Elementary Examples of Nonaffine Schemes

If only all schemes were affine!

Today we look at the simplest examples of non-affine schemes. How do we show that they are not affine?

 These schemes $X$ will be obtained by gluing affine pieces. Thus we can compute their global sections $O_X(X)$: these correspond to compatible sections on each affine piece. Using the fact that the functor Spec gives an equivalence of categories, we argue that if $X$ is affine, it must equals $\text{Spec}(O_X(X)).$ By comparing number of points, we argue that this is impossible.

Example 1. The Punctured Plane 

Let $U = \mathbb{A}^2 - \{(0,0)\}$,  be equipped with the structure of open subscheme  of $\mathbb{A}^2=: X$.

We compute its global section. To this end, we think of $U $ as $D(x) \cup D(y)$. So global sections of $U$ correspond to $(s, t) \in O_X(D(x)) \times O_X(D(y))$ that agree on intersection $O_X(D(xy))$.

Notice that the corresponding restriction maps are just inclusion $k[x,y]_x \hookrightarrow k[x,y]_{x,y}$ and $k[x,y]_y \hookrightarrow k[x,y]_{(x, y)}.$ So $s$ and $t$ are compatible iff $s = t$. Thus $O_U(U) = O_X(D(x)) \cap O_X(D(y)) = k[x,y]_x \cap k[x,y]_y = k[x, y].$

If $U$ is affine, it must therefore be $X = \mathbb{A}^2.$

However, the function $x$ and $y$ have common zeroes on $X$, but they not have any common zeroes in $U$.

Another way to phrase it is that we should have a bijection:
$ \{ \text{Prime ideals of $O_U(U)$} \} \leftrightarrow U$ given by
$\mathfrak{p} \mapsto \text{generic point of } \mathbb{V}(\mathfrak{p})$
$\mathbb{I}(x) \leftarrow x.$
So in particular the prime ideal $(x, y)$ should cut out a point in $U$. But it does not.

Example 2. Affine line with doubled origin

We glue $X = \text{Spec} k[t]$ and $Y = \text{Spec}  k[u]$ together along $X \supset U = \text{Spec} k[t, t^{-1}]$ and $Y \supset V = \text{Spec} k[u, u^{-1}]$ via the isomorphism

$ k[t, t^{-1}] \to k[u, u^{-1}]$ given by $t \mapsto u$.

To show that this scheme is not affine, compute its global section in the same way as above.

Example 3. Projective Line 

We glue $X = \text{Spec} k[t]$ and $Y = \text{Spec}  k[u]$ together along $X \supset U = \text{Spec} k[t, t^{-1}]$ and $Y \supset V = \text{Spec} k[u, u^{-1}]$ via the isomorphism

$ k[t, t^{-1}] \to k[u, u^{-1}]$ given by $t \mapsto 1/u$.

We compute the global sections of $\mathbb{P}^1$. These correspond to pairs of sections $(s_1, s_2 ) \in O_X(X) \times O_Y(Y)$ that agree on intersection $U \cong V$.

Notice that corresponding restriction maps  are injections, so $s_1|_U = s_1$ and $s_2|_V = s_2$. We use the map $O_X(U) \to O_Y(V)$ to identify $s_1 = f(t)$ as the section $f(1/u)$ on $V$.  So must have $f(1/u) = s_2$ as elements of $k[u, u^{-1}]$. As $s_2 \in k[u]$, this means $s_2 \in k$, and so is $s_1$, and $s_1 = s_2$.

So $\Gamma(\mathbb{P}^1, O_{\mathbb{P}^1}) = k$. If $\mathbb{P}^1$ were affine, it must equal $\text{Spec} k$ which consists of a single point.

Example 4. $\mathbb{P}^n$ for $n>0$.