1 Affine algebraic varieties

MATH32062 Algebraic Geometry

1 Affine algebraic varieties

1.1 Definition of affine algebraic varieties

We want to define an algebraic variety as the solution set of a collection of polynomial equations, or equivalently, a set where every element of a set of polynomials vanishes. Then every combination of those polynomials vanishes on the set, so we can use the ideal generated by those polynomials to define the algebraic variety. This motivates the definition below. Using ideals instead of concrete polynomials also turns out to be more useful for proving most of the theorems about algebraic varieties.

In the following, K denotes an arbitrary field, unless specified otherwise.

Definition. An affine algebraic variety is a set of the form

V(I) = {(a

, a

, . . . , a

) ∈ K

| f (a

, a

, . . . , a

) = 0, ∀f ∈ I}, where I is an ideal I / K[x

, x

, . . . , x

].

It is clear from the definition that if I

⊆ I

, then V(I

) ⊇ V(I

), but I

$ I

does not imply V(I

) % V(I

), see examples 7 and 8 below. Conversely, V(I

) % V(I

) does not imply I

$ I

. (Try to think of an example where I

6⊂ I

, but V(I

) ⊃ V(I

).)

Examples:

1. If I = {0}, then V(I) = K

.

2. If I = K[x

, x

, . . . , x

], then V(I) = ∅.

3. Let I = hx

+ y

− 1i / R[x, y], then V(I) is the circle of radius 1 centred at the origin defined by the equation x

+ y

= 1 in R

.

4. Let f

, f

, . . . , f

be polynomials of degree 1 in x

, x

, . . . x

, and let I = hf

, f

, . . . f

i be the ideal generated by them. Then V(I) is the set of solutions of the system of linear equations f

= f

= . . . = f

= 0. Such a variety is called an affine subspace of K

. This, in particular, includes sets containing a single point. If P = (b

, b

, . . . , b

) ∈ K

, then the ideal I = hx

− b

, x

− b

, . . . x

− b

i defines the variety V(I) = {P }.

5. Let I = hy

− x

i / R[x, y], then V(I) is the cuspidal cubic curve, also

called Neile’s semicubical parabola after William Neile, who computed its

arc length in 1657. It is defined by the equation y

= x

in R

, shown below

on the left.

-0.5 0.5 1.0 1.5 2.0 2.5

-4 -2 2 4

-1 1 2

-4 -2 2 4

6. Let I = hy

− x

− x

i / R[x, y], then V(I) is the nodal cubic curve defined by the equation y

= x

+ x

in R

, shown above on the right.

7. If I = hx, yi / R[x, y], then V(I) = {(0, 0)} ⊂ R

. 8. If I = hx

, y

i / R[x, y], then V(I) = {(0, 0)} ⊂ R

.

9. If I = hx

+ y

+ z

− 1, x

− x + y

i / R[x, y, z], then V(I) ⊂ R

is a “figure of eight” curve, shown below.

-2 -1

0 1

2 -2

-1 0

1 2

-2 -1

0 1 2

10. The special linear group SL(n, R) consisting of all n × n real matrices

with determinant 1 is also an affine algebraic variety. The set of all n × n

real matrices is R

, the determinant is a polynomial in the entries of the matrix, therefore SL(n, R) = V(hdet(A) − 1i).

Note: Examples 7 and 8 show that different ideals can define the same variety.

The exact nature of the correspondence between ideals and varieties will be investigated later in Section 1.3.

What do affine algebraic varieties look like? There is not a way of telling whether a set is an affine algebraic variety or not just by looking at it. Her- wig Hauser’s gallery of algebraic surfaces https://homepage.univie.ac.

at/herwig.hauser/bildergalerie/gallery.html contains a large number of algebraic surfaces and shows that affine algebraic varieties can have lots of different shapes.

