**Proof.**
If $I = A$, then $X$ is affine and the statements are trivial. Hence we may and do assume $I \not= A$. Thus $Y$ and $X$ are nonempty schemes.

Let us prove (1) and (2) by induction on $n$. The base case $n = 1$ is our definition of $q_0$ as $Y_1 = Y$. Recall that $\mathcal{O}_ X(1) = \mathcal{O}_ X(-Y)$. Hence we have a short exact sequence

\[ 0 \to \mathcal{O}_{Y_ n}(1) \to \mathcal{O}_{Y_{n + 1}} \to \mathcal{O}_ Y \to 0 \]

Hence for $i > 0$ we find

\[ H^ i(X, \mathcal{O}_{Y_ n}(q + 1)) \to H^ i(X, \mathcal{O}_{Y_{n + 1}}(q)) \to H^ i(X, \mathcal{O}_{Y}(q)) \]

and we obtain the desired vanishing of the middle term from the given vanishing of the outer terms. For $i = 0$ we obtain a commutative diagram

\[ \xymatrix{ 0 \ar[r] & I^{q + 1}/I^{q + 1 + n} \ar[d] \ar[r] & I^ q/I^{q + 1 + n} \ar[d] \ar[r] & I^ q/I^{q + 1} \ar[d] \ar[r] & 0 \\ 0 \ar[r] & H^0(X, \mathcal{O}_{Y_ n}(q + 1)) \ar[r] & H^0(X, \mathcal{O}_{Y_{n + 1}}(q)) \ar[r] & H^0(Y, \mathcal{O}_ Y(q)) \ar[r] & 0 } \]

with exact rows for $q \geq q_0$ (for the bottom row observe that the next term in the long exact cohomology sequence vanishes for $q \geq q_0$). Since $q \geq q_0$ the left and right vertical arrows are isomorphisms and we conclude the middle one is too.

We omit the proofs of (3) and (4) which are similar. In fact, one can deduce (3) and (4) from (1) and (2) using the theorem on formal functors (but this would be overkill).
$\square$

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