $Y$:

\[ \begin{matrix} C_1 & : & x = 0 \\ C_2 & : & y = 0 \end{matrix} \]

Note that $C_1 \cap C_2 = \emptyset $. We choose the isomorphism $\varphi : C_1 \to C_2$, $(0, y) \mapsto (y^{-1}, 0)$. We claim there is a unique morphism $\pi : Y \to X$ as above such that

\[ \xymatrix{ C_1 \ar@<1ex>[rr]^{\text{id}} \ar@<-1ex>[rr]_{\varphi } & & Y \ar[r]^\pi & X } \]

is a coequalizer diagram in the category of varieties (and even in the category of schemes). Accepting this for the moment let us show that such an $X$ cannot be quasi-affine. Namely, it is clear that we would get

\[ \Gamma (X, \mathcal{O}_ X) = \{ f \in k[x, y] \mid f(0, y) = f(y^{-1}, 0)\} = k \oplus (xy) \subset k[x, y]. \]

In particular these functions do not separate the points $(1, 0)$ and $(-1, 0)$ whose images in $X$ (we will see below) are distinct (if the characteristic of $k$ is not $2$).

To show that $X$ exists consider the Zariski open $D(x + y) \subset Y$ of $Y$. This is the spectrum of the ring $k[x, y, 1/(x + y)]$ and the curves $C_1$, $C_2$ are completely contained in $D(x + y)$. Moreover the morphism

\[ C_1 \amalg C_2 \longrightarrow D(x + y) \cap Y = \mathop{\mathrm{Spec}}(k[x, y, 1/(x + y)]) \]

is a closed immersion. It follows from More on Algebra, Lemma 15.5.1 that the ring

\[ A = \{ f \in k[x, y, 1/(x + y)] \mid f(0, y) = f(y^{-1}, 0)\} \]

is of finite type over $k$. On the other hand we have the open $D(xy) \subset Y$ of $Y$ which is disjoint from the curves $C_1$ and $C_2$. It is the spectrum of the ring

\[ B = k[x, y, 1/xy]. \]

Note that we have $A_{xy} \cong B_{x + y}$ (since $A$ clearly contains the elements $xyP(x, y)$ any polynomial $P$ and the element $xy/(x + y)$). The scheme $X$ is obtained by glueing the affine schemes $\mathop{\mathrm{Spec}}(A)$ and $\mathop{\mathrm{Spec}}(B)$ using the isomorphism $A_{xy} \cong B_{x + y}$ and hence is clearly of finite type over $k$. To see that it is separated one has to show that the ring map $A \otimes _ k B \to B_{x + y}$ is surjective. To see this use that $A \otimes _ k B$ contains the element $xy/(x + y) \otimes 1/xy$ which maps to $1/(x + y)$. The morphism $Y \to X$ is given by the natural maps $D(x + y) \to \mathop{\mathrm{Spec}}(A)$ and $D(xy) \to \mathop{\mathrm{Spec}}(B)$. Since these are both finite we deduce that $Y \to X$ is finite as desired. We omit the verification that $X$ is indeed the coequalizer of the displayed diagram above, however, see (insert future reference for pushouts in the category of schemes here). Note that the morphism $\pi : Y \to X$ does map the points $(1, 0)$ and $(-1, 0)$ to distinct points in $X$ because the function $(x + y^3)/(x + y)^2 \in A$ has value $1/1$, resp. $-1/(-1)^2 = -1$ which are always distinct (unless the characteristic is $2$ – please find your own points for characteristic $2$). We summarize this discussion in the form of a lemma.

Lemma 109.22.1. Let $k$ be a field. There exists a variety $X$ whose normalization is quasi-affine but which is itself not quasi-affine.

Proof. See discussion above and (insert future reference on normalization here). $\square$

Comments (2)

Comment #3580 by Neeraj Deshmukh on

In the paragraph preceding lemma 0272, it says "" in two places instead of "".

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0271. Beware of the difference between the letter 'O' and the digit '0'.