# Lang's Introduction to Algebraic Geometry conventions

As I mention in my question regarding Lang's definition of a generic point of the variety, the conventions he follows in his book Introduction to Algebraic Geometry seem weird to me. He considers a universal domain $$\Omega$$ which is algebraically closed and shares much of the properties of that of complex numbers. In particular, $$\Omega$$ has infinite degree of transcende over its prime field. So far so good.

The problems begin when he considers other fields $$K,L,\dots$$ such that their corresponding prime fields are exactly that of $$\Omega$$ and the latter is a field extension of infinite degree of transcende of any of them. Then, all the algebraic objects will be defined over $$K,L,\dots$$. These are stated at the beginning of Section 1 in Chapter II.

For example, a variety over $$K$$, as defined on page 25, is the set

$$V=\{(x)\in \Omega^n \mid f(x)=0 \quad \forall f\in \mathfrak p\},$$ where $$\mathfrak p$$ is a prime ideal of $$K[X_1,\dots,X_n]$$, and the topology is the Zariski topology given by $$K[X_1,\dots,X_n]$$ (that is why Theorem 4 on page 25 remains true, even if $$\mathfrak p$$ is no a prime ideal of $$\Omega[X_1,\dots,X_n]$$, e.g. $$K=\mathbb Q$$, $$\Omega=\mathbb C$$, $$n=1$$ and $$\mathfrak p$$ generated by the polynomial $$f(X)=X^2+1$$).

Notice that if $$\Omega=\mathbb C$$ as usual in most practical scenarios, his assumptions forbid considering the field of real numbers.

Why does he make such assumptions? And how much of his theory remain true when the subfields of $$\Omega$$ are finite algebraic extensions?

• I don't know what your specific circumstances for using this book are, but there's a good reason that basically nobody uses these methods anymore. If you have any choice at all in the matter, I would highly recommend following a more modern presentation. Commented Dec 17, 2021 at 16:17
• @KReiser, The part about Jacobian and Albanese varieties in his book Abelian Varieties looked good to me, and of course the reader is assumed to be falimiar with the concepts explain in IAG, that is why I was checking the basic definitions. What is that good reason that you mention? Commented Dec 17, 2021 at 16:49
• All this business about universal domains is an approach (a crutch) from Weil's foundations designed to rigorously introduce generic points. It's basically all been superseded by schemes and almost nobody works in this setting anymore - so almost nothing truly requires this language, and if you dive in and get stuck you're kind of on your own. The combination of those two situations is a pretty good reason, imo - if you'd like to read a more forceful articulation, check Brian Conrad's site (search for "With all due respect to the role of Andre Weil..."). Commented Dec 17, 2021 at 18:31
• @KReiser But is there no relation between Weil's generic points and the modern notion of a generic point then? Commented Dec 17, 2021 at 18:40
• What? No, Weil's generic points are trying to get at the same thing that scheme-theoretic generic points are doing - the issue is with the implementation, not the intent. Weil made a valiant effort, but Grothendieck's approach to generic points has turned out to be much more understandable. Commented Dec 17, 2021 at 18:48

So in some sense the requirement that all fields considered be subfields of the universal domain such that the universal domain be of infinite transcendence degree over the subfield is just a matter of convenience... but I would say it is not possible to accommodate in Weil's framework varieties defined over subfields such that the universal domain is algebraic over the subfield. The reason is simple: generic points (in the sense of Weil) of varieties of (strictly) positive dimension defined over a field $$k$$ necessarily have coordinates in a transcendental extension of $$k$$. So, in Weil's framework, the only varieties defined over $$\mathbb{R}$$ in the universal domain $$\mathbb{C}$$ are 0-dimensional, i.e. points.
I have the impression that the transcendence degree condition is not something that bothers algebraic number theorists, though. After all, $$\mathbb{C}$$ is of infinite transcendence degree over $$\bar{\mathbb{Q}}$$.