# Intuition behind generic points in a scheme

In a scheme, each point is a generic point of its closure. In particular each closed point is a generic point of itself (the set containing it only), but that's perhaps of little interest. A point that's not closed, is probably more interesting, and there are no such thing in ordinary varieties.

What I have been wondering about is that there must be a good reason they are called generic points. Here are what I have got so far:

• A non-closed generic point is not closed, so it cannot be cut out from the scheme by any polynomial equations in any affine patch, and thus it does not posses any extra algebraic property that's not shared by others.
• A non-closed generic point is not a specialization of the scheme.

Are these correct? If not, what's the right intuition? Also, can the following statement be make more precise using the language of generic points?

• A degree $n$ and a generic degree $m$ algebraic curves intersect at $n \cdot m$ distinct points in $\mathbb{P}^2$ (the planar Bezout's theorem)
• Common solutions of $n$ polynomial systems in $n$ variables with generic complex coefficients in $\mathbb{C}^\ast$ are all isolated. (corollary of the Cheater's homotopy theorem)
• The number of common isolated solutions of $n$ polynomial systems in $n$ variables with generic complex coefficients in $\mathbb{C}^\ast$ equals the mixed volume of the Newton polytopes of the system. (Bernshtein's theorem)
• Generic points on a nonreduced scheme have isosingular structure (i.e., at all such points the local ring fail to be reduced in exactly the same way).

In each case I am familiar with their original meaning of the word generic, but I'm wondering if we can state the genericity conditions using the concept of generic points of schemes.

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There are two meanings of a property being generically true: (1) over the generic point and (2) over a nonempty open subset. One important thing about generic points is that in many cases (1) implies (2). See also mathoverflow.net/questions/28496/… –  user2035 Feb 24 '12 at 6:30
To apply generic points to the first one, we take the ring of projective space, or at least the cone on it: $\mathbb C[x,y,z]$, and then quotient out by a degree-$n$ and degree-$m$ homogeneous polynomial. What are the coefficients? The degree-$n$ polynomial is any one you specify, but the coefficients of the degree $m$ polynomial are new invariants. The scheme of possible polynomials is an affine space. The generic point corresponds to looking at things over the field generated by those invariants. –  Will Sawin Feb 24 '12 at 6:41
I'm not sure you actually will get a nice answer that looks like $n\cdot m$ points or the cone on it, though. –  Will Sawin Feb 24 '12 at 6:42
That is indeed the point of view I would like to explore. Could you give me more detail? How do you form such a "scheme of possible polynomials"? I can see that a degree-$m$ polynomial is determined by a list of coefficients, and the space of such polynomials actually form a projective space since scaling of coefficients does not change the polynomial equation. How does this connect to the scheme you were talking about? Thanks. (P.S., your comment is more like an answer than a comment) –  ssquidd Feb 24 '12 at 16:12
I decided to delete my answer since it may not be getting to the heart of you are after. Although, and I don't mean to be impolite, I'm not quite sure what that is. –  Donu Arapura Feb 24 '12 at 17:09
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My favorite view on generic points is found in Mumford's book: Complex Projective Varieties, on page 2. It goes as roughly as follows:

Definition: Let $k \subset \mathbb{C}$ be a subfield of the complex numbers and $V$ an affine complex variety. A point $x \in V$ is $k$-generic if every polynomial with values in $k$ that vanishes on $x$, vanishes on all of $V$.

Proposition: If $\mathbb{C}/k$ has infinite transcendental degree, then every variety $V$ has a $k$-generic point.

Proof: Extend $k$ by all coefficients of a finite set of equations for $V$. Note that $\mathbb{C}/k$ has still infinite transcendental degree. But now $V$ becomes a variety over $k$ in a canonical way. The function field $L$ of $V$ is an $k$-extension of finite transcendental degree, and therefore can be embedded into $\mathbb{C}$. The images of the coordinate function $X_i \in L$ in $\mathbb{C}$ give the coordinates of a $k$-generic point.

For the relation to the generic point $\eta \in V$ from scheme theory note the following:

• If we define a $k$-Zariski topology on $V(\mathbb{C})$ defined by polynomials over $k$, a $k$-generic point $x$ will be non-closed with closure $V$.
• The field $K(x)/k$ generated by the coefficients of $x$, is canonically isomorphic to the function field $L=K(V)$, which is the residue field of the generic point $K(\eta)$.

From the abstract perspective the function field $L=K(\eta)$ is as good as $K(x)$, if not better, because it does not depend on choices. Moreover, all $k$-linear algebraic operations cant tell a difference between $\eta$ and $x$.

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As far as Bezout's theorem stating that a generic poynomial of degree $n$ have $n$ roots, this should be understood as follows: polynomials in a Zariski open subset of the set degree $n$-polynomials have exactly $n$ roots. The complement of this open set is Zariski closed so it is cut-out by a finite number of algebraic equations. For example for degree $2$ polynomials $az^2+bz+c$, the condition $b^2- 4ac\neq 0$ describes a Zariski open set of quadratic polynomials with two distinct roots. Note that measure theoretically, Zariski closed subsets have (Lebesgue) measure zero, so that, with "probability" $1$, a degree $n$-polynomial has exactly $n$-roots.