In a first course on rings and fields we met the concept of field extensions, especially algebraic ones. The presentation of the material was very algebraic and felt a little lifeless. I was wondering whether there is some geometric way to think of (different types) of field extensions. I am familiar with the basic formalism of schemes and varieties, but I don't know algebraic geometry. In particular, I am curious how to think of splitting fields in geometric terms.

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    $\begingroup$ If you want to think about Galois theory in geometric terms then the geometric side is the theory of covering spaces. Although the technical details involve scheme theory, the intuition comes from algebraic topology. $\endgroup$
    – Zhen Lin
    Feb 14, 2016 at 14:11
  • $\begingroup$ FWIW, Bamboo English can be considered a 'field extension' of English. $\endgroup$
    – Mike Jones
    Aug 11, 2017 at 11:01

1 Answer 1


Galois theory.

A more elaborate version of Zhen Lin's comment is the following: Galois theory studies certain types of finite field extensions (and you can also treat certain types of algebraic field extensions, as a limit of the finite case). The philosophy is that a finite field extension is the same thing as a finite morphism $\operatorname{Spec} L \to \operatorname{Spec} K$.

In algebraic geometry, finite morphisms of (say smooth projective) varieties correspond in the complex manifold world to proper maps $X \to Y$ with finite fibres. Such a map is close to being a covering space, but this is not always the case. For example, the map $\mathbb A^1 \to \mathbb A^1$ given by $x \mapsto x^2$ is not a covering space, because it is not a local homeomorphism near the origin.

It turns out that there is a super general algebraic notion of étale morphisms, and finite étale morphisms correspond to covering spaces. General étale morphisms include open immersions as well, which are neither finite nor covering spaces; this is a point that leads to a bit of confusion for the novice.

Then one could say that a finite Galois extension is a field extension such that the map $\operatorname{Spec} L \to \operatorname{Spec} K$ is finite étale. This is historically very inaccurate, and for most people this is also the wrong order to learn the material. Moreover, it requires a bit of work to show that a field extension is Galois if and only if it is étale; most courses on Galois theory do not touch on this. However, given that you indicated to know some algebraic geometry, this might be a useful way for you to think about it geometrically.

In this language, the analogue of the fundamental group $\pi_1(X)$ is the absolute Galois group $\operatorname{Gal}(\bar K/K)$, in the very precise sense that $\operatorname{Gal}(\bar K/K) \cong \pi_1^{\operatorname{alg}}(\operatorname{Spec} K)$. Galois theory can then be viewed as the study of finite covering spaces of $\operatorname{Spec} K$, and their deck transformations.

However, I should point out that this is a very beautiful and unexpected analogy, that was only pointed out by Grothendieck. Galois theory takes place centuries earlier, and is a very rich and well-developed theory in itself. It is crucial to number theorists, and the explicit knowledge of Galois cohomology is very important for the development of the much harder theory of étale cohomology.

Splitting fields.

Let me specifically address splitting fields because you ask about them.

Suppose $f \in k[x]$ is a separable polynomial (no repeated roots). Then we get a set $V(f) \subseteq \mathbb A^1_k$. The size of this set should equal $\deg f$: a polynomial of degree $n$ without multiple roots has $n$ roots.

However, if $k$ is not algebraically closed, it may happen that $f$ is irreducible, in which case set-theoretically $V(f)$ is just a point. However, over the splitting field $\ell$ of $f$, we know that $f$ factors as a product of linear factors, so $f$ really does have $n$ roots.

Geometrically: if $f$ is separable of degree $n$, then we get a finite morphism $$X = \operatorname{Spec} k[x]/(f) \to \operatorname{Spec} k$$ of degree $n$. Then $X$ might have fewer than $n$ points; however $X \times_{\operatorname{Spec} k} \operatorname{Spec} \ell$ always splits into $n$ distinct points. This is of course not what the name splitting field comes from, but it is another way to look at it.

Transcendence theory.

Of course, Galois theory does not nearly cover all the field extensions; however, it is a very nice case because there is so much you can say about it.

On the other hand, there is also a direct use of transcendental extensions in algebraic geometry. The field $k(x_1,\ldots,x_n)$ has transcendence degree $n$ over $k$, and the variety $\mathbb A^n_k$ is $n$-dimensional. This is no coincidence: one can prove that for an integral scheme of finite type over a field $k$, the dimension equals the transcendence degree of the function field.

Moreover, the category of algebraic varieties with dominant rational maps as morphisms turns out to be equivalent to the category of fields of finite type over $k$. A lot of questions in algebraic geometry (especially birational geometry) have been motivated by field theory.

For example, the only way I know how to prove that the field $\operatorname{Frac} k[x,y]/(y^2 - x^3 - x)$ is not isomorphic to $k[t]$ is to use the genus in algebraic geometry.

