Lecture Notes 18.721 Algebraic Geometry

Michael Artin

September 15, 2009

Lecture 1

Introductory Remarks Scalars will for the most part be taken in C. The emphasis

will therefore be on affine spaces of the form An = Cn. In particular we will study

affine varieties, subsets of An corresponding to a locus of the form

f1(x) = . . . = fr(x) = 0

with fi(x) ∈ C [x1, . . . , xn].

Example 1.

• Point: (a1, a2) in An such that x1 − a1 = 0 and x2 − a2 = 0.

• Line: a1x1 + a2x2 + a3 = 0

• Curve: f(x1, x2) = 0 f a polynomial.

• SL2 : In A4 x11x22 − x12x21 = 1

Example 2. Conics: ax2 + bxy + cy2 + dx+ ey + f = 0

transformation ellipse hyperbola parabola

isometry (a,b > 0) ax2 + by2 = 1 ax2 − by2 = 1 ax2 = y

affine transformation (GLn, translation) x2 + y2 = 1 x2 − y2 = 1 x2 = y

complex numbers x2 + y2 = 1 x2 + y2 = 1 x2 = y

projection transformation x2 = y x2 = y x2 = y

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Projective Line and Plane The projective line P is defined on A − (0, 0) using

the rule (u, v) = (λu, λv), that is, taking vectors modulo scaling. Normalizing

(u, v)→ (u

v, 1 ) if v 6= 0

(u, v)→ (1, 0) if v = 0

we find that P can be identified with A ∪ {pt. at∞}. The projective plane P2 is

defined with the same rule but on A3 − (0, 0, 0). Normalizing

(u, v, w)→ (u

w,v

w, 1) if w 6= 0

(u, v, w)→ (u, v, 0)→ A if w = 0

we find that P2 can be identified with A2 ∪ P or equivalently A2 ∪ A ∪ {pt. at∞}.

Transformations of the general linear group operate on projective space– GL2 on Pand GL3 on P2. In particular(

a b

c d

)·

(u

v

)=

(au+ bv

cu+ dv

)

t =u

v→ au+ bv

cu+ dv=at+ b

ct+ d

showing that the operation of GL2 reduces to fractional linear transformations. Notice

that a matrix M and δ ·M for δ ∈ C give the same transformation. This prompts

the following definition,

Definition 3. PGLn = GLn / {scalar matrices}

For GLn begin by setting uv

= x and vw

= y. Using the transformation 1

1

1

uvw

=

wuv

2

and the fact that x2 − y = u2

w2 − vw

u2 − vw → w2 − uv → 1− xy.

Since any two lines intersect on the line at infinity the two arcs of the parabola are

bent together forming an ellipse. So using projective transformations there exist only

on class of conics.

Cubic Polynomials For a cubic polynomial there are ten coefficients corresponding

to 1, x3, y3, x2y, xy2, xy, y, x, x2, y2 and so, setting equal to zero, nine parameters.

There are nine coefficients for GLn and 9− 8 parameters. And so

Corollary 4. Cubics up to projection depend on ≥ 1 parameters.

Morphisms A morphism between affine varieties should be given by polynomial

functions. Let X ⊂ An and y ⊂ An be varieties. A mapping φ : X → Y is called

a polynomialmap if there are polynomials T1, . . . , Tm ∈ C [X1, . . . , Xn] such that

φ(a1, . . . , an) = (T1(a1, . . . , an), . . . , Tm(a1, . . . , an)) for all (a1, . . . , an) ∈ X. Two

polynomials f and g represent the the same map on X iff f − g ≡ 0 on X. In this

way the polynomial maps can be associated with the ideals,

Definition 5. For an algebraic set X let I(X) = {f | f ≡ 0 on X} be an ideal of

[X1, . . . , Xn]. The affine coordinate ring of X is defined as the quotient Γ(X) =

[X1, . . . , Xn] /I.