Proposition 1.1 Let K be a field. Let f

, f

, . . . , f

∈ K[x

, x

, . . . , x

] and let I = hf

, f

, . . . , f

i. Then

V(I) = {(a

, a

, . . . , a

) ∈ K

| f

(a

, a

, . . . , a

) = 0, ∀i, 1 ≤ i ≤ m}.

Proof. f

∈ I for each i, 1 ≤ i ≤ m, so if (a

, a

, . . . , a

) ∈ V(I), then f

(a

, a

, . . . , a

) = 0 for every i, 1 ≤ i ≤ m.

Conversely, assume that f

(a

, a

, . . . , a

) = 0 for every i, 1 ≤ i ≤ m. Any f ∈ I can be written as f =

m

P

i=1

f

g

for suitable g

∈ K[x

, x

, . . . , x

], 1 ≤ i ≤ m, therefore

f (a

, a

, . . . , a

) =

m

X

i=1

f

(a

, a

, . . . , a

)g

(a

, a

, . . . , a

) = 0, so (a

, a

, . . . , a

) ∈ V(I).

Definition. A ring R is Noetherian if and only if all of its ideals can be generated by finitely many elements. (Equivalently, R is Noetherian if and only if for any increasing sequence of ideals I

⊆ I

⊆ I

⊆ . . . ⊆ I

⊆ I

. . . there exists an N such that I

= I

for all n ≥ N .)

Theorem 1.2 (Hilbert Basis Theorem) (Not examinable) If K is a field, then the polynomial ring K[x

, x

, . . . , x

] is Noetherian for any n ≥ 0.

Note: The Hilbert Basis Theorem justifies why we only considered ideals generated by a finite set of polynomials in Proposition 1.1. The combination of Proposition 1.1 and Theorem 1.2 shows that any affine algebraic variety is the set of solutions of finitely many polynomial equations.

The proposition below describes some methods which can be used to con-

struct further affine algebraic varieties from known ones.

Proposition 1.3 (i) Let V

= V(I

), V

= V(I

), . . . , V

= V(I

) be affine algebraic varieties in K

. Then

V

∪ V

∪ . . . ∪ V

= V(I

∩ I

∩ . . . ∩ I

) = V(I

I

. . . I

) is also an affine algebraic variety.

(ii) Let V

= V(I

), α ∈ A be affine algebraic varieties in K

. Then

\

α∈A

V

= V

 X

α∈A

I



is also an affine algebraic variety.

(iii) Let V

= V(I

) ⊆ K

, V

= V(I

) ⊆ K

be affine algebraic varieties.

Then V

× V

⊆ K

× K

∼ = K

is also an affine algebraic variety.

Proof. (i) Let’s assume first that k = 2. Let P ∈ V

∪ V

and let f ∈ I

∩ I

. If P ∈ V

, then f (P ) = 0 because f ∈ I

, if P ∈ V

, then f (P ) = 0 because f ∈ I

. Therefore in either case we have f (P ) = 0, so P ∈ V(I

∩ I

), V

∪ V

⊆ V(I

∩ I

). We also have V(I

∩ I

) ⊆ V(I

I

), therefore

V

∪ V

⊆ V(I

∩ I

) ⊆ V(I

I

) (1.1) (I

I

⊆ I

∩ I

, since all the elements of the form i

i

, i

∈ I

, i

∈ I

are contained in both I

and I

by the definition of an ideal, and if all the generators of I

I

are elements of I

∩ I

, then necessarily I

I

⊆ I

∩ I

. This means that the elements of V(I

∩ I

) have to satisfy all the polynomials in I

I

, therefore V(I

∩ I

) ⊆ V(I

I

).)

Let now P ∈ K

\ (V

∪ V

). Then P / ∈ V

, so there exists f

∈ I

such that f

(P ) 6= 0. Similarly, P / ∈ V

, so there exists f

∈ I

such that f

(P ) 6= 0.

Then f

f

∈ I

I

and (f

f

)(P ) = f

(P )f

(P ) 6= 0, therefore P / ∈ V(I

I

).