As for a much harder example, consider the following question: let $K/k$ be a finitely generated field extension, and suppose that $K(x_1,\ldots,x_n) \cong k(y_1,\ldots,y_m)$ for certain $m,n \in \mathbb Z_{\geq 0}$. Is it true that $K \cong k(z_1,\ldots,z_{m-n})$?

The answer is, somewhat surprisingly, no. What you can prove is that there exist varieties which are stably rational but not rational, and this settles the algebra problem by taking the function field.

These are just examples of the interplay between field theory and algebraic geometry; there are many more things one could say. I would certainly say that a good command of field theory is essential to a modern algebraic geometer.

Some references:

A popular introductory reference to the analogy between Galois groups and fundamental groups seems to be the lecture notes Galois theory of schemes by Hendrik Lenstra. These notes assume familiarity with Galois theory and algebraic geometry (it seems that for a large portion, one can get away with only knowing commutative algebra). On the other hand, no knowledge of étale morphisms or harder topics like étale cohomology is assumed. Another reference is the chapter Fundamental groups of schemes of the stacks project (online or pdf). This also gives further references, e.g. to books written on the subject. A great book is Szamuely's Galois groups and fundamental groups.

Transcendence theory belongs to the realm of (commutative) algebra. Three useful references are Chapter VIII of Lang's Algebra, the chapter on fields in the stacks project (online or pdf), and Appendix A1 of Eisenbud's Commutative algebra with a view towards algebraic geometry. However, the latter contains some mistakes (he gets confused between separable and separably generated extensions in the course of trying to prove the relation between them).

I am not aware myself of any reference other than the original research papers for the non-rationality of the cubic threefold (which is the example of a stably rational variety that is not rational); I would be very interested if someone else knows one. The original papers are:

  • C. A. Clemens, P. A. Griffiths, The intermediate Jacobian of the cubic threefold (1972). Available on jstor here.
  • J.P. Murre, Reduction of the proof of the non-rationality of a non-singular cubic threefold to a result of Mumford (1973). Available on EUDML here.

Both contain a proof of the non-rationality of the cubic threefold, but I'm not sure they prove stable rationality. Murre's proof is supposed to be a simplification of the proof by Clemens—Griffiths.

  • $\begingroup$ Thank you for the incredible answer! Where can I learn about field extensions this way? Any books you can recommend? $\endgroup$
    – Arrow
    Feb 18, 2016 at 15:04
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    $\begingroup$ I wrote an exceedingly long math.se post a while ago (found here: math.stackexchange.com/questions/1392519/…) which might complement Remy's nice answer. Also, I would HIGHLY HIGHLY recommend the text "Galois Groups and Fundamental Groups" for all things related to this subject. Seriously—go read it now. $\endgroup$ Feb 19, 2016 at 11:22
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    $\begingroup$ @Remy Also, viz. your comment about the elliptic curve, I think you can argue that if $k(t)$ were isomorphic to $\text{Frac}(k[x,y]/(y^2-x^3-x)$ then they should be isomorphic after base changing to some uncountable field extension, so I think you can assume that $k$ is uncountable. Then, I think it's not too difficult to see that every Dedekind subdomain of $\text{Frac}(k[x,y]/(y^2-x^3-x))$ (with fraction field $\text{Frac}(k[x,y]/(y^2-x^3-x))$) has uncountable Picard group. So, there is no where for $k[t]$ to go under an isomorphism $k(t)\to\text{Frac}(k[x,y]/(y^2-x^3-x))$. $\endgroup$ Feb 19, 2016 at 11:27
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    $\begingroup$ Of course, even if one susses this out, one is really cheating. As you probably know, the only reasonable way (at least I can think of) to prove the claim I made about Dedekind subdomains of $\text{Frac}(k[x,y]/(y^2-x^3-x))$ is secretly using that it's associated to an elliptic curve (e.g. the Dedekind subdomain should differ from the e.c. by finitely many points, which means that it's Picard group differs from the full group by a countable group. But, since $\text{Pic}^0(E)\cong E(k)$ and the latter is uncountable, one can conclude. $\endgroup$ Feb 19, 2016 at 11:29
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    $\begingroup$ Also, last one, I swear—you have a misplaced $\text{op}$. Namely, $G_K$ IS isomorphic to $\pi_1(\text{Spec}(K))$. The op comes in because one defines (or at least one definition) is $\pi_1(X)=\varprojlim \text{Gal}(Y/X)$ where $Y/X$ ranges over all Galois covers. Here though $\text{Gal}(Y/X)=\text{Aut}(Y/X)^\text{op}$. In particular, tracing through this one sees that one has actual equality $G_K\cong\pi_1(\text{Spec}(K))$. $\endgroup$ Feb 19, 2016 at 11:32

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