Generic zeros of a polynomial according to Lang's IGA

Solution 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.