This implies V(I

I

) ⊆ V

∪ V

. By combining this with (1.1), we obtain V

∪ V

= V(I

∩ I

) = V(I

I

).

For k > 2, use induction on k.

(ii) Assume that P ∈ T

α∈A

V

. Then P ∈ V

= V(I

) for every α ∈ A, so f

(P ) = 0 for every α ∈ A and f

∈ I

. Any f ∈ P

α∈A

I

can be written as f = P

α∈A

f

with f

∈ I

for every α ∈ A and f

= 0 for all but finitely many α. Hence f (P ) = P

α∈A

f

(P ) = P

α∈A

0 = 0, so P ∈ V

 P

α∈A

I



. This implies

\

α∈A

V

⊆ V

 X

α∈A

I



. (1.2)

Assume now that P ∈ K

\ T

α∈A

V

. Then there exists α

∈ A such that P / ∈ V

, therefore there exists f ∈ I

such that f (P ) 6= 0. Now f ∈ P

α∈A

I

, therefore P / ∈ V( P

α∈A

I

). This implies that

V

 X

α∈A

I



⊆ \

α∈A

V

.

By combining this with (1.2), we obtain V

 X

α∈A

I



= \

α∈A

V

.

(iii) Let x

, x

, . . . , x

be co-ordinates on K

, and y

, y

, . . . , y

co- ordinates on K

. Let J

= hI

i / K[x

, x

, . . . , x

, y

, y

, . . . , y

]. (I

is an ideal in K[x

, . . . , x

], J

is the ideal generated by the elements of I

in the bigger ring K[x

, x

, . . . , x

, y

, y

, . . . , y

].) We claim that V(J

) = V

× K

. Let P = (a

, a

, . . . , a

, b

, b

, . . . , b

) ∈ V(J

). f (P ) = 0 for every f ∈ I

, since I

⊂ J

. As f only involves the variables x

, x

, . . . , x

,

0 = f (P ) = f (a

, a

, . . . , a

, b

, b

, . . . , b

) = f (a

, a

, . . . , a

).

Hence (a

, a

, . . . , a

) ∈ V(I

) = V

, so P ∈ V

× K

, therefore V(J

) ⊆ V

× K

.

Let now P = (a

, a

, . . . , a

, b

, b

, . . . , b

) ∈ V

× K

. Let f ∈ J

. f can be written as f =

r

P

i=1

f

g

, where f

∈ I

and g

∈ K[x

, x

, . . . , x

, y

, y

, . . . , y

].

Now f

(P ) = f

(a

, a

, . . . , a

, b

, b

, . . . , b

) = f

(a

, a

, . . . , a

) = 0. (The second equality holds because f

only involves on the variables x

, x

, . . . , x

, and the last equality holds because P ∈ V

× K

.) Therefore f (P ) =

r

P

i=1

f

(P )g

(P ) = 0, so P ∈ V(J

). This means V

×K

⊆ V(J

), by combining this with V(J

) ⊆ V

× K

proved above, we obtain V(J

) = V

× K

. Similarly, if we define J

= hI

i/K[x

, x

, . . . , x

, y

, y

, . . . , y

], then V(J

) = K

× V

.

Hence

V

× V

= (V

× K

) ∩ (K

× V

) = V(J

) ∩ V(J

) = V(J

+ J

)

by (ii).

In practical calculations, the ideals defining the varieties are given by sets of generators.

To find a generating set for an ideal defining the union of the varieties, take all the possible products of the generators, one factor from each ideal.

To find a generating set for an ideal defining the intersection of the varieties, take the union of the generating sets of the ideals.

To find a generating set for an ideal defining the Cartesian product of the varieties, take the union of the generating sets of the ideals, but remember that this ideal will be in a different ring.

