INTRODUCTION TO ALGEBRAIC GEOMETRY, CLASS 1

RAVI VAKIL

Contents

1. Commutative algebra 2 2. Algebraic sets 2 3. Nullstellensatz (theorem of zeroes) 4

I’m going to start by telling you about this course, and about the field of algebraic geometry.

Goals:

• geometric insight

• concrete examples (geometric and arithmetic)

• hands on calculations (no fear of commutative algebra)

• no cohomology, flatness, differentials

Modern algebraic geometry lies somewhere between differential geometry, number theory, and topology. In a loose sense, it is polynomial equations, and sets defined by polynomial equations. This seems to be extremely narrow and low-tech, but it surprisingly ends up being extremely broad, powerful, and abstract.

Some of the philosophy — get at geometry via algebra, algebra via “pictures”. High school reference. Here’s a high-powered example of the link between geometry and arithmetic. x^{n} + y^{n} = z^{n}. Finite number of solutions for each n > 1: the Mordell Conjecture, Faltings’ Theorem. Vojta’s conjecture. Weil conjectures.

Give out handout of motivating problems.

This will be a tools course: examples and pictures, but with generality.

• fast-moving, but grounded by intuition

• exercises are important

• concepts really generalize, but become more abstract.

Date: September 9, 1999.

Objects: smooth varieties over ℂ (over k) varieties over ℂ (over k) schemes (stacks) We won’t be shy about schemes in this course. Administrative stuff. Give out handout. Go through it. Headcount. Theme: curves. Examples: ℤ and \overline{k}[t].

1. Commutative algebra

Don’t worry about it. See books on handout. Ideas you should know: ring (commutative, has 1), field, integral domain, has

quotient field, prime ideal, maximal ideal.

Sample problem (to appear on problem set): Let A be a (commutative) ring. An element a ∈ A is nilpotent (that is, a^{n} = 0

for some n > 0) if and only if a belongs to every prime ideal of A.

2. Algebraic sets

Throughout this course: k is a field. \overline{k} is an algebraically closed field. For now we work over \overline{k}. Feel free to think of this as ℂ for now. \overline{k}^{n} will be rewritten 𝔸^{n}(\overline{k}), affine n-space; we’ll often just write 𝔸^{n} when there’s

no confusion about the field. Coordinates x_{1} to x_{n}.

Algebraic geometry is about functions on the space, which form a ring. The only functions we will care about will be polynomials, i.e. \overline{k}[x_{1}, …, x_{n}]. We’ll eventually think of that ring as being the same thing as 𝔸^{n}.

We’ll next define subsets of 𝔸^{n} that we’ll be interested in. Because we’re being very restrictive, we won’t take any subsets, or even analytic subsets; we’ll only think of subsets that are in some sense defined in terms of polynomials.

Let S be a set of polynomials, and define V (S) to be the locus where these polynomials are zero. (“Vanishing set”.) Definition: Any subset of 𝔸^{n}(\overline{k}) of the form V (S) is an algebraic set.

Exercise (to appear on problem set): prove that the points of the form (t, t^{2}, t^{3}) in 𝔸^{3} form an algebraic set. In other words, find a set of functions that vanish on these points, and no others.

Example/definition: hypersurface, defined by 1 polynomial.

Facts.

• If I = (S), then V (I) = V (S). So we usually will care only about ideals. Hence: subsets of \overline{k}[x_{1}, ..., x_{n}] give us subsets of 𝔸^{n}; specifically, ideals give us algebraic sets

• V (∪I_{a}) = ∩V (I_{a}) (Say it in english.)

• I ⊂ J, then V (I) ⊃ V (J)

• V (F G) = V (F) ∪ V (G)

Note: Points are algebraic. Finite unions of points are algebraic.

Definition. A radical of an ideal I ⊂ R, denoted \sqrt{I}, is defined by

\sqrt{I} = \{r ∈ R|r^{n} ∈ I for some n\}.

Exercise. Show that \sqrt{I} is an ideal.

Definition. An ideal I is radical if I = \sqrt{I}.

Claim. V (\sqrt{I}) = V (I). (Explain why.)

Conversely, subsets of 𝔸^{n} give us a subset of \overline{k}[x_{1}, ..., x_{n}] For each subset X, let I(X) be those polynomials vanishing on X.

Claim. I(X) is a radical ideal. (Explain.)

