Serre duality over a non-algebraically closed field 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)
 A: 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.
