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If $L$ is algebraically closed, fields of transcendence degree one over $L$ correspond to algebraic curves over $L$ up to birational equivalence, and finite extensions correspond to finite Galois coverings. A common example is $L(\sqrt{t})/L(t)$, which defines a two-sheeted covering ramified at one point. Furthermore, if $K$ is a subfield of $L$, then fields of transcendence degree one over $K$ whose intersection with $\bar{K}$ is equal to $K$ correspond to curves defined over $K$. If we consider a covering of such curves, we are considering an extension which does not add anything algebraic over $K$. Such a covering is what I mean by 'geometric' in the title. The fact that $L(\sqrt{t})/L(t)$ is a lift of the extension $K(\sqrt{t})/K(t)$ corresponds to the fact that we can define this covering morphism over $K$.

Now, my question is, if we also include field extensions which add elements algebraic over $K$ ('constant' ones), what is the geometric picture? For example, if $F/K$ is a finite non-trivial (Galois) extension, what is the geometry behind the extension $F(t)/K(t)$? (If you'd like, take more concrete example, such as $\mathbb{Q}(i)(t)/\mathbb{Q}(t)$.) It should be something like, the covering of the projective line over $K$ by the projective line over $F$. Or, what happens if we act both geometrically and arithmetically at the same time, and consider an extension like $\mathbb{Q}(\sqrt[3]{2},\sqrt{-3})(\sqrt{t})/\mathbb{Q}(t)$? Or, to be a little more geometric (i.e. a case which also counts as a manifold), an extension like $\mathbb{C}(\sqrt{t})/\mathbb{R}(t)$ or even $\mathbb{C}(t)[x]/(x^2-t^4-t)/\mathbb{C}(t)$? Furthermore, the Galois group of $F/K$ is the Galois group of $F(t)/K(t)$, so can we view the Galois group as a set of geometric transformations in the same way that we view $\mathrm{Gal}(L(t)/L(t^2))$ as the monodromy group? And in the case where we have both 'geometric' and 'arithmetic' components in our extension, how do we interpret the Galois group then?

What's interesting is that if we view $K[t]$ as the affine coordinate ring of the variety associated to $K(t)$, and similarly $F[t]$, then $F/K$ is separable iff our extension $F[t]/K[t]$ is unramified, and separability might be viewed in light of ramification theory. In addition, might we combine class field theory for number fields and function fields by considering abelian extensions of fields like $K(t)[x]/(f(x))$, where $K$ is a number field?

I believe the better way to view this is to consider not just the locus of points corresponding to the projective line over $K$ (i.e. $K \cup \{\infty\}$), but the affine scheme $Spec(K[x])$ (or its projective completion?). Then we are, in either case, looking at a covering of one scheme by another. (Since there is not in fact a nice map $\mathbb{P}^1(\mathbb{C}) \to \mathbb{P}^1(\mathbb{R})$ corresponding to the extension $\mathbb{C}(t)/\mathbb{R}(t)$, for example! But there is if we consider schemes...)

I'm aware that we can draw a geometric picture for an extension $E/F$ of number fields by looking at Spec of their integer rings. This might be sufficient when looking at affine lines, but I'm also interested in the more general case, with other curves and even possibly higher-dimensional varieties.

Edit: I changed 'arithmetic' to 'constant' to reflect more standard terminology.

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It sounds like your real question is: "How do I visualize curves and their finite covers over non-algebraically closed fields?" –  S. Carnahan Jul 28 '10 at 17:23
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It sounds like you answered your question yourself. You visualize finite extensions $E/F$ by looking at the morphism between the Specs of their integer rings. So you do the same thing with curves over $E$ and $F$; instead of looking at $Spec F[t]$ you look at the surface $Spec \mathcal O_F[t]$ etc. Then your geometric pictures becomes a finite covering of this surface (which will of course have some ramification in general). It will also involve the choice of a model over $\mathcal O_F$ for every curve over $F$. So I am not quite sure what else you are looking for. –  Arend Bayer Jul 28 '10 at 19:03
    
