Definition of algebraic variety In general, the definition of an algebraic variety differs from one reference to other. The definition that I was used to is to consider an algebraic variety as an integral scheme of finite type. However this definition differs from the one in the book of Masayoshi Miyanishi ("Algebraic Geometry", in page 89). 
The author defines an algebraic variety over a field $k$ as an integral $k$-scheme $X$ (this means that there exists a quasi-compact locally finite generated morphism $f: X \longrightarrow Spec(k)$) of finite type such that the base change by $\overline{k}$ ( $X_{\overline{k}} =X \times_{Spec(k)} Spec(\overline{k})$) is integral of finite type and the structure morphism is a $f: X \longrightarrow Spec(k)$ separated morphism ($\Delta_{X/Spec(k)}: X \longrightarrow X \times_{Spec(k)} X$ is a closed immersion). 
Could please someone explains why there is these additional assumptions in the above definition? The author just explains briefly (in page 92) why the separateness is useful in this case (he says that $f$ being separated is equivalent to two specializations of a point along a valuation ring $\mathscr{O}$ coincide under the assumption that the induced morphisms $k(x) \longrightarrow Q(\mathscr{O})$ are identical), however I didn't understand this too.
Thanks in advance.
 A: Let $X$ be integral and of finite type over $k$. Then we have two additional assumptions:


*

*$X$ separated. This is needed in some theorems, in the same vein that some theorems about topological spaces need the Hausdorff separation axiom. Projective varieties are always separated. Note that the gluing of two copies of $\mathbb{A}^1$ at $\mathbb{A}^1 \setminus \{0\}$, i.e. the affine line with double origin, is not separated, but it is of finite type and integral.

*$X$ is geometrically integral, i.e. $X \otimes_k \overline{k}$ is integral. Basically this is needed in order to reduce everything to the case of algebraically closed fields. For example, if $X,Y$ are geometrically integeral, then $X \times_k Y$ is geometrically integral. But this does not hold just for integral, i.e. we really need the base change to the algebraic closure here. For example $\mathrm{Spec}(\mathbb{C}) \times_{\mathbb{R}} \mathrm{Spec}(\mathbb{C}) \cong \mathrm{Spec}(\mathbb{C}) \sqcup \mathrm{Spec}(\mathbb{C})$ is reducible. Again, some theorems only work for geometrically integral varieties, that's why many authors include this into the definition. We have just seen that $\mathrm{Spec}(\mathbb{C})$ is not geometrically integral over $\mathbb{R}$. Notice that $\mathbb{A}^n_k$ and $\mathbb{P}^n_k$ are geometrically integral.