Examples:

1. Find an ideal I / R[x, y] such that V(I) = {(1, 1), (2, −3)}.

Let I

= hx − 1, y − 1i and I

= hx − 2, y + 3i. Then V(I

) = {(1, 1)} and V(I

) = {(2, −3)}. By Proposition 2.3 (i),

I = I

I

= h(x − 1)(x − 2), (x − 1)(y + 3), (y − 1)(x − 2), (y − 1)(y + 3)i is a suitable ideal.

2. Let V

= V(hx

+ x

− 1i) ⊂ R

and let V

= V(hy

− y

i) ⊂ R

. Then V

× V

= Vhx

+ x

− 1, y

− y

i ⊂ R

is a cylinder.

Definition. Let K be a field and let V ⊆ K

be an affine algebraic variety.

W is a subvariety of V if and only if W ⊆ V and W is also an affine algebraic variety.

Remark. By Proposition 1.3 (i) and (ii), affine algebraic varieties behave

like closed sets in a metric space. This motivates the introduction of the

Zariski topology on an affine algebraic variety, in which the closed sets are the

subvarieties and the open sets are the complements of subvarieties. There

are, however, certain differences from metric spaces, for example, any two

non-empty Zariski open sets in R

have a non-empty intersection.

1.2 Affine spaces

Affine spaces are the simplest algebraic varieties, they are the solutions sets of system of linear equations, i. e., they are defined by ideals generated by degree 1 polynomials.

Definition. Let K be a field, and let V be a finite dimension vector space over K. U ⊆ V is an affine subspace of V if and only if U = ∅ or U = u

+ W = {u

+ w | w ∈ W }, where u

∈ U and W is a linear subspace of V . (Linear subspace or vector subspace of a vector space means subspace as defined in MATH10202 or MATH10212.)

Examples: K

as a subspace of itself, ∅, any set containing a single point are all affine subspaces of K

. The line y = 2x + 5 is an affine subspace of R

, in this case we can take W = span{(1, 2)} and u

= (0, 5).

Proposition 1.4 Let U = u

+W be a non-empty affine subspace in a vector space V . Then U = u + W for any u ∈ U , in other words, u

can be chosen to be an arbitrary element of U , but W is uniquely determined by U . Proof. Let u ∈ U , then u = u

+ (u − u

), so u − u

∈ W . We have

u + W = {u + w | w ∈ W } = {u

+ (u − u

) + w | w ∈ W }.

(u − u

) + w ∈ W as W is a vector space, therefore u + w ∈ u

+ W for every w ∈ W , hence u + W ⊆ u

+ W . Similarly,

u

+ W = {u

+ w | w ∈ W } = {u + (u

− u) + w | w ∈ W } ⊆ u + W, therefore u + W = u

+ W = U .

W can be obtained as U − u = {v − u | v ∈ U } for any u ∈ U .

Definition. The dimension of an affine space U = u

+ W is defined to be dim W . (Sometimes it is convenient to define dim ∅ = −1.)

Definition. Let K be a field. K

, considered as an affine subspace of itself is called an n-dimensional affine space over K, and is denoted by A

(K) or A

if the field is understood.

Definition. Let V

, V

be vector spaces. An affine map from V

to V

is

defined to be a function Φ : V

→ V

which can be written in the form

Φ(x) = T (x) + b, where T : V

→ V

is a linear transformation and b ∈ V

.

It is easily checked that the composition of affine maps is also an affine

map. The invertible affine maps V → V form a group. Within this group

the translations form a normal subgroup and the quotient by this normal

subgroup is the group of invertible linear transformations on V .

Proposition 1.5 (Not examinable) Let V

, V

be vector spaces and let Φ : V

→ V

be an affine map.

(i) For any affine subspace U

⊆ V

, the image Φ(U

) is also an affine subspace of V

.

(ii) For any affine subspace U

⊆ V

, the preimage Φ

(U

) is also an affine subspace of V

.

