Hartshorne assumes we are working over an algebraically closed field $k$ and we simply define algebraic sets to be any set of the form $V(I)\subset k^n$, where $I\subset k[X_1,\ldots,X_n]$ is an ideal.
Silverman's Arithmetic of Elliptic Curves extends the definition to non-algebraically closed fields, by looking at a field $k$ and its algebraic closure $\overline{k}$ and distinguishing between sets $V(I)\subset \overline{k}^n$, where $I$ can be generated by polynomials in $k[X_1,\ldots,X_n]$ vs. only by a family of polynomials in $\overline{k}[X_1,\ldots,X_n]$. In the former case he says that $V(I)$ is defined over $K$. He needs this to do arithmetic and work over $k=\mathbb{Q}$.
Now another interesting case I found is given by the notes of Steven Shatz and Jean Gallier (http://www.cis.upenn.edu/~jean/algeom/steve01.html). Here the definition is the following:
Let $k$ be any field and $\Omega/k$ an extension of $k$ s.t.
$\Omega$ is algebraically closed.
$\textrm{tr.deg}_k \Omega =\aleph_0$.
They define $\mathbb{A}^n=\Omega^k$. And say that the $k$-topology on $\mathbb{A}^n$ is the topology where closed sets are of the form $V(I)$ with $I\subset k[X_1,\ldots,X_n]$ an ideal and the Zariski topology on $\mathbb{A}^n$ is the topology with $I\subset \overline{k}[X_1,\ldots,X_n]$.
My question is the following: What is the gain of the more complicated definition in the Shatz, Gallier book? Why is it not enough to just take the algebraic closure?