Right, but that's still a set of algebraic objects - it doesn't help me visualize. I was including the example of Spec of a number ring as something that isn't particularly geometric (the topology is almost essentially the finite complement topology...). Since curves are actually sets of points, I figured there might be a better way to think of it. –  David Corwin Jul 28 '10 at 20:30
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Davidac897: I think your notion of "visualize" is getting unreasonable. The marvel is that we can develop a language to discuss "geometry" over finite fields and trick our minds into applying intuition from manifolds when working with these manifestly algebraic objects. Don't take the Zariski topology so seriously; close your eyes and think of Riemann surfaces to guide your intuition, even when the rigorous details rest on commutative algebra (which is what replaces the role of multivariable calculus for differential geometry). That's the magic and difficulty of algebraic geometry. –  BCnrd Jul 28 '10 at 22:45
    
Hah! Don't take the Zariski topology too seriously! I will never let you forget that! –  JBorger Jul 29 '10 at 9:29

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I think it's possible that non-geometric extensions are indeed not as directly visualizable as geometric ones.

Some terminology: let $k$ be a field, and either assume $k$ has characteristic $0$ or beware that some separability issues are being omitted in what follows. A (one variable) function field over $k$ is a finitely generated field extension $K/k$ of transcendendence degree one. This already allows for the possibility of a nontrivial constant extension, which is often excluded in geometric endeavors: for instance, according to this definiton, $\mathbb{C}(t)$ is a function field over $\mathbb{R}$, but a sort of weird[1] one: e.g. it has no $\mathbb{R}$-points.

One says a function field $K/k$ is regular if $k$ is algebraically closed in $K$; i.e., any element of $K$ which is algebraic over $k$ already lies in $k$ [plus separability stuff in positive characteristic]. Any function field can be made regular just by enlarging the constant field to be the algebraic closure of $k$ in $K$; e.g., the previous example is a regular function field over $\mathbb{C}$.

Regularity is what one needs to think about function fields as geometric objects: namely, there is a bijective correspondence between regular function fields $K/k$ and complete, nonsingular algebraic curves $X_{/k}$.

Now, on to covers. Let $L/K$ be a finite degree extension of function fields over $k$. One says (often; this is slightly less standard terminology) that the exension $L/K$ is geometric over $k$ if both $L$ and $K$ are regular function fields. And again, there is a bijective correspondence between geometric extensions of function fields and finite $k$-rational morphisms of algebraic curves $Y \rightarrow X$.

Assuming that the bottom function field $K$ is regular, every extension $L/K$ may be decomposed into a tower of a constant extension $lK/K$ followed by a geometric extension $L/lK$. Constant extensions have a role to play in the theory -- see for instance the chapter on constant extensions in Rosen's Number theory in function fields, but I think it is fair to describe their role as algebraic rather than geometric: at least that's the standard view.

In fact, the issue that not all extensions of regular function fields are geometric is an important technical one in the subject, because sometimes natural algebraic constructions do not preserve the class of geometric extensions.

Here is an example very close to my own heart: let $p$ be an odd prime. The elliptic modular curves $X(1)$ and $X_0(p)$ have canonical models over $\mathbb{Q}$ and there is a natural "forgetful modular" covering $X_0(p) \rightarrow X(1)$. This corresponds to a geometric extension of function fields $\mathbb{Q}(X_0(p)) / \mathbb{Q}(X(1))$. This is not a Galois extension: what is the Galois closure and what is its Galois group? If -- as was classically the case -- our constant field were $\mathbb{C}$ -- then the Galois closure is the function field of the modular curve $X(p)$ and the Galois group of the covering $X(p)/X(1)$ is $\operatorname{PSL}_2(\mathbb{Z}/p\mathbb{Z})$. However, over $\mathbb{Q}$ the Galois closure also contains the quadratic field $\mathbb{Q}\left(\sqrt{(-1)^{\frac{p-1}{2}} p}\right)$ so is an extension of a cyclic group of order $2$ by $\operatorname{PSL}_2(\mathbb{Z}/p\mathbb{Z})$ (in fact it is $\operatorname{PGL}_2(\mathbb{Z}/p\mathbb{Z})$). Thus the extension is not geometric. This is unfortunate, because Hilbert's Irreducibility Theorem says that if one has a geometric Galois extension $L/k(t)$ with $k$ a number field, then one can realize $\operatorname{Aut}(L/k(t))$ as a Galois group over $k$. So in this case, this obtains $\operatorname{PSL}_2(\mathbb{Z}/p\mathbb{Z})$ as a Galois group over not $\mathbb{Q}$ but over the variable quadratic field given above. K.-y. Shih found a brilliant way to "tweak" this construction to realize $\operatorname{PSL}_2(\mathbb{Z}/p\mathbb{Z})$ over $\mathbb{Q}$ in certain (infinitely many) cases, and other mathematicians -- e.g. Serre, myself, my graduate student Jim Stankewicz -- have put a lot of thought into extending Shih's work, but with only very limited success.

