Lemma 38.38.1. Let $Z, X, X', E$ be an almost blow up square as in Example 38.37.11. Then $H^ p(X', \mathcal{O}_{X'}) = 0$ for $p > 0$ and $\Gamma (X, \mathcal{O}_ X) \to \Gamma (X', \mathcal{O}_{X'})$ is a surjective map of rings whose kernel is an ideal of square zero.
Proof. First assume that $A = \mathbf{Z}[f_1, f_2]$ is the polynomial ring. In this case our almost blow up square is the blowing up of $X = \mathop{\mathrm{Spec}}(A)$ in the closed subscheme $Z$ and in fact $X' \subset \mathbf{P}^1_ X$ is an effective Cartier divisor cut out by the global section $f_2T_0 - f_1 T_1$ of $\mathcal{O}_{\mathbf{P}^1_ X}(1)$. Thus we have a resolution
\[ 0 \to \mathcal{O}_{\mathbf{P}^1_ X}(-1) \to \mathcal{O}_{\mathbf{P}^1_ X} \to \mathcal{O}_{X'} \to 0 \]
Using the description of the cohomology given in Cohomology of Schemes, Section 30.8 it follows that in this case $\Gamma (X, \mathcal{O}_ X) \to \Gamma (X', \mathcal{O}_{X'})$ is an isomorphism and $H^1(X', \mathcal{O}_{X'}) = 0$.
Next, we observe that any diagram as in Example 38.37.11 is the base change of the diagram in the previous paragraph by the ring map $\mathbf{Z}[f_1, f_2] \to A$. Hence by More on Morphisms, Lemmas 37.72.1, 37.72.2, and 37.72.4 we conclude that $H^1(X', \mathcal{O}_{X'})$ is zero in general and the surjectivity of the map $H^0(X, \mathcal{O}_ X) \to H^0(X', \mathcal{O}_{X'})$ in general.
Next, in the general case, let us study the kernel. If $a \in A$ maps to zero, then looking on affine charts we see that
\[ a = (f_1x - f_2)(a_0 + a_1x + \ldots + a_ rx^ r)\text{ in }A[x] \]
for some $r \geq 0$ and $a_0, \ldots , a_ r \in A$ and similarly
\[ a = (f_1 - f_2y)(b_0 + b_1y + \ldots + b_ s y^ s)\text{ in }A[y] \]
for some $s \geq 0$ and $b_0, \ldots , b_ s \in A$. This means we have
\[ a = f_2 a_0,\ f_1 a_0 = f_2 a_1,\ \ldots ,\ f_1 a_ r = 0, \ a = f_1 b_0,\ f_2 b_0 = f_1 b_1,\ \ldots ,\ f_2 b_ s = 0 \]
If $(a', r', a'_ i, s', b'_ j)$ is a second such system, then we have
\[ aa' = f_1f_2a_0b'_0 = f_1f_2a_1b'_1 = f_1f_2a_2b'_2 = \ldots = 0 \]
as desired. $\square$
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