Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|cite|improve this question
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

2 Answers 2

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 $Spec L \to X$.

share|cite|improve this answer

Jacob has already given a good answer. I just want to add something that convinces me that the definition of geometric fibre is indeed a good one.

Let's fix a field $k$, and consider the concept of smoothness of a variety $X$ over $k$ (or more generally, schemes locally of finite type over $k$). It's quite natural to call some (closed) point $x \in X$ a smooth point if $\Omega_{X/k,x}$ if free of rank equal the local dimension of $X$ at $x$. Geometrically, this simply states there are enough independent differentials locally.

We would like to express this condition in terms of the local ring $O_{X,x}$. In the case $k$ is algebraically closed, this condition is (as is well known) that $O_{X,x}$ is a regular local ring. However, in general, the equivalent condition is $O_{X,x}\otimes_k \bar{k}$ is regular. (This means each localization is regular, as in general this ring is only semi-local). Here geometric fibre arises naturally.

The reason for this is somehow technical, there is a result characterizing regular local rings in terms of Kaehler differentials, see for example Hartshorne Thm II.8.8 (notice that there the assumption that $k$ being perfect is superfluous).

Theorem: Let $B$ be a local ring containing a field $k$ isomorphic to its residue field. Assume furthermore that $B$ is a localisation of a finitely generated $k$-algebra. Then $\Omega_{B/k}$ is a free $B$-module of rank equal to dim $B$ if and only if $B$ is a regular local ring.

The first assumption is clearly satisfied for varieties over an algebraically closed field, but fails in general. So a base extension to the algebraic closure arises quite naturally.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.