Proof. Let x

, x

, . . . , x

be the co-ordinates on V

, y

, y

, . . . , y

the co-ordinates on V

. Let Φ(x) = T (x) + b, where T : V

→ V

is a linear transformation and b = (b

, b

, . . . , b

)

∈ V

is a vector. Let {t

}

be the matrix of T with respect to the co-ordinates on V

and V

.

(i) Let U

⊆ V

be an affine subspace. If U

= ∅, Φ(U

) = ∅, too. Otherwise, we can write U

= u

+ W

, where u

∈ U

and W

is a linear subspace of V

. Then

Φ(U

) = {Φ(u

+ w) | w ∈ W

} = {T (u

+ w) + b | w ∈ W

}

= {(T (u

) + b) + T (w) | w ∈ W } = (T (u

) + b) + T (W )

As T (W ) is a linear subspace of W

, it follows that Φ(U

) is an affine sub- space, as claimed.

(ii) Let now U

be an affine subspace of V

.

There exist degree 1 polynomials f

, f

, . . . , f

in y

, y

, . . . , y

such that U

= {(y

, y

, . . . , y

) ∈ V

| f

(y

, y

, . . . , y

) = 0, ∀i, 1 ≤ i ≤ k}.

Then

Φ((x

, x

, . . . , x

)

) =

 b

+

m

X

j=1

t

x

, b

+

m

X

j=1

t

x

, . . . , b

+

m

X

j=1

t

x



,

therefore Φ((x

, x

, . . . , x

)

) ∈ U

if and only if

f

 b

+

m

X

j=1

t

x

, b

+

m

X

j=1

t

x

, . . . , b

+

m

X

j=1

t

x



= 0

for every i, 1 ≤ i ≤ k. These are linear equations in the x

, 1 ≤ i ≤ m, therefore Φ

(U

) is an affine subspace of V

, as claimed.

Definition. Let X, Y be subsets of a vector space V . X and Y are affine

equivalent if and only if there exist mutually inverse affine maps Φ, Ψ : V →

V such that Φ(X) = Y and Ψ(Y ) = X. It is easy to verify that affine

equivalence is an equivalence relation.

Examples:

1. All affine subspaces of the same dimension are affine equivalent. This can be proved by showing that any k-dimensional affine subspace is affine equivalent to the linear subspace spanned by the first k standard unit vectors.

2. Any two triangles in R

are affine equivalent. An easy way to prove it is to show that any triangle is affine equivalent to the triangle with vertices (0, 0), (1, 0) and (0, 1). See https://personalpages.manchester.ac.uk/

staff/gabor.megyesi/teaching/MATH32062/triangles.pdf for an exam- ple of calculating an explicit affine equivalence between two triangles.

3. Similarly, any two parallelograms in R

are because all parallelograms affine equivalent to the unit square.

4. Plane conics

A conic (short for a conic section) is the set of points satisfying an equation of the form ax

+ bxy + cy

+ dx + ey + f = 0 with at least one of a, b, c not 0.

By a sequence of change of co-ordinates corresponding to affine transforma- tions and by multiplying the equation by a non-zero scalar, every such equa- tion can be reduced to one of finitely many standard forms, which shows that there are finitely many affine equivalence classesc of conics. The tranforma- tion of the points is given inverse of the change of co-ordinates, for example, if the new co-ordinates are x

= 2x and y

= y + 5, then its effect on the points is the map (x, y) 7→ (x/2, y − 5).

The procedure consists of four steps. The co-ordinates after the ith step will be denoted by x

and y

, the coefficients by a

, b

, . . . , f

.

Step 1. If a 6= 0, let x

= x, y

= y. If a = 0 but c 6= 0, let x

= y, y

= x.

If a = c = 0, then let x

= (x + y)/2, y

= (x − y)/2. In this case b 6= 0 and bxy = b(x

− y

). We have achieved that a

, the coefficient of x

in the new equation is not 0. If necessary, we can multiply the equation by −1 to make a

positive.