Added: Brian's example in the comments is very nice. Maybe another remark to make is that in the arithmetic theory of coverings of curves (an active branch of arithmetic geometry) the distinction between a Galois extension and a geometrically Galois extension of fields (i.e., one which becomes Galois after base change to $\overline{k}$) is a key one: it's certainly something that many arithmetic geometer think a lot about. It just doesn't come with an obvious "visualization", at least not to me. Not everything in algebraic or arithmetic geometry can be visualized, or at least not visualized in a way common to different workers in the field. For instance, an inseparable field extension $l/k$ is by definition ramified, but I have never seen anyone describe this visually. (There are things you can say to justify that this is not a "covering map", e.g. by pointing to the nonreducedness of $l \otimes_k l$, but I don't think this is direct visualization either. Maybe some would disagree?) What you do is think of the case of a ramified cover of Riemann surfaces, and take away the (key) piece of intuition that an inseparable field extension -- which is, visually speaking, just one closed point mapping to another -- behaves like a ramified cover of Riemann surfaces in many ways. So, as Brian says, in this subject a lot of geometric reasoning proceeds by analogy. Unlike in, say, certain branches of low-dimensional topology, one does not prove a theorem by referring to (allegedly) visually apparent features of one's constructions.

[1]: Those who know me well know that I certainly don't think that a curve is weird just because it has no degree one closed points. More accurate is to say that this curve doesn't have any degree one closed points for a "weird reason".

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Pete's example has Gal. closure with non-triv. "constant" part and some more. Here's a purely constant example. Consider ell. curve $E$ over field $k$ and $n > 1$ with $E(k)[n]=0$. For the finite etale geometrically conn'd covering of curves $[n]:E\rightarrow E$, over $k_s$ it's a conn'd Gal. cover with Gal. gp $E(k_s)[n]$. The corresponding nonzero translations over $k_s$ aren't def'd over $k$, so this cover is "geometrically Galois" but the function field ext'n over $k$ has no nontrivial auts. Exercise: deduce that the fn field ext'n has Gal. closure inside $k_s(E)$. –  BCnrd Jul 28 '10 at 22:15
    
"but I think it is fair to describe their role is algebraic rather than geometric: at least that's the standard view." is definitely an answer to my question. –  David Corwin Jul 28 '10 at 22:50
    
Maybe this will answer my question: If $\mathbb{R}(x)[y]/(f(x,y))/\mathbb{R}$ is regular (transcendence degree $1$), and $\mathbb{C}(x)[y]/(g(x,y))/\mathbb{C}$ is regular, it would seem that there should be a geometric picture, since these fields relate to geometric objects, i.e. the loci of $f(x,y)$ and $g(x,y)$ in $\mathbb{R}^2$ (respectively $\mathbb{C}^2$). Is the reason this doesn't work because we must consider the whole scheme, not just the variety? I.e. we must work with all of $Spec(\mathbb{R}[x,y]/(f(x,y)))$? And this latter is conjugate pairs of $\mathbb{C}$-rational points of it? –  David Corwin Jul 28 '10 at 22:51

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