Lemma 108.22.1. Let $k$ be a field. There exists a variety $X$ whose normalization is quasi-affine but which is itself not quasi-affine.

## 108.22 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 $Y$:

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

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

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

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

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

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.

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

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## Comments (2)

Comment #3580 by Neeraj Deshmukh on

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