Step 2. Complete the square with respect to x

, let x

= √

a



x

+ b

2a

y

+ d

2a



and y

= y

. After this step, the equation will be of the form x

+ c

y

+ e

y

+ f

= 0.

Step 3. If c

6= 0, complete the square with respect to y

, let x

= x

and y

= p|c

|(y

+e

/(2c

)), then the equation will have the form x

±y

+f

= 0.

If c

= 0 but e

6= 0, let x

= x

and y

= e

y

+ f

, so the equation becomes

x

+ y

= 0. If c

= e

= 0, then let x

= x

and y

= y

, so the equation will have the form x

+ f

= 0.

Step 4. If f

6= 0, then let x

= x

/p|f

|, y

= y

/p|f

| and after the change of variable divide the whole equation by |f

|. After this, the form of the equation will be the same, but f

will be ±1. If f

= 0, then let x

= x

, y

= y

.

At the end of this step, either the equation is x

+ y

= 0, or it contains x

, possibly ±y

and possibly ±1 and no other term. x

− y

− 1 = 0 can be tranformed into x

− y

+ 1 = 0 by swapping x

and y

, so these two are affine equivalent.

Therefore any variety in R

defined by a degree 2 equation is affine equivalent to one of the following:

Case 1. x

+ y

− 1 = 0 (ellipses) Case 2. x

+ y

= 0 (single point) Case 3. x

+ y

+ 1 = 0 (∅)

Case 4. x

− y

− 1 = 0 (hyperbolas) Case 5. x

− y

= 0 (two intersecting lines) Case 6. x

+ y = 0 (parabolas)

Case 7. x

− 1 = 0 (two parallel lines) Case 8. x

= 0 (“double line”)

Case 9. x

+ 1 = 0 (∅)

It is also true, but requires a separate proof, that these cases, apart from the two empty sets are not affine equivalent. See https://personalpages.

manchester.ac.uk/staff/gabor.megyesi/teaching/MATH32062/conic.pdf for examples of transforming the equation of a conic to one of the above forms.

Over C we can use essentially the same method to reduce the equation of a conic to one of the above forms. The difference is that we do not need to multiply the equation to ensure a

> 0 in Step 1 and there is no need to take the modulus of c

and f

in Steps 2 and 3, resp. If a term occurs in the final equation, it will always have coefficient +1, so we will end up with one of Cases 3, 6, 8 or 9.

Over C, the change of co-ordinates x

= x, y

= iy converts Case 1 into Case 4 and vice versa, so all ellipses and hyperbolas are affine equivalent over C.

By a similar argument, Cases 1, 3 and 4 are affine equivalent over C, similarly

Cases 2 and 5, and Cases 7 and 9.

1.3 The correspondence between varieties and ideals

Definition. Let X ⊆ A

. The ideal of X is defined as

I(X) = {f ∈ K[x

, x

, . . . , x

] | f (a

, a

, . . . , a

) = 0, ∀(a

, a

, . . . , a

) ∈ X}.

I(X) is an indeed ideal of K[x

, x

, . . . , x

].

It is clear from the definition that X

⊆ X

implies I(X

) ⊇ I(X

).

Proposition 1.6

(i) If J / K[x

, x

, . . . , x

], then J ⊆ I(V(J )).

(ii) X ⊆ V(I(X)) for any X ⊆ A

(K). Equality holds if and only if X is an affine algebraic variety.

Proof. (i) If f ∈ J and P ∈ V(J ) then f (P ) = 0 by the definition of V, therefore f ∈ I(V(J )) by the definition of I, so J ⊆ I(V(J )) as claimed.

(ii) It is also clear from the definitions that X ⊆ V(I(X)) for any set X ⊆ A

and that if equality holds then X is an affine algebraic variety.

The only thing that remains to be proved that if X is an algebraic va- riety, then X = V(I(X)). Assume that X = V(J ) for some ideal J / K[x

, x

, . . . , x

]. By (i), J ⊆ I(V(J )), therefore X = V(J ) ⊇ V(I(V(J ))) = V(I(X)). We have already seen that X ⊆ V(I(X)), therefore X = V(I(X)), as required.

