Lemma 54.16.3. Let $X$ be a Noetherian scheme. Let $E \subset X$ be an exceptional curve of the first kind. Let $E_ n = nE$ and denote $\mathcal{O}_ n$ its structure sheaf. Then
is a complete local Noetherian regular local ring of dimension $2$ and $\mathop{\mathrm{Ker}}(A \to H^0(E_ n, \mathcal{O}_ n))$ is the $n$th power of its maximal ideal.
Proof. Recall that there exists an isomorphism $\mathbf{P}^1_ k \to E$ such that the normal sheaf of $E$ in $X$ pulls back to $\mathcal{O}(-1)$. Then $H^0(E, \mathcal{O}_ E) = k$. We will denote $\mathcal{O}_ n(iE)$ the restriction of the invertible sheaf $\mathcal{O}_ X(iE)$ to $E_ n$ for all $n \geq 1$ and $i \in \mathbf{Z}$. Recall that $\mathcal{O}_ X(-nE)$ is the ideal sheaf of $E_ n$. Hence for $d \geq 0$ we obtain a short exact sequence
Since $\mathcal{O}_ E(-(d + n)E) = \mathcal{O}_{\mathbf{P}^1_ k}(d + n)$ the first cohomology group vanishes for all $d \geq 0$ and $n \geq 1$. We conclude that the transition maps of the system $H^0(E_ n, \mathcal{O}_ n(-dE))$ are surjective. For $d = 0$ we get an inverse system of surjections of rings such that the kernel of each transition map is a nilpotent ideal. Hence $A = \mathop{\mathrm{lim}}\nolimits H^0(E_ n, \mathcal{O}_ n)$ is a local ring with residue field $k$ and maximal ideal
Pick $x, y$ in this kernel mapping to a $k$-basis of $H^0(E, \mathcal{O}_ E(-E)) = H^0(\mathbf{P}^1_ k, \mathcal{O}(1))$. Then $x^ d, x^{d - 1}y, \ldots , y^ d$ are elements of $\mathop{\mathrm{lim}}\nolimits H^0(E_ n, \mathcal{O}_ n(-dE))$ which map to a basis of $H^0(E, \mathcal{O}_ E(-dE)) = H^0(\mathbf{P}^1_ k, \mathcal{O}(d))$. In this way we see that $A$ is separated and complete with respect to the linear topology defined by the kernels
We have $x, y \in I_1$, $I_ d I_{d'} \subset I_{d + d'}$ and $I_ d/I_{d + 1}$ is a free $k$-module on $x^ d, x^{d - 1}y, \ldots , y^ d$. We will show that $I_ d = (x, y)^ d$. Namely, if $z_ e \in I_ e$ with $e \geq d$, then we can write
where $a_{e, j} \in (x, y)^{e - d}$ and $z_{e + 1} \in I_{e + 1}$ by our description of $I_ d/I_{d + 1}$. Thus starting with some $z = z_ d \in I_ d$ we can do this inductively
with some $a_{e, j} \in (x, y)^{e - d}$. Then $a_ j = \sum _{e \geq d} a_{e, j}$ exists (by completeness and the fact that $a_{e, j} \in I_{e - d}$) and we have $z = \sum a_{e, j} x^{d - j} y^ j$. Hence $I_ d = (x, y)^ d$. Thus $A$ is $(x, y)$-adically complete. Then $A$ is Noetherian by Algebra, Lemma 10.97.5. It is clear that the dimension is $2$ by the description of $(x, y)^ d/(x, y)^{d + 1}$ and Algebra, Proposition 10.60.9. Since the maximal ideal is generated by two elements it is regular. $\square$
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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)