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.
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.