Definition. A field K is algebraically closed if and only if every polynomial of degree at least 1 in K[x] has a root in K.

Examples: C and ¯ Q (the field of algebraic numbers in C) are algebraically closed. Q, R and the finite fields are not algebraically closed.

Remark. For any field K, there exists a minimal algebraically closed field containing it, which is unique up to isomorphism, this field is called the algebraic closure of K. For example, the algebraic closure of R is C and the algebraic closure of Q is ¯ Q

Definition. Let R be a (commutative) ring. Let I /R be an ideal. The radical of I, denoted by √

I or rad I, is

√

I = {a ∈ R | ∃n ∈ N such that a

∈ I}.

I is called a radical ideal if and only if √ I = I.

Algebraic facts: For any ideal I, √

I is also an ideal, I ⊆ √

I and p√

I = √ I, i. e., √

I is a radical ideal.

Theorem 1.7 (Hilbert’s Nullstellensatz) Let K be an algebraically closed field.

(i) Every maximal ideal of K[x

, x

, . . . , x

] is of the form hx

− a

, x

− a

, . . . , x

− a

i, where a

∈ K, 1 ≤ i ≤ n.

(ii) If J is an ideal of K[x

, x

, . . . , x

], J 6= K[x

, x

, . . . , x

] , then V(J ) 6= ∅.

(iii) I(V(J )) = √

J for any J / K[x

, x

, . . . , x

].

Proof. (i) Not proved here.

Note that hx

−a

, x

−a

, . . . , x

−a

i is a maximal ideal of K[x

, x

, . . . , x

] for any field K. We can define a ring homomorphism K[x

, x

, . . . , x

] → K, f 7→ f (a

, a

, . . . , a

), whose kernel is exactly hx

− a

, x

− a

, . . . , x

− a

i.

Since the image is a field, the kernel is a maximal ideal. The substance of this part is that over an algebraically closed field the converse is also true.

(ii) As J 6= K[x

, x

, . . . , x

], there exists a maximal ideal m containing J . By (i), m = hx

− a

, x

− a

, . . . , x

− a

i for some a

∈ K, 1 ≤ i ≤ n.

Therefore {(a

, a

, . . . , a

)} = V(m) ⊆ V(J ), so V(J ) 6= ∅.

(iii) If f ∈ √

J , then f

∈ J for some positive integer k. Hence f

(P ) = (f (P ))

= 0 for every P ∈ V(J ), therefore f (P ) = 0 for every P ∈ V(J ), i. e., f ∈ I(V(J )). This means that √

J ⊆ I(V(J )) without assuming that K is algebraically closed.

We need to prove the other inclusion, I(V(J )) ⊆ √

J . Let f ∈ I(V(J )).

Let’s introduce a new variable y. Let f

, f

, . . . , f

be a set of generators for J , and let J

= hf

, f

, . . . , f

, f y − 1i / K[x

, x

, . . . , x

, y].

We claim that V(J

) = ∅. Assume that P = (a

, a

, . . . , a

, b) ∈ V(J

) ⊆ A

. As f

∈ J

for every i, 1 ≤ i ≤ r, we have f

(a

, a

, . . . , a

) = 0 for every i, 1 ≤ i ≤ r, so (a

, a

, . . . , a

) ∈ V(J ). Therefore f (a

, a

, . . . , a

) = 0, too, but then (f y − 1)(P ) = −1, so P / ∈ V(J

) after all. Therefore V(J

) = ∅, as claimed.

