Step 1a: Explicit construction for $\mathbb{P}^n$.

Let $X = \mathbb{P}^n_k$ with homogeneous coordinates $[x_0 : \cdots : x_n]$. The canonical sheaf is

$\omega_X = \Omega^n_X = \mathcal{O}_X(-n-1)$

We have $H^n(\mathbb{P}^n, \mathcal{O}(-n-1)) \cong k$ by direct computation using Čech cohomology.

Step 1b: Čech computation.

Cover $\mathbb{P}^n$ by standard open sets $U_i = \{x_i \neq 0\}$. A generator of $H^n(\mathbb{P}^n, \mathcal{O}(-n-1))$ can be represented by the Čech cocycle on $U_0 \cap \cdots \cap U_n$ given by

$\omega_0 = \frac{dx_1 \wedge \cdots \wedge dx_n}{x_0^{n+1}}$

in the local coordinate system on $U_0$, or more invariantly as the residue of a differential form.

Step 1c: Definition of the trace.

The trace map $t: H^n(\mathbb{P}^n, \omega_{\mathbb{P}^n}) \to k$ sends the generator $\omega_0$ to $1 \in k$. This is well-defined and independent of the choice of coordinates.

Step 1d: Verification of non-degeneracy.

This map is an isomorphism, so $H^n(\mathbb{P}^n, \omega_{\mathbb{P}^n}) \cong k$ via the trace map.

Step 2a: Construction of the pairing.

Let $\mathcal{F}$ be a coherent sheaf on $X$. We define a pairing

$\langle \cdot, \cdot \rangle: H^i(X, \mathcal{F}) \times H^{n-i}(X, \mathcal{F}^\vee \otimes \omega_X) \to k$

as follows. Given $\alpha \in H^i(X, \mathcal{F})$ and $\beta \in H^{n-i}(X, \mathcal{F}^\vee \otimes \omega_X)$, we form their cup product

$\alpha \cup \beta \in H^n(X, \mathcal{F} \otimes \mathcal{F}^\vee \otimes \omega_X)$

Step 2b: Natural evaluation map.

There is a natural evaluation map

$\mathcal{F} \otimes \mathcal{F}^\vee \to \mathcal{O}_X$

given by $(s, \phi) \mapsto \phi(s)$. This induces

$H^n(X, \mathcal{F} \otimes \mathcal{F}^\vee \otimes \omega_X) \to H^n(X, \omega_X)$

Step 2c: Composition with trace.

Composing with the trace map gives

$\langle \alpha, \beta \rangle = t(\text{ev}(\alpha \cup \beta))$

This defines the desired pairing.

Step 2d: Bilinearity.

The pairing is clearly $k$-bilinear by the properties of cup product, evaluation, and the trace map.

Step 3a: Embedding into projective space.

Since $X$ is projective, there exists a closed embedding $i: X \hookrightarrow \mathbb{P}^N$ for some $N$. We can work locally and reduce dimension by dimension, so it suffices to prove the theorem for $\mathbb{P}^n$.

Step 3b: Using the projection formula.

For a general $X$, we use the projection formula and base change to relate cohomology on $X$ to cohomology on $\mathbb{P}^N$. Specifically, if $i: X \hookrightarrow \mathbb{P}^N$, then

$i_* \mathcal{F}$

is a coherent sheaf on $\mathbb{P}^N$, and we have

$H^i(X, \mathcal{F}) = H^i(\mathbb{P}^N, i_* \mathcal{F})$

Step 3c: Dimension reduction.

The proof proceeds by induction on dimension. The case $\dim X = 0$ is trivial (points have no higher cohomology). Assuming the result for varieties of dimension $< n$, we prove it for dimension $n$.

Step 3d: Hyperplane section.

Choose a hyperplane $H \subset \mathbb{P}^N$ not containing $X$ such that $Y = X \cap H$ is smooth of dimension $n-1$. We obtain an exact sequence

$0 \to \mathcal{F}(-Y) \to \mathcal{F} \to \mathcal{F}|_Y \to 0$

This induces a long exact sequence in cohomology which, combined with the five lemma and inductive hypothesis, proves the result.

Step 4a: Line bundles on projective space.

First consider $\mathcal{F} = \mathcal{O}(m)$ for some integer $m$. We need to show

$H^i(\mathbb{P}^n, \mathcal{O}(m)) \cong H^{n-i}(\mathbb{P}^n, \mathcal{O}(-m-n-1))^\vee$

Step 4b: Cohomology computation.

