Suppose $X$ is a projective smooth variety over a non-algebraically closed field , do we still have $Ext^i(F,\omega)\to H^{n-i}(X,F)^{\vee}$? (Hartshorne's proof Thm III 7.6 requires $k$ to be algebraically closed)



Serre duality over a non-algebraically closed field
Ok, here's a proof via specializing Grothendieck duality. It is probably useful for people to see this worked out. Say $f: X \to \text{Spec }k$ is the structural map and its proper and $X$ is just a scheme of finite type over $k$, say $F$ is a coherent sheaf on $X$ (for simplicity). Then Grothendieck duality says that 
$$R f\_\* R \mathcal{H}\text{om}\_{O\_X}(F, f^! k) \simeq R\mathcal{H}\text{om}\_{k}(R f\_\* F, k).$$
In particular, this is an isomorphism in the derived category.


### Analyzing the Left side


If $X$ is Cohen-Macaulay, then $f^! k$ is a canonical module on $X$ with a shift (locally by the dimension, note that Cohen-Macaulay means it is locally equidimensional). So lets work with one connected component at a time, where it is of some dimension $d$. Then $f^! k \simeq \omega\_X[d]$ (take it as a definition if you aren't comfortable with it). We also notice that $\mathcal{H}\text{om}(F, \bullet)$ takes injectives to flasques and so we get the composition of derived functors: 
$$
Rf\_\* R\mathcal{H}\text{om}(F, \bullet) = R\text{Hom}(F, \bullet).
$$
(ie, global sections of sheafy hom are non-sheafy-hom)
We take the $(i-d)$th cohomology of the left side and get
$$
\begin{array}{rl}
& \mathcal{H}^{i-d}(R f\_\* R \mathcal{H}\text{om}\_{O\_X}(F, f^! k) ) \\
= & \mathcal{H}^{i-d} R \text{Hom}\_{\mathcal{O}\_X}(F, \omega\_X[d]) \\
= & \mathcal{H}^{i} R \text{Hom}\_{\mathcal{O}\_X}(F, \omega\_X) \\
= & \text{Ext}^i(F, \omega\_X).
\end{array}
$$


### Analyzing the right side


This is even easier, $R\mathcal{H}\text{om}\_{k}(\bullet, k)$ is just the derived functor of $k$-vector space duality, which we identify with itself (its already exact). So I'll just write it as $(\bullet)^{\vee}$. Again we take the $(i-d)$th cohomology and get 
$$
\mathcal{H}^{i-d} R\mathcal{H}\text{om}\_{k}(R f\_\* F, k) = \mathcal{H}^{i-d}(( R f\_\* F)^{\vee}) = (\mathcal{H}^{d-i} Rf\_\* F)^{\vee} = (H^{d-i}(X, F))^{\vee}. 
$$


In conclusion
-------------


Plugging these two things in we recover the desired Serre duality. Notice we never used the fact that $k$ was algebraically closed. We only used the fact that it was a field when noticing that $\mathcal{H}\text{om}\_k(\bullet, k)$ was exact already. You can still say similar things (that look a bit more complicated) when $k$ is a Gorenstein ring instead of a field.


