Let $i:X \to \mathbb{P}^n$ be a smooth projective variety defined by the vanishing locus of polynomials $(\underline{f}) = (f_1,\ldots,f_k)$ which have degrees $>0$ and are pairwise coprime, meaning $gcd(f_i,f_j) = 1$ for $i \neq j$. Recall that we can compute the sheaf cohomology of $\mathcal{O}_X$ using the pushforward $$ H^i(X,\mathcal{O}_X) = H^i(\mathbb{P}^n,i_*(\mathcal{O}_X)) $$ If $X$ is a complete intersection, then this can be computed using the koszul complex $$ H^i(X,\mathcal{O}_X) = H^i(\mathbb{P}^n,K_\bullet(\underline{f})) $$ Does this result extend even to the general case that $X$ is smooth? That is, does $$ H^i(X,\mathcal{O}_X) = \mathbb{H}^i(\mathbb{P}^n, K_\bullet(\underline{f})) $$
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$\begingroup$ I suggest that you compute what happens if you take $X$ equal to all of $\mathbb{P}^n$ and then you take $\underline{f}$ to be $(0,\dots,0)$, where the $\text{i}^{\text{th}}$ component is the zero homomorphism $\mathcal{O}_{\mathbb{P}^n}(d_i)\to \mathcal{O}_{\mathbb{P}^n}$, for whatever degree $d_i$ that you like. $\endgroup$ – Jason Starr Jun 7 '17 at 20:35

$\begingroup$ Sure. I see how that is a problem, but if I restrict to the case of having polynomials of degrees $> 0$, my question still seems interesting. $\endgroup$ – 54321user Jun 7 '17 at 20:59

$\begingroup$ The Koszul complex of $(f,f)$ is equivalent to the Koszul complex of $(f,0)$, and that is going to be a tensor product with a complex like the one in my previous comment. This issue is persistent. $\endgroup$ – Jason Starr Jun 7 '17 at 21:04

$\begingroup$ and nonequal (or maybe even projectively equivalent) polynomials $\endgroup$ – 54321user Jun 7 '17 at 21:07

$\begingroup$ Okay, form the Koszul complex of $(f,f^2)$. Those polynomials are not projectively equivalent, but you have the same problem. $\endgroup$ – Jason Starr Jun 7 '17 at 21:10