By the standard computation of cohomology of line bundles on $\mathbb{P}^n$:

• $H^i(\mathbb{P}^n, \mathcal{O}(m)) = 0$ for $0 < i < n$ and all $m$
• $H^0(\mathbb{P}^n, \mathcal{O}(m)) = k[x_0, \ldots, x_n]_m$ (homogeneous polynomials of degree $m$) if $m \geq 0$, and $0$ if $m < 0$
• $H^n(\mathbb{P}^n, \mathcal{O}(m)) \neq 0$ only if $m \leq -n-1$

Step 4c: Explicit pairing for $i=0$.

For $i=0$, we need to show

$H^0(\mathbb{P}^n, \mathcal{O}(m)) \times H^n(\mathbb{P}^n, \mathcal{O}(-m-n-1)) \to k$

is a perfect pairing. Given $f \in H^0(\mathbb{P}^n, \mathcal{O}(m))$ (a degree $m$ polynomial) and $\omega \in H^n(\mathbb{P}^n, \mathcal{O}(-m-n-1))$, their product $f \cdot \omega$ lies in $H^n(\mathbb{P}^n, \mathcal{O}(-n-1))$, which we map to $k$ via the trace.

Step 4d: Verification of perfect pairing.

To verify this is a perfect pairing, we must show:

1. If $\langle f, \omega \rangle = 0$ for all $f$, then $\omega = 0$
2. If $\langle f, \omega \rangle = 0$ for all $\omega$, then $f = 0$

For (1): Suppose $\omega \neq 0$. Then multiplication by a suitable monomial gives a non-zero element of $H^n(\mathbb{P}^n, \mathcal{O}(-n-1))$, which pairs non-trivially with $1$ under the trace.

For (2): The statement is vacuous if $m < 0$ (since then $H^0 = 0$). If $m \geq 0$, we use the fact that $H^n(\mathbb{P}^n, \mathcal{O}(-m-n-1))$ has dimension $\binom{m+n}{n}$, equal to the dimension of $H^0(\mathbb{P}^n, \mathcal{O}(m))$.

Step 4e: Extension to coherent sheaves.

Any coherent sheaf $\mathcal{F}$ on $\mathbb{P}^n$ admits a finite resolution by direct sums of line bundles:

$0 \to \bigoplus_j \mathcal{O}(a_j) \to \cdots \to \bigoplus_i \mathcal{O}(b_i) \to \mathcal{F} \to 0$

Using this resolution and the fact that duality holds for line bundles, we deduce duality for $\mathcal{F}$ by a spectral sequence argument.

Step 6a: Ext interpretation.

By the adjunction between $\otimes$ and $\mathcal{H}om$, we have

$\mathcal{F}^\vee \otimes \omega_X = \mathcal{H}om(\mathcal{F}, \omega_X)$

Thus Serre Duality can be stated as

$H^i(X, \mathcal{F}) \cong H^{n-i}(X, \mathcal{H}om(\mathcal{F}, \omega_X))^\vee$

Step 6b: Yoneda product.

The pairing comes from the Yoneda product

$\text{Ext}^i(\mathcal{O}_X, \mathcal{F}) \times \text{Ext}^{n-i}(\mathcal{F}, \omega_X) \to \text{Ext}^n(\mathcal{O}_X, \omega_X)$

Since $\text{Ext}^i(\mathcal{O}_X, \mathcal{F}) = H^i(X, \mathcal{F})$ and $\text{Ext}^n(\mathcal{O}_X, \omega_X) = H^n(X, \omega_X) \cong k$, this gives the duality pairing.

Step 6c: Compatibility with derived categories.

In the language of derived categories, Serre Duality states that the functor

$R\mathcal{H}om(\cdot, \omega_X): D^b_{\text{coh}}(X) \to D^b_{\text{coh}}(X)$

is a dualizing functor, meaning it is an anti-equivalence with $R\mathcal{H}om(R\mathcal{H}om(\mathcal{F}, \omega_X), \omega_X) \cong \mathcal{F}$.

Step 9a: Cup product compatibility.

The cup product

$H^i(X, \mathcal{F}) \times H^j(X, \mathcal{G}) \to H^{i+j}(X, \mathcal{F} \otimes \mathcal{G})$

is compatible with Serre Duality in the following sense: if we dualize and use the duality isomorphisms, the diagram

$H^i(X, \mathcal{F}) \times H^j(X, \mathcal{G}) \to H^{i+j}(X, \mathcal{F} \otimes \mathcal{G})$

corresponds under duality to

$H^{n-i-j}(X, (\mathcal{F} \otimes \mathcal{G})^\vee \otimes \omega_X) \to H^{n-i}(X, \mathcal{F}^\vee \otimes \omega_X) \times H^{n-j}(X, \mathcal{G}^\vee \otimes \omega_X)$

Step 9b: Associativity.

This compatibility ensures that the pairing structure is associative in the appropriate sense, making cohomology with Serre Duality into a Frobenius algebra structure.