Facts. If X ⊂ Y, then I(X) ⊃ I(Y). I(∅) = \overline{k}[x_{1}, ..., x_{n}]. I(𝔸^{n}) = (0).

Question. What’s I((a_{1}, ..., a_{n}))?

(Discuss.)

Notice: ideal is maximal. Quotient is field. Quotient map can be interpreted as “value of function at that point”.

Exercise. (a) Let V be an algebraic set in 𝔸^{n}, P a point not in V. Show that there is a polynomial F in \overline{k}[x_{1}, ..., x_{n}] such that F (Q) = 0 for all Q in V, but F (P) = 1. Hint: I(V) ≠ I(V ∪ P).

(b) Let \{P_{1}, ..., P_{2}\} be a finite set of points in 𝔸^{n}(\overline{k}). Show that there are polynomials F_{1}, ..., F_{r} ∈ \overline{k}[x_{1}, ..., x_{n}] such that F_{i}(P_{j}) = 0 if i ≠ j, and F_{i}(P_{i}) = 1.

Exercise. Show that for any ideal I in \overline{k}[x_{1}, ..., x_{n}], V (I) = V (\sqrt{I}), and \sqrt{I} is contained in I(V (I)).

3. Nullstellensatz (theorem of zeroes)

Earlier, we had: algebraic sets → radical ideals and ideals → algebraic sets.

This theorem makes an equivalence. In the literature, the word “nullstellensatz” is used to apply to a large number of results, not all of them equivalent.

Nullstellensatz Version 1. Suppose F_{1}, …, F_{m} ∈ \overline{k}[x_{1}, …, x_{n}]. If the ideal (F_{1}, …, F_{m}) ≠ (1) = \overline{k}[x_{1}, …, x_{n}] then the system of equations F_{1} = ⋯ = F_{m} = 0 has a solution in \overline{k}.

Proof next day. (There is a better version for fields that are not necessarily algebraically closed, but we’re not worrying about that right now.)

Nullstellensatz Version 2. Supopse 𝔪 is a maximal ideal of \overline{k}[x_{1}, …, x_{n}]. Then

𝔪 = (x_{1} − a_{1}, …, x_{n} − a_{n})

for some a_{1}, …, a_{n} ∈ \overline{k}.

Show that this is equivalent to version 1, modulo fact that ideals are finitely generated.

Nullstellensatz Version 3 (sometimes called the “Weak Nullstellensatz”). If I is a proper ideal in \overline{k}[x_{1}, ..., x_{n}], then V (I) is nonempty. (From Version 2.)

Nullstellensatz Version 4. Let I be an ideal in \overline{k}[x_{1}, ..., x_{n}]. Then I(V (I)) = \sqrt{I}. Equivalently: Radical ideals are in 1-1 correspondence with algebraic sets: If I is a radical ideal in \overline{k}[x_{1}, ..., x_{n}] then I(V (I)) = I. So there is a 1-1 correspondence between radical ideals and algebraic sets.

Nullstellensatz Version 5. A radical ideal of \overline{k}[x_{1}, …, x_{n}] is the intersection of the maximal ideals containing it. This is the geometric rewording of 4. By version 4, a radical ideal is I(X) for some algebraic set X. Functions vanishing on X are precisely those functions vanishing on all the points of X.

Nullstellensatz Version 6. If F_{1}, ..., F_{r}, G are in \overline{k}[x_{1}, ..., x_{n}], and G vanishes wherever F_{1}, ..., F_{r} vanish, then there is an equation G^{N} = A_{1}F_{1} + ... + A_{r}F_{r} for some N > 0 and some A_{i} in \overline{k}[x_{1}, ..., x_{n}].

This has a cute proof, with a useful trick in it.

Proof. The case G = 0 is obvious, so assume G ≠ 0. Introduce a new variable U, and consider the polynomials

F_{1}, …, F_{m}, and UG − 1 ∈ \overline{k}[x_{1}, …, x_{n}, U].

They have no common solutions in \overline{k}, so by Version 1 they generate the unit ideal, so there are polynomials P_{1}, …, P_{m}, Q ∈ \overline{k}[x_{1}, …, x_{n}, U] such that

P_{1}F_{1} + ⋯ + P_{m}F_{m} + Q(UG − 1) = 1.

Now set U = 1/G in this formula, and multiply by some large power G^{N} of G to clear denominators. Then the right side is G^{N}, and the left side is in (F_{1}, …, F_{m}).