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.

isn'tparticularly 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. $\endgroup$languageto 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. $\endgroup$1more comment