# To what extent can fields be classified?

The study of algebraic geometry usually begins with the choice of a base field $k$. In practice, this is usually one of the prime fields $\mathbb{Q}$ or $\mathbb{F}_p$, or topological completions and algebraic extensions of these. One might call such fields $0$-dimensional. Then one could say that a field $K$ is $d$-dimensional if it has transcendence degree $d$ over a $0$-dimensional field.

But is there a way to make this less ad hoc? Is there a reason I've never seen any such definition of $0$-dimensional fields? Am I missing something?

To what extent has the classification of abstract fields been considered?

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I think classifying fields of finite transcendence degree over a base field $k$ is equivalent to classifying varieties over $k$ up to birational equivalence, so that in sense algebraic geometers have done a lot of work on this question. –  Qiaochu Yuan Jul 9 '13 at 23:08
Qiaochu, I think you should make your comment an answer. I think it really answers the question. –  Joël Jul 10 '13 at 2:32
@QiaochuYuan: Yes, I'm aware that classifying higher dimensional fields is the same as classifying varieties up to birational equivalence. But what do you think about the problem of classifying "zero-dimensional" fields? I.e. What are all the possible ground fields? –  Drew Armstrong Jul 10 '13 at 17:36
@Drew: I don't understand what you mean by "ground field." Do you want a list of fields such that every field is a field of finite transcendence degree over a field in that list or something? That doesn't seem easier than listing all fields. If you don't care about finiteness, then every field is a field of some transcendence degree over $\mathbb{Q}$ or $\mathbb{F}_p$. –  Qiaochu Yuan Jul 10 '13 at 22:41

## 2 Answers

I think you are lumping too many disparate kinds of fields together under the heading "zero-dimensional". As Jason says in his answer, there are some precise definitions of dimensions of fields (e.g. cohomological dimension but also other definitions of a field-arithmetic nature).

Another important comment is that in modern algebraic / arithmetic geometry there is no restriction on the kind of field that can be taken as a "ground field", i.e., over which to define algebraic varieties. Although you have indeed listed some common ground fields, some people think about other special cases as well as the general case. (For instance I have recently been thinking about elliptic curves defined over transfinitely iterated function fields.)

The basic classification of fields, as I understand/view it, is as follows:

Step 1: Every field has a characteristic, either $0$ or a prime number. There are no homomorphisms between fields of different characteristics, so fields of a given characteristic are somehow different worlds (one can think of the characteristic as a connected component in the category of fields, essentially).

Step 2: For a fixed characteristic $p \geq 0$ there is a unique minimal field, called the prime subfield, say $k_0$. This is precisely the initial object in the category of fields of characteristic $p$: it's $\mathbb{Q}$ in characteristic $0$ and $\mathbb{F}_p$ in characteristic $p$. This means that the absolute theory of fields is reduced to the theory of field extensions, and one can define "absolute invariants" on a field $K$ by giving invariants of $K/k_0$. In particular:

Step 3: The absolute transcendence degree of a field $K$ is the cardinality of a transcendence basis for $K/k_0$. This means that there is a subextension $k_0 \subset F \subset K$ such that $F/k_0$ is purely transcendental: a rational function field, perhaps in infinitely many variables, and $K/F$ is algebraic.

Step 4: The algebraic extension $K/F$ is in many ways the most interesting part. For instance, as Qiaochu Yuan mentions, algebraic geometry in dimension $n$ is the study of finite degree field extensions of $\mathbb{C}(t_1,\ldots,t_n)$.

When $n = 1$ we get precisely the compact Riemann surfaces, or (replacing $\mathbb{C}$ with any field $k$ and looking at finitely generated field extensions of transcendence degree $1$) of complete, regular, integral algebraic curves over $k$. So a classification here is the (rich) classical story of the genus, the moduli spaces $\mathcal{M}_g$, and so forth. When $k$ is not algebraically closed, Galois cohomology enters the picture.

When $n \geq 2$ we are studying birational algebraic geometry only, but that is still very rich. When $n = 2$ there is a "classification of algebraic surfaces" which ends up tossing most of them into a very large box called "general type". To the best of my knowledge there is no explicit description of the connected components of the (infinite type) moduli space of all complex algebraic surfaces like there is in dimension one.

In fact, I believe that probably starting even in dimension $2$ it is in some sense hopeless to try to give an algorithmic classification, although I cannot recall having seen a precise impossibility theorem along these lines analogous e.g. to the algorithmic impossibility of classifying compact $4$-manifolds because this problem -- via the fundamental group -- contains the word-problem for finitely presented groups which Novikov proved is algorithmically unsolvable. Bjorn Poonen discussed similar (open) problems in the last of a series of three lectures on undecidability that he gave at UGA a few years ago; notes are available on his webpage.

Step 5: A vaguely dual role to purely transcendental extensions of the prime subfield is played by the algebraically closed fields. Here there is a precise classification: the characteristic and the absolute transcendence degree determine an algebraically closed field up to isomorphism. If the field is uncountable, then the absolute transcendence degree is just its cardinality, so that classifies the field: e.g. the only algebraically closed field of characteristic $0$ and continuum cardinality is $\mathbb{C}$. This has been used for some sneaky purposes: e.g. the algebraic closure of $\mathbb{Q}_p$ must then also be isomorphic to $\mathbb{C}$. (However for countable fields cardinality is not enough: e.g. $\overline{\mathbb{Q}}$ is not isomorphic to $\overline{\mathbb{Q}(t)}$.) This uncountable categoricity means that the first order theory of algebraically closed fields of given characteristic $p \geq 0$ is complete, which has various pleasant consequences. Algebraically closed fields of infinite transcendence degree have some further nice properties which makes them suitable for use as "ground fields" in algebraic geometry, as was exploited by Weil in his pre-(scheme-theoretic) foundations. From a model-theoretic perspective, these large algebraically closed fields enjoy good saturation properties.

Fields which are "close" to being algebraically closed tend to be better understood than fields which are farther away from being algebraically closed. One can measure this (at least in characteristic $0$) by the size / complexity of the absolute Galois group $\operatorname{Aut}(\overline{K}/K)$. Thus for instance the absolute Galois group of $\mathbb{Q}_p$ is known completely -- i.e., it is a topologically finitely generated compact totally disconnected Hausdorff group, and explicit generators and relations are known, which in some sense means we understand every algebraic extension of $\mathbb{Q}_p$. This needs to be taken with a grain of salt: a certain filtration on the absolute Galois group provides much coarser information -- e.g. you can break the group up into three pieces, two of which are commutative and one of which is pro-$p$, so the group is pro-solvable. That's useful: the pro-$p$-part is the hard part and knowing generators and relations for it does not in practice seem to take away the mystery. For instance, the Local Langlands Correspondence is a deep theorem about representations of the absolute Galois group of a $p$-adic field, and it was certainly not proved by looking at its explicit structure as a topologically finitely presented group!

Let me note finally that every compact totally disconnected topological group occurs up to isomorphism as the automorphism group of an algebraic Galois extension of fields (the Leptin-Waterhouse Theorem; I discovered this independently as a graduate student, and at the time I knew neither that others had published the result before nor even that such a result would be publishable)...so algebraic field extensions can be awfully complicated.

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The usual definition of "dimension" for fields is "cohomological dimension". The list of fields you gave have quite different cohomological dimensions. You can read much more about this in Serre's "Galois Cohomology".

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That just means that the term “dimension” was an unfortunate choice on OP’s part. Maybe a less loaded word is called for, like “grandiosity”. –  Lubin Jul 10 '13 at 1:54