Why do I expect $\operatorname{Spec}(\Bbb Q(i)\otimes_{\mathbb Q} \Bbb Q(i))$ contains 2 points? This is related to Iitaka's Algebraic Geometry Chpt 1, Sec 20, Example 1.27. 
Let $X=\operatorname{Spec}(\Bbb Q(i))$ where $\Bbb Q$ is rational numbers. Then $X\times_{\operatorname{Spec}(\Bbb Q)}X=\operatorname{Spec}(\Bbb Q(i)\otimes_{\Bbb Q}\Bbb Q(i))$. 
$\textbf{Q:}$ If I am a normal person, I would expect fiber product of one point with another point over a point is a point.(i.e. If $X=\{pt\}$ instead and given $X\to\star$ map where $\star$ is another point, then $X\times_\star X=\{(\text{pt},\text{pt})\}$.) For the scheme category, this yields 2 points as can be seen $\Bbb Q(i)\otimes_{\Bbb Q}\Bbb Q(i)\cong \Bbb Q(i)\times \Bbb Q(i)$. Why does this happen or what is the intuitive reason that I would expect the fiber product have 2 points? It seems this is due to the fact that scheme category does not behave like set category or topological space category as it has extra information contained in sheaves. 
 A: From what I understand the fiber product for affine schemes $Y=Spec(R),Z=Spec(S)$ is the affine scheme $Spec(S\otimes_T R)$ for some subring $T$ of both $S,R$ (more generally two subrings $T_1,T_2$ together with an isomorphism $T_1\to T_2$)
Here
$$X = Spec(\Bbb{Q}[t]/(t^2+1))$$
$$X\times_{Spec(\Bbb{Q})}X = Spec(\Bbb{Q}[t]/(t^2+1) \otimes_\Bbb{Q} \Bbb{Q}[s]/(s^2+1)) = Spec(\Bbb{Q}[t,s]/(t^2+1,s^2+1))$$
$$= Spec(\Bbb{Q}[t,s]/(t^2+1,(s+t)(s-t))$$
It is not hard to see that $\Bbb{Q}[t,s]/(t^2+1,(s+t)(s-t))$ has two prime ideals $(s+t)$ and $(s-t)$.
A: It's just not very accurate to think of $\text{Spec } \mathbb{Q}(i)$ as being "a point," for exactly this kind of reason. In this context a more accurate picture is that it's "two points related by a Galois action." 
More generally, if $X$ is a scheme over $\mathbb{Q}$, rather than look at its Zariski spectrum we can consider the set of points $X(\overline{\mathbb{Q}})$ over the algebraic closure of $\mathbb{Q}$, equipped with the action of the absolute Galois group $\text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$. IMO this is a much more geometrically sensible notion of "point" to consider; among other things, unlike the Zariski spectrum it preserves finite products.
If we apply this recipe to $X = \text{Spec } K$ where $K$ is a number field, say $\mathbb{Q}[x]/f(x)$, then $X(\overline{\mathbb{Q}})$ consists of $\dim K$ points, one for each root of $f$ over $\overline{\mathbb{Q}}$, together with the Galois action that factors through the action of the Galois group of $f$ on its roots. Since $f$ is irreducible this Galois action is transitive, and the Zariski spectrum can only see the orbits of this action, rather than the action itself. 
I mentioned that this functor preserves finite products, so now let's apply it to the product of two copies of $X = \text{Spec } \mathbb{Q}(i)$. We can think of $X$ as being the two points $i, -i \in \overline{\mathbb{Q}}$, related by complex conjugation. (Here I'm secretly thinking about the embedding of $X$ into the affine line given by the quotient $\mathbb{Q}[x] \to \mathbb{Q}[x]/(x^2 + 1)$.) So $X \times X$ is four points $(\pm i, \pm i)$, which are divided into two orbits by complex conjugation: $\{ (i, i), (-i, -i) \}$ and $\{ (i, -i), (-i, i) \}$. Again, the Zariski spectrum only sees the orbits of this action, which is less information. 