The correspondence is made precise in the following,

Proposition 6. Let X ⊂ An and Y ⊂ Am be affine varieties. There is a natural

one-to-one correspondence between the polynomial maps φ : Y 7→ X and the homo-

morphisms φ : Γ(X) 7→ Γ(Y ).

pf. See [Ful] 2.2�

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Lecture 2

What is Geometry? The usual notion of geometry is the study of points, lines,

curves, etc. One has the concept of incidence, which suggests a sort of inclusion, for

instance p ∈ L, M ∈ SL2 and Y ⊂ X. The tangent line to a curve can be made

to obey an incidence relation, but this requires some fudging. Moreover one has the

concept of bitangent lines, that is, lines which intersect a curve in preciously two

places. A result in this direction is the fact that a quartic curve has twenty eight

bitangents.

Basic Notions Let X denote an affine variety in An, that is, the locus correspond-

ing to a set of polynomials. Let I = {f | f ≡ 0 on X} be the associated ideal. By

the restriction to X one means the quotient C[X1, . . . , Xn]/I.

The point p = (a1, a2) corresponds to the ideal J = (X1 − a1, X2 − a2). The line

l = b1x1 + b2x2 + c = 0

corresponds to the principal ideal (l) = I. To say that the point p lies on the line

means that l = b1a1 + b2a2 + c = 0 or more precisely that

l = b1(x1 − a1) + b2(x2 − a2) + c = 0.

Let A = C[X]/I and B = C[X]/J ' C. Since p is incident to the line it follows that

I ⊂ J and that A � B.

The notion of a morphism introduced in Lecture 1 is analogous to the incidence

relation and generalize the previous example. Let V , W denote two varieties and

A = C[X]/I(V ), B = C[X]/I(W ) their respective coordinate rings. Given a polyno-

mial map s : V 7→ W one gets a homomorphism 1 φ : B 7→ A given by φ(f) = fs,

1Here and in what follows homomorphism will refer to an algebra homomorphism, namely, onethat restricts to the identity on C

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Vs−→ Wfs

↘f

↓C

(FuncW ) ←− (FuncV )

∪ ∪B

morphism←− A

And so composing with the canonical homomorphism π : C[X] 7→ A,

C[X]π

↓φπ

↘A

φ−→ B

B-valued points For an ideal I ∈ A, let A = C[X]/I. Given a homomorphism

ψ : A 7→ B, where B is an arbitrary ring, with xi 7→ ui ∈ B, it follows that f(u) = 0

for f ∈ I. Note that by the Basis Theorem [Ful] 1.4 the ideal I is finitely generated

since it lies in C[X] with C obviously Noetherian. So if I is generated by {f1, . . . , fn}then the ui’s provide B-valued solutions to the system of equations.

Example 7.

Let Y = A, B = C[t] and Ys→ X. Assume the map corresponds to the zeros of {f},

that is, it assigns xi → ui ∈ B where the polynomial is such that f(x(t)) ≡ 0. So for

instance

X : f = 0 in C2 y(t) = t3 − t

f : y2 − x3 − x2 x(t) = t2 − 1

then it follows that f(x(t), y(t)) = 0. Selected values of the locus are,

+ -1 0 +1 > 1 < −1

x 0 -1 0 + +

y 0 0 0 + -

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A way of going about solving the question of whether the curve f : y2 − x3 − x2 − ccan be parameterized in this was is by counting coefficient. Knowing that the leading

terms have to cancel one obtains something of the form,

x(t) = t2k + (2k terms )

y(t) = t3k + (3k terms ).

Lecture 3

Weak Nullstellensatz Some notation before proceeding. If X denotes a subset

of An, then we let I(x) = {f | f ≡ 0 on X}. If I is a subset of C[X] then V (I) =

{p | f(x) = 0 ∀ f ∈ I}. It follows immediately from the definitions that

I(V (I)) ⊃ I V (I(X)) ⊃ X

with equality holding in the latter if X is an affine variety.

