How to show holomorphic de Rham complex is exact? Supppse $X$ is a smooth algebraic manifold, how does one show the holomorphic de Rham complex is exact in characteristic $0$?
I knew a method is to resolve it by double complex $A^{p,q}$, then the column exactness shows it is quasi isomorphic to differential de Rham complex, thus exact. But this method is transcendental, is there a algebraic way of showing this?
 A: I can think of a lot of interpretations of this question. Let $X$ be an algebraic manifold. Here are things which I could imagine "the holomorphic de Rham complex is exact" meaning.
The complex $\mathcal{O} \to \Omega^1 \to \Omega^2 \to \cdots$ is exact, where $\Omega^p$ is the algebraic $p$-forms. This isn't true. Take $X = \{ z : z \neq 0 \}$. Then $dz/z$ is a closed $1$-form which is not exact.
The complex $\mathcal{O} \to \Omega^1 \to \Omega^2 \to \cdots$ is exact, where $\Omega^p$ is the holomorphic $p$-forms. This is also false; the same counterexample works.
The complex of sheaves $\mathcal{O} \to \Omega^1 \to \Omega^2 \to \cdots$ is exact in the Zariski topology, for either of the options above Again false: $dz/z$ does not become exact on any Zariski open.
The complex of sheaves $\mathcal{O} \to \Omega^1 \to \Omega^2 \to \cdots$ is exact in the analytic topology, where $\Omega^p$ is the holomorphic $p$-forms. This is a true theorem. It's pretty intrinsically an analytic result though; it uses analytic topology and holomorphic functions. You don't need to go to $(p,q)$ forms to do it though. Here is a proof that stays purely in the holomorphic world. On a formal level, this is exactly the same proof (or a same proof, anyway) as in the smooth real function case.
We will show that $\mathcal{O} \to \Omega^1 \to \Omega^2 \to \cdots$ is exact on any polydisc $\{ (z_1, \ldots, z_n) : |z_i| < r_i \}$, where $\Omega^p$ is the holomorphic $p$-forms. Any complex manifold obviously has a basis of open sets which are polydiscs, so this is enough.
We prove the following statement by induction on $k$: Let 
$$\omega = \sum_{1 \leq i_1 < \cdots < i_p \leq k} f_{i_1 i_2 \cdots i_p}(z_1, \ldots, z_n)\  d z_{i_1} \wedge \cdots \wedge d z_{i_p}$$
be a closed holomorphic $p$-form. Then $\omega$ is $d \eta$ for some holomorphic $p$-form $\eta$. Note that $k$ appears in the subscript of the summation, and the number of variables is $n$.
The base case $k=0$ (or $k < p$ more generally) is trivial; $\omega$ must be zero.
Now, we do the inductive case. The equation $d \omega=0$ implies that, for $j>k$, we have $\partial f_{i_1 \cdots i_p}(z_1, \ldots, z_n)/\partial z_j=0$. So the $f_{i_1 \cdots i_p}$ functions only depend on $(z_1, \ldots, z_k)$ and we will write them as $f_{i_1 \cdots i_p}(z_1, \ldots, z_k)$ from now on.
Set $\phi = \sum_{1 \leq i_1 < \cdots < i_{p-1} < k} dz_{i_1} \wedge \cdots \wedge d z_{i_{p-1}} \int_{t=0}^{z_k} f_{i_1 \cdots i_{p-1} k}(z_1, \ldots, z_{k-1}, t) dt.$ (The integral is along any path through the disc $|t|<r_k$.)
Then $\omega - d \phi$ has no $dz_k$ terms. So, by induction, $\omega - d \phi = d \psi$ for some $\psi$, and $\omega = d(\phi+\psi)$.
The de Rham complex of formal power series, or on polynomials, is exact This is true. You can copy the above proof, but my preferred argument is the following: Put a $\mathbb{Z}^n$ grading on polynomial differential forms, where $z_i$ and $dz_i$ are in degree $(0,0,\ldots,0,1,0,\ldots,0)$, with the $1$ in the $i$-th place. So $d$ preserves the degree. Then the complex breaks up into a direct sum (direct product for power series) of finite dimensional complexes, one each degree. 
The complex in degree $(k_1, k_2, \ldots, k_n)$ is the tensor product of $n$ complexes that look like $\mathbb{R} \overset{k_i}{\longrightarrow} \mathbb{R}$ if $k_i > 0$, or $\mathbb{R} \to 0$ if $k_i=0$. This complex is exact if any of the $k_i$ are $>0$. So the only cohomology comes from the degree $(0,0,\ldots,0)$ complex, which is all in $H^0$.
