# Geometric fibers of schemes.

I would like to get an understanding of the notion of geometric fibers of scheme morphisms: If $f: X \rightarrow Y$ is a morphism of schemes, then its geometric fiber is defined to be $X \times_{Y} \overline{k(p)}$ for the quotient field $k(p)$ at $p \in Y$. I would like to know, why this is a good choice for the notion of "fiber". Why does one pick such an abstract notion and why does exactly this the right job? Every help will be appreciated.

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Basically because this reduces to the classical case (variety over alg cl field). –  Martin Brandenburg Feb 15 '13 at 19:20
"Why does exactly this the right job?" For what? I'd argue that this doesn't do the right thing if what you are interested in is arithmetic properties of the fiber. This gives you "geometric" information about the fiber and hence is the "geometric fiber." Where did you see that this did the "right job?" We need this information if we want to answer why it did it. –  Matt Feb 16 '13 at 5:09

If $U \subset Y$ then the fibre product $U \times_Y X$ should be thought of as the preimage $f^{-1}(U)$. Similarly, if you are happy about thinking of points of $Y$ as being morphisms from Specs of fields then the fibre at a point $x: Spec k \to Y$ is the fibre product $Spec k \times_Y X$, often denoted $X_x$ or $X \otimes_Y k$.
Now, if you are working with varieties (or schemes of locally of finite type) over the complex numbers (or an algebraically closed field) then that's that. However in the general setting it is important to understand how your variety behaves under field extensions, and often a property is true for every finite extension if and only if it's true over the algebraic closure, that's a good reason to define the geometric fibres to be $X \otimes_Y \overline{k}$.
I guess the reason they are called geometric is that, first of all there is no Galois group acting, and second there are no phantom points appearing in the following sense. If you take $x^2 + y^2 = -1$ over the reals, then it won't have any real solutions but will have complex solutions.
If $X = Spec R = k[x_1,\ldots,x_n]/(f_1,\ldots,f_r)$ is an affine variety then points of X should correspond to simultaneous solutions to the $f_i$. If you want the solutions to be in some field extension $L$, then these correspond to morphisms of k-algebras $R \to L$. Reversing arrows, these correspond to scheme morphisms $Spec L \to X$.