Fix a field $K$ and consider a family of nonsingular projective curves $X\to S$ over an irreducible scheme $S$. Suppose that $\eta$ is the generic point of $S$, then the generic fiber of the family is the curve $X_s\to K(S)$ obtained by base change.
Now consider a concrete example, but not in the category of schemes. We work with "equations" in the category of classical algebraic varieties.
Let
$$X: aX^2+bYZ=0\quad a,b\in K$$
$$S=\mathbb A^2_K$$
Then $X$ is a family of plane projective curves over $\mathbb A^2_K$ parameterized by $a$ and $b$ which can be seen as a morphism:
$$X\subset \mathbb P_K^4\to\mathbb A^2_K$$
$$[a:b:x:y:z]\to (a,b)$$
Now: can I see the generic fibre of this map from the equations? Usually books says that the generic fibre is the curve defined over the field $K(a,b)$ by the same equations of $X$. So
$$X_\eta: aX^2+bYZ=0\quad a,b\in K(a,b)\,.$$
This makes perfectly intuitive sense because we have a curve over $K(S)$ as we require in the category of schemes.
The problem is that in the passage from the category of schemes to the category of varieties, we recall only the closed points of objects, so things like generic fibers shouldn't have any geometric naive sense.
Here my question:
In presence of any family of curves, why do we take the generic fibre as explained before (we take the same equations but in another field)? What is the formal explanation of this fact?
Best Answer
I am not completely sure that I understand what it is you are looking for, but let me give an answer in the hope that you find it helpful.
Remember that base change is constructed (e.g. in Hartshorne) first in the case of morphisms of affine schemes, and the general case then follows by a rather involved glueing process. So to avoid the complications, I will modify your example to the case of affine schemes instead, by taking the open subscheme $\{ Z \neq 0 \}$ inside your family of curves.
Let me also change the names of your fields so that $k$ is the field we start with, and $K=k(a,b)$.
We then have a family $$ X \subset \mathbf A^4_k$$ cut out by the equation $$ax^2+by=0;$$ in other words $$X = \operatorname{Spec} \left( \frac{k[x,y,a,b]}{I} \right) $$
where $I$ is the ideal generated by $ax^2+by$.
Now the projection morphism $$\pi : X \rightarrow \mathbf A^2 \\ (x,y,a,b) \mapsto (a,b)$$ corresponds to the map of rings $$ k[a,b] \rightarrow \frac{k[x,y,a,b]}{I}.$$
On the other hand, the inclusion of the generic point
$$ \iota: \operatorname{Spec} K \rightarrow \mathbf A^2_k$$
corresponds to the ring map
$$ k[a,b] \hookrightarrow K$$ embedding the polynomial ring in its field of fractions.
Finally, the crux of the matter: the base change of $\pi$ along $\iota$ is (by definition) given by the tensor product
$$ K \otimes_{k[a,b]} \frac{k[x,y,a,b]}{I}$$ which is isomorphic to $$\frac{K[x,y]}{J}$$ where now $J$ is the ideal in $K[x,y]$ generated by the polynomial $ax^2+by$.
So the generic fibre is indeed defined by the same equation, but now viewed as a polynomial over a different field.
I hope that helps.