Transcendental Galois Theory

Is there a good reference on transcendental Galois Theory?

More precisely, if $K/k$ admits a separating transcendence basis (or maybe if it is a separably generated extension) it seems to me that many of the usual theorems of Galois theory go through. Moreover, the group $\text{Aut}(K/k)$ seems to have additional structure; namely it should be an algebraic group over $k$.

For example, it seems to me that $k(x_1, ..., x_n)/k$ has automorphism group $GL_n(k)$. (EDIT: As Qiaochu Yuan points out, this is incorrect; the automorphism group at least must contain $PGL_{n+1}(k)$, acting via its action on the function field of $\mathbb{P}_k^n$.) This sort of thing must be well-studied; if so, what are the standard references on the subject?

I have seen Pete L. Clark's excellent (rough) notes on related subjects here but they seem not to address quite these sorts of questions.

-
Isn't the automorphism group more something like $\text{PGL}_{n+1}(k)$? – Qiaochu Yuan Apr 13 '11 at 6:11
How do you figure? I could just be being silly but I don't even see an action of $PGL_{n+1}$... – Daniel Litt Apr 13 '11 at 6:16
I think the action is by generalized fractional linear transformations, e.g. when $n = 1$ we can send $x_1$ to any $\frac{ax_1 + b}{cx_1 + d}$ and this has inverse the corresponding inverse fractional linear transformation. – Qiaochu Yuan Apr 13 '11 at 6:22
I'm pretty sure you're right; as I've remarked in my edit, this action should arise via the natural action of $PGL_{n+1}$ on the function field of $\mathbb{P}^n_k$. – Daniel Litt Apr 13 '11 at 6:45

For every $n \geq 1$, there is a natural effective action of $\operatorname{PGL}_{n+1}(k)$ on $k(x_1,\ldots,x_n)$. In fact $\operatorname{PGL}_{n+1}(k)$ is the automorphism group of $\mathbb{P}^n_{/k}$, the action being the obvious one induced by the action of $\operatorname{GL}_{n+1}(k)$ on the vector space $k^{n+1}$ in which $\mathbb{P}^n$ is the set of lines.
However, no one said this was the entire automorphism group of $k(x_1,\ldots,x_n)$! It is when $n = 1$ -- for instance because every rational map from a smooth curve to a projective variety is a morphism ("valuative criterion for properness"). However, $\operatorname{PGL}_{n+1}(k)$ is known not to be the entire automorphism group of $k(x_1,\ldots,x_n)$ when $n > 1$. Rather, the full automorphism group is called the Cremona group. For $n = 2$ we have a problem in the geometry of surfaces, and it was shown (by Max Noether when $k = \mathbb{C}$) that the automorphism group here is generated by the linear automorphisms described above together with a certain set of simple, well-understood birational maps, called quadratic maps or indeed Cremona transformations. But even when $n = 2$ this automorphism group is not an algebraic group: it's bigger than that.
When $n \geq 3$ it is further known that the linear automorphisms and the Cremona transformations do not generate the whole automorphism group, and apparently no one has even a decent guess as to what a set of generators might look like. I had the good fortune of hearing a talk by James McKernan on (in part) this subject within the last few months, so I am a bit more up on this than I otherwise would be. Anyway, he gave us the sense that this is a pretty hopeless problem at present. For instance, see this recent preprint in which a rather eminent algebraic geometer works rather hard to prove a seemingly rather weak result about finite subgroups of the three dimensional Cremona group!
Thanks! I should have realized that we could allow birational automorphisms of $\mathbb{P}^n$ as well; I wonder if there is some way of functorially identifying an algebraic subgroup (e.g. look at automorphisms that fix some basepoint on $\mathbb{P}^n$...). – Daniel Litt Apr 13 '11 at 7:12