By (ii), this implies that J

= K[x

, x

, . . . , x

, y], in particular 1 ∈ J

. This means that we can write

1 = (f y − 1)g

+

r

X

i=1

g

f

for suitable g

, g

, . . . , g

∈ K[x

, x

, . . . , x

, y]. Let y

be the highest power

of y occurring in any of g

, g

, . . . , g

. Let’s multiply the above equality

by f

. Whenever there is a power y, say y

, occurring in g

for some i,

0 ≤ i ≤ r, we can write y

f

as (f y)

f

, so for each i, 0 ≤ i ≤ r,

f

g

(x

, x

, . . . , x

, y) = G

(x

, x

, . . . , x

, f y) for a suitable polynomial G

.

Therefore

f

= (f y − 1)G

(x

, x

, . . . , x

, f y) +

r

X

i=1

f

G

(x

, x

, . . . , x

, f y).

By substituting y = 1/f into the above equality, we obtain

f

=

r

X

i=1

f

G

(x

, x

, . . . , x

, 1).

The right-hand side is an element of J , therefore f ∈ √

J as required.

The function V from ideals of K[x

, x

, . . . , x

] to affine algebraic varieties

in A

is not injective, and the function I from subsets of A

to ideals of

K[x

, x

, . . . , x

] is neither injective nor surjective. However, the combina-

tion of Proposition 1.6 and the Nullstellensatz implies that V and I create

a bijection between radical ideals of K[x

, x

, . . . , x

] and affine algebraic

varieties in A

.

1.4 Irreducibility

Definition. An affine algebraic variety W is reducible if and only if it can be written as W = V

∪V

, where W

, W

are also affine algebraic varieties, W

6=

V 6= V

. If W is not reducible, it is called irreducible. (The irreducibility of W is equivalent to saying that if W = V

∪ V

, where W

, W

are also affine algebraic varieties, then W

= V or W

= V .)

Examples:

1. The variety V(hx

−y

i) ⊂ R

is reducible, since it is the union of V(hx−yi) and V(hx + yi), which are the lines y = x and y = −x.

2. Any variety consisting of a single point is obviously irreducible.

3. Warning: Graphs can be useful in finding the irreducible components, but they can also be misleading. The graph below shows the the curve

y

+ (x

− 4)(x

− 1) = 0

in R

. Topologically it consists of two disjoint connected components, but it is irreducible, the two parts cannot be separated algebraically.

-2 -1 1 2

-2 -1 1 2

Theorem 1.8 Every affine algebraic variety V can be decomposed into a union V = V

∪ V

∪ . . . ∪ V

such that every V

, 1 ≤ i ≤ k, is an irreducible affine algebraic variety and V

6⊆ V

for i 6= j. The decomposition is unique up to the ordering of the components.

Definition. The V

, 1 ≤ i ≤ k, in the above theorem are called the irreducible components of V .

Proof. We call an affine algebraic variety good if it can be written as the union of finitely many irreducible affine algebraic varieties and bad otherwise. We shall prove first that all affine algebraic varieties are good.

Assume that there exists a bad affine algebraic variety.

Claim: There exists a bad affine algebraic variety W such that for all affine

algebraic varieties W

, I(W ) $ I(W

) implies that W

is good.

Assume that this claim is false. Let V

be a bad affine algebraic variety, then there exists a bad variety V

such that I(V

) $ I(V

). Similarly, there exists a bad variety V

such that I(V

) $ I(V

). Continuing like this, we can construct an infinite sequence of bad varieties V

, V

, V

, . . . , such that

I(V

) $ I(V

) $ I(V

) $ . . . $ I(V

) $ I(V

) $ . . . .

The I(V

), m = 1, 2, 3, . . . are ideals in the polynomial ring K[x

, x

, . . . , x

] for some n, which is Noetherian by the Hilbert Basis Theorem (Theorem 1.2), so it does not contain a strictly increasing infinite sequence of ideals.

This contradiction shows that the claim is true.

Let now W be a bad affine algebraic variety with the property described in the claim. W cannot be irreducible, since any irreducible variety is good, because it is the union of one irreducible variety. Therefore W is reducible, there exist affine algebraic varieties W

and W

such that W

6= W 6= W

and W = W

∪ W

. W

⊂