I have problems proving the following assertion:
Let $X=\mathbb{P}^n$ be the projective $n$-scheme over some field $k$, and $Y \hookrightarrow X$ a complete intersection, i.e., pure $q$-dimensional closed subscheme of $X$ isomorphic to $Proj(k[x_0,...,x_n]/(f_1,...,f_{n-q}))$. We also assume that $q$ is at least 1. Now $H^i(X,\mathcal{O}_Y(m))=0$ for all $0<i<q$ and for all $m \in \mathbb{Z}$.
If we assume that the $f_i$ form a regular sequence, then this can be easily proved by induction: denote by $Y'$ the scheme $Proj(k[x_0,...,x_n]/(f_1,...,f_{n-q-1}))$. By regularity of the sequence, we get the short exact sequence \begin{equation*} 0 \to \mathcal{O}_{Y'}(m-d_{n-q}) \to \mathcal{O}_{Y'}(m) \to \mathcal{O}_Y(m) \to 0, \end{equation*} where $d_i$ denotes the degree of $f_i$. (We can treat the structure sheaves as if they were sheaves on $X$, as it does not change cohomology groups.) Now the claim follows from the long exact sequence associated to sheaf cohomology.
I haven't been able to make the above argument to work in the more general case where the polynomials $f_i$ don't form a regular sequence.
One idea, that initially was promising but ultimately lead to nowhere, was to consider the kernel of $\mathcal{O}_{Y'}(m) \to \mathcal{O}_Y(m)$. If we denote by $R_\bullet$ the graded ring $k[x_0,...,x_n]/(f_1,...,f_{n-q-1})$, then the kernel is just the sheaf associated to $K_\bullet = f_{n-q}R_\bullet$ twisted by $m$. This $K_\bullet$ is just $R_\bullet / J$ shifted by the degree of $f_{n-q}$, for some ideal $J$. As $f_{n-q}$ cannot be in any of the minimal primes of $R_\bullet$, any element annihilated by $f_{n-q}$ must be in all of them, i.e., $J$ must consist entirely of nilpotents. However, I couldn't deduce anything useful from that. I had some other ideas as well, but they also failed.
Am I missing something simple here? I haven't been able to find any help anywhere thus far, and I have pretty much ran out of ideas. I would be very grateful if someone could show, or at least give useful ideas, how to prove the case where the $f_i$ don't form a regular sequence.
Note: although this is really just the exercise III.5.5c I have not done III.5.5a, as c) seemed to be a more natural way to approach the problem.
You do not have to worry about the ideal being globally generated by a regular sequence. It suffices, that it is locally generated by a regular sequence by the definition of a complete intersection.
Let $Y=H_1 \cap \dotsc \cap H_r$ (with $r=n-q$) and $Z=H_1 \cap \dotsc \cap H_{r-1}$. By definition, the $H_i$ are each cut out by locally principal ideals. Let the degree of $H_r$ be $d$.
All you need for a proof by induction is an exact sequence
$$0 \to \mathcal O_Z(-d) \to \mathcal O_Z \to \mathcal O_Y \to 0$$
, i.e. we need $\mathcal I_{Y/Z} = \mathcal O_Z(-d)$.
In general, if we have a cartesian square of closed immersions
$\require{AMScd} \begin{CD} Y @>>> H_r\\ @VVV @VVV \\ Z @>i>> \mathbb P^n \end{CD}$
, we have $\mathcal I_{Y/Z} = i^* I_{H_r/\mathbb P^n}$, thus in our case:
$$\mathcal I_{Y/Z} = i^* \mathcal O_{\mathbb P^n}(-d) = \mathcal O_Z(-d).$$