Generic zeros of a polynomial according to Lang's IGA

Question

The concept of generic zero of a prime ideal is defined in Lang's Introduction to Algebraic Geometry at page 27 (1972 ed.).

Because the setting is quite weird in my opinion (I have posted another question asking exactly about that), let's provide a concrete example which should do. We consider a prime ideal $\mathfrak p$ of $\mathbb Q[X_1,X_2]$. (I am trying to give concrete examples for the case when $\mathfrak p$ is generated by the irreducible polynomial $f(X_1,X_2)=X_1^2+X_2^2-1$). Then, a generic zero of $\mathfrak p$ is a point $(x_1,x_2)\in \mathbb C^2$ such that

$$ \mathbb Q[X_1,X_2]/\mathfrak p \cong \mathbb Q[x_1,x_2]. $$

Actually, a generic zero of $\mathfrak p$ is defined as a pair $(x_1,x_2)\in\mathbb C$ such that such that $\mathbb Q[X_1,X_2]/\mathfrak p \cong \mathbb Q[x_1,x_2]$ AND the following holds:

A polynomial $g\in\mathbb Q[X_1,X_2]$ belongs to $\mathfrak p$ if and only if $g(x_1,x_2)=0$.

However, I think the first part already implies the second one, does not it?

My goal is to make the link between this definition and the usual definition for schemes. The setting makes things more confusing in my opinion, but the idea is that if the variety is defined by $\mathfrak p$, then a generic point of $V$ is a generic zero of $\mathfrak p$. Here $V$ is regarded as a subset of $\mathbb C^n$ ($n=2$ in my example).

I expect the following:

If $W$ is the set of generic points of $V$, then $W$ is a sub-variety defined by a minimal prime ideal in $\mathbb Q[X_1,X_2]/\mathfrak p$.

In scheme notation, I am thinking about this:

$$ \text{The generic zeros of $\mathfrak p$ are the closed points of the generic point (in the sense of schemes) $\eta\in\operatorname{Spec} \mathbb Q[X_1,\dots,X_n]/\mathfrak p$.} \tag{$\ast$} $$

However, because I am dealing with $\mathbb Q$ and $\mathbb C$ at the same time, I do not know what to expect exactly. Also, I would expect generic zeros to be ''generic'' (dense), but that cannot happen if ($\ast$) holds.

So, to sum up: What is the relation between ''Lang's'' generic points and generic points in the modern sense?

Addendum. A similar definition can be found on pages 19--20 of the book Differential Algebra & Algebraic Groups, by Ellis Robert Kolchin (Academic Press, 1973).

EDIT (17/12/2021)

Thinking a bit more, I have realised that ($\ast$) cannot be true. An easy counterexample to that is provided by an Exercise of the Rising Sea (3.6.M.): Let $\mathfrak p$ the prime ideal of $\mathbb Q[X_1,X_2]$ generated by the polynomial

$$ f(X_1,X_2)=X_2-X_1^2. $$

Then,the generic point of $\operatorname{Spec} \mathbb Q[X_1,X_2]/\mathfrak p$ is precisely $\mathfrak p$, so that, according to ($\ast$), the variety $W=V=V(\mathfrak p)$, which is not true, since $(i,-1)\in V$ but it is clearly not a generic zero of $\mathfrak p$.

Another easy example example is when $n=1$ and $f=X^2+1$. Then $V$ consists of two points, namely $x=\pm i$. Since $\mathbb Q[X]/(X^2+1)\cong \mathbb Q(i)$, both points are generic, i.e. every point of the variety is a generic point. What does it mean?

Answer 1

Lang's concept of generic zero seems to be an ideals-first way of thinking about Weil's concept of generic points, which is defined in terms of specialisation (related to but different from the scheme-theoretic notion). So let me explain that first.

Let $K$ be a field and let $k$ be a subfield of $K$. (Actually, for the purposes of this definition, we only need $k$ to be a commutative ring and $K$ to be a $k$-algebra.) Let $x \in K^n$. A specialisation of $x$ over $k$ (in the sense of Weil) is any $x' \in K^n$ with the following property:

• For every polynomial $F$ in $n$ variables with coefficients in $k$, $F (x) = 0$ implies $F (x') = 0$.

In other words, $x'$ satisfies all the algebraic relations (with coefficients in $k$) that $x$ satisfies (and perhaps more besides).

A generic specialisation of $x$ over $k$ is any $x' \in K^n$ such that $x'$ is a specialisation of $x$ over $k$ and, vice versa, $x$ is a specialisation of $x'$ over $k$. In other words, $x$ and $x'$ satisfy exactly the same algebraic relations (with coefficients in $k$).

Notice that "$x'$ is a specialisation of $x$ over $k$" defines a preorder on $K^n$, i.e. $x$ is a specialisation of $x$ over $k$, and if $x''$ is a specialisation of $x'$ over $k$ and $x'$ is a specialisation of $x$ over $k$, then $x''$ is a specialisation of $x$ over $k$. I haven't read Lang, but I am sure that a generic zero is a zero that is maximal with respect to this preorder. In the case where $K$ is an algebraically closed field and $k$ is a subfield, if we are given a prime ideal $I$ of $k [X_1, \ldots, X_n]$, $x \in K^n$ is a generic zero of $I$ if and only if $I$ is the kernel of the evaluation homomorphism $k [X_1, \ldots, X_n] \to K$ sending each $X_i$ to $x_i$. (This uses the Nullstellensatz.) This implies that $k [x_1, \ldots, x_n] \cong K [X_1, \ldots, X_n] / I$ but the converse may not hold if you are not careful about what isomorphisms you allow. (There is a tendency for modern mathematicians to be careless about what they mean by $\cong$, but I digress.)

To connect this to the scheme-theoretic conception of generic points, we should restrict attention to the case where $K$ is an algebraically closed field and $k$ is a subfield. Then the Nullstellensatz is available, so there is a natural bijection between prime ideals of $k [X_1, \ldots, X_n]$ and equivalence classes of points of $K^n$ under the relation of generic specialisation over $k$. Another way of looking at it is to consider the projection morphism $\mathbb{A}^n_K \to \mathbb{A}^n_k$. We can identify $K^n$ with the closed points of $\mathbb{A}^n_K$. A generic zero of a prime ideal $\mathfrak{p}$ is then seen to be the same thing as a closed point of $\mathbb{A}^n_K$ whose image in $\mathbb{A}^n_k$ is $\mathfrak{p}$. It is in fact true that the set of generic zeros is dense in the set of (not necessarily generic) zeros. This is true even with the scheme-theoretic conception of generic points: $\{ \mathfrak{p} \}$ is a dense subset of $V (\mathfrak{p})$.

It can indeed happen that every zero is generic. That means the ideal is maximal.