If $Y$ is closed in $X$, and $X(F) \cap Y$ is dense in $Y$, then $Y$ is defined over $F$

Question

I've been trying to understand why this result in algebraic geometry is true for a long time.

In the language of classical algebraic geometry, this is what I want to prove: Let $k$ be a universal field, and let $F \subseteq k$ be a subfield. Let $X \subseteq \mathbb{A}^n$ be an affine (not necessarily irreducible) variety which is defined over $F$. Let $Y$ be a closed subset of $X$. If $Y \cap X(F)$ is dense in $Y$, then $Y$ is defined over $F$. Here $X(F) = X \cap \mathbb{A}^n(F)$ is the set of $F$-rational points.

In the language of commutative algebra: Let $A$ be a finitely generated, reduced $k$-algebra, and let $A_0$ be an $F$-subalgebra of $A$ such that $A = k \otimes_F A_0$. Let $I \subseteq A$ be a radical ideal. Suppose that for every $a \in A$, not in $ I$, there exists a $k$-algebra homomorphism $\phi: A \rightarrow k$ such that $\phi(A_0) \subseteq F$, $I \subseteq \textrm{Ker } \phi$, and such that $\phi(a) \neq 0$. Then $I$ is the extension of an ideal from $A_0$.

This is 11.2.4(ii) in Springer, Linear Algebraic Groups. I do not think the explanation given in the textbook is satisfactory, and was hoping someone may know a complete proof or a reference. Also, I was wondering if there is a version of this result in modern algebraic geometry.

Answer 1

In scheme language, the reformulation of your question (which we'll see really does achieve what you want, and unsurprisingly plays just as essential a role in the scheme version of the theory of linear algebraic groups as it does in the older language) is to show that if $X$ is a scheme of finite type over a field $F$ and if $S$ is a subset of $X(F)$ then the Zariski closure $Z_S \subset X$ of $S$ equipped with its reduced structure is geometrically reduced (over $F$) with the further (crucial) property that for any extension field $F'/F$, the reduced closed subscheme $(Z_S)_{F'} \subset X_{F'}$ is the Zariski closure of $S \subset X(F) \subset X(F')$.

To see that this really settles what you want, if $Y \subset X_{F'}$ is a reduced closed subscheme that is the Zariski closure of a subset $S \subset X(F) \cap Y(F')$ (intersection inside $X(F')$) then clearly $(Z_S)_{F'} \subset Y$ as closed subschemes of $X_{F'}$ yet both are reduced and by design a dense subset $S \subset Y(F')$ lies inside $(Z_S)_{F'}$, so $Y = (Z_S)_{F'}$ is "defined over $F$" (relative to the $F$-descent $X$ of the ambient $X_{F'}$ containing $Y$).

The entire theory of linear algebraic groups is developed from scratch using the scheme framework from the beginning in at least two places: Milne's notes at http://www.jmilne.org/math/CourseNotes/iAG200.pdf, and the two sets of lecture notes at the top of http://math.stanford.edu/~conrad/249BW16Page/handouts.html. In the first of the two sets of lecture notes at the latter link, Proposition 3.2.4(ii) is exactly the above reformulated version of your question (used there for purposes exactly akin to the role played by the result in Springer's book).

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I recommend to forget about the Weil-style language of algebraic geometry used in Springer's book and always work with schemes. My experience as a student was that it is better when reading the standard books on linear algebraic groups to translate all proofs into scheme language and not do anything in the older language. Don't worry about trying to decipher whatever is unclear in Springer's proof of his 11.2.4.

One of the (many) great virtues of the scheme approach is that instead of making smooth constructions over $\overline{F}$ and then having to work hard to descend things to be smooth $F$-schemes (especially tricky at times when $F$ is not perfect, so we cannot use Galois descent), one makes all constructions directly over $F$ and instead the burden of work shifts to proving such $F$-schemes are smooth (using infinitesimal criteria, or flatness arguments, etc.). In particular, with the scheme approach nearly all of Chapter 11 of Springer's book becomes moot.

Springer's book is a wonderful resource over algebraically closed fields, especially for its treatment of the root-system aspects of the structure theory of reductive groups over algebraically closed fields, but you have to be extremely careful with anything done there involving geometric arguments over a ground field that is not algebraically closed; Borel's textbook is more reliable on such matters. A prototypical illustration of the subtle errors that creep in over imperfect ground fields due to the avoidance of schemes in Springer's book is seen in the proof of 15.2.3 in Springer's book: near the end it has to be proved that a certain morphism between $F$-varieties (= geometrically integral $F$-schemes) has finite fibers and one easily reduces to the case $F=F_s$. Springer seems to be misled by the density of $F$-points in an $F$-variety because he checks that fibers over $F$-points are finite and then declares the proof done, completely missing the possibility (which one sees vividly from the scheme perspective, though of course it can also be noticed in the Weil framework too) for imperfect $F$ that the reduced Zariski-closed subset of the target over which fibers are positive-dimensional may not be generically smooth and so may be non-empty but have no $F$-points at all!

Springer's method can't handle fibers not over $F$-points there, so that seems to be a genuine error in the proof of his 15.2.3. A correct proof of a stronger result than 15.2.3 (involving Tits' notion of "pseudo-completeness over $F$" that goes beyond Springer's notion called "($P_F$)") is given as Proposition C.1.6 in the book Pseudo-reductive Groups (using a corrected notion of pseudo-parabolicity since the definition in Springer's book is incorrect, as illustrated by the fact that both parts of the crucial 15.1.2 in his book are false for this notion, the error being in the first sentence when he says one can reduce to the case $\mathscr{R}_{u,F}(G)=1$; this step is fine with the definition of pseudo-parabolicity used in Pseudo-reductive Groups).