The Stacks project

Example 26.14.4 (Projective line). Let $k$ be a field. Let $X_1 = \mathop{\mathrm{Spec}}(k[x])$, let $X_2 = \mathop{\mathrm{Spec}}(k[y])$. Let $0 \in X_1$ be the point corresponding to the maximal ideal $(x) \subset k[x]$. Let $\infty \in X_2$ be the point corresponding to the maximal ideal $(y) \subset k[y]$. Let $U_{12} = X_1 \setminus \{ 0\} = D(x) = \mathop{\mathrm{Spec}}(k[x, 1/x])$ and let $U_{21} = X_2 \setminus \{ \infty \} = D(y) = \mathop{\mathrm{Spec}}(k[y, 1/y])$. Let $\varphi _{12} : U_{12} \to U_{21}$ be the isomorphism coming from the isomorphism of $k$-algebras $k[y, 1/y] \to k[x, 1/x]$ mapping $y$ to $1/x$. Let $\mathbf{P}^1_ k$ be the scheme obtained from the glueing data $(X_1, X_2, U_{12}, U_{21}, \varphi _{12}, \varphi _{21} = \varphi _{12}^{-1})$. Via the slight abuse of notation introduced above the example we think of $X_ i \subset \mathbf{P}^1_ k$ as open subschemes. In this case we see that $\Gamma (\mathbf{P}^1_ k, \mathcal{O}) = k$ because the only polynomials $g(x)$ in $x$ such that $g(1/y)$ is also a polynomial in $y$ are constant polynomials. Since $\mathbf{P}^1_ k$ is infinite we see that $\mathbf{P}^1_ k$ is not affine.

We claim that there exists an affine open $U \subset \mathbf{P}^1_ k$ which contains both $0$ and $\infty $. Namely, let $U = \mathbf{P}^1_ k \setminus \{ 1\} $, where $1$ is the point of $X_1$ corresponding to the maximal ideal $(x - 1)$ and also the point of $X_2$ corresponding to the maximal ideal $(y - 1)$. Then it is easy to see that $s = 1/(x - 1) = y/(1 - y) \in \Gamma (U, \mathcal{O}_ U)$. In fact you can show that $\Gamma (U, \mathcal{O}_ U)$ is equal to the polynomial ring $k[s]$ and that the corresponding morphism $U \to \mathop{\mathrm{Spec}}(k[s])$ is an isomorphism of schemes. Details omitted.

Comments (0)

There are also:

  • 7 comment(s) on Section 26.14: Glueing schemes

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Your email address will not be published.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 01JE. Beware of the difference between the letter 'O' and the digit '0'.