Theorem 8 (Weak Nullstellensatz). The maximal ideals of C[X1, . . . , Xm] are in

bijective correspondence with points of An. The correspondence is given by

(X1 − a1, . . . , Xn − an) = Ker(C[X]→ C)bij←→ p = (a1, . . . , an).

pf. See [Art] 11.10.1�

Note the fact that C is algebraically closed is crucial in the result. For instance while

(x2 + 1) in R[x] is a maximal ideal it has no solutions in R. Recall the following

statement about rings,

Theorem 9 (Correspondence Theorem). If Rφ� S with Kerφ = I then

( Maximal ideals of R that contain I)bij←→ ( Maximal ideals of S)

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So if A = C[X]/I for some ideal I then the maximal ideals of C[X] containing I

are bijective correspondence with the maximal ideals of A. Suppose I = (f1, . . . , fn)

when does a maximal ideal MP corresponding to the point p = (a1, . . . , an) contain

I? This happesn if and only if f1, . . . , fn ∈MP , that is, fi(p) = 0 for i = 1, . . . , n. So

in other words if and only if p ∈ V (I). We have

MaxA ↪→ Max C[X]

l lX ⊂ An

So as a result of the Weak Nullstellensatz we have the following

Corollary 10. For an ideal I, V (I) = ∅ ↔ 0 = C[X1, . . . , Xm]/I, that is, I = (1).

Inverting Elements Suppose g ∈ C[X] and we want to invert g. We take C[X;Y ]

and mod out by gy−1, that is, we let C[X][g−1] = C[X;Y ]/(gy−1). Let Y = V (gy−1)

where y ⊂ An+1(X,Y ). As shown there is a bijective correspondence between maximal

ideals of C[X, g−1] and Y . Now what does it mean for a point (a1, . . . , an, b) = (a; b)

to lie in Y . We need g(a)b = 1 so there are two cases

g(a) 6= 0→ ∃!b

g(a) = 0→ b does not exist

So we have a bijective correspondence between Y and An − V (g). Suppose we take

A = C[X]/I with I = (f1, . . . , fr) and we want to invert the residue g of g. We

take A[Y ]/(gy − 1) = C[X, Y ]/(f1, . . . , fr; gy − 1) and for Y = V (f1, . . . , fr; gy − 1)

similarly ask what it means for a point (a1, . . . , an, b) = (a; b) to be in Y . We need

fi(a) = 0 ∀i and g(a)b = 1. The same two cases as before apply to solutions of the

latter. So if V (f1, . . . , fr) ∩ V (g) = C then Y = X − C.

Strong Nullstellensatz We use the above facts to strengthen the Nullstellensatz

following a method of Rabinowitsch.

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Theorem 11 (Strong Nullstellensatz). Let I = (f1, . . . , fr) denote an ideal of C[X1 . . . Xn].

For g ∈ C[X1 . . . Xn], g ≡ 0 on V (I) implies that gn ∈ I for some n, in other words

gn =∑hifi .

pf. Since V (I) ⊂ V (g), it follows that V (I)−(V (I)∩V (g)) = ∅. As shown this implies

that C[X1, . . . , Xm]/(f1, . . . , fr, gy−1) = 0, or in other words (f, gy−1) = (1). Hence

H1(x, y)f1(x) + . . .+Hr(x, y)fr(x) +K(x, y)(g(x)y − 1) = 1

Since gy ≡ 1 mod(gy− 1), multiplying by a suitable power of g removes the y depen-

dence modulo (gy − 1) in the above statement, that is,

h1(x)f1(x) + . . .+ hr(x)fr(x) ≡ gn mod(gy − 1)

Therefore gn ∈ (f, gy − 1).�

Example 12. Let f1 = x21, f2 = x2

2 − x31 and g = x2. Since f1, f2 vanish at the

origin, it follows that a power of g can be expressed as a combination of f1 and f2.

In particular g2 = f2 + x1f1.

We review the following definition, before restating the result more succintly.

Definition 13. Let A be a ring and I an ideal. The radical of I is defined as rad(I) =

{r | rn ∈ I for some n}.

So it follows immdediately from theorem that

Corollary 14. If I = (f1, . . . , fr) is an ideal C[X1 . . . Xn] then I(V (I)) = rad(I).

References

[Ful] Fulton, W., Algebraic Curves: An Introduction to Algebraic Geometry,(2008).

[Art] Artin, M., Algebra, 2nd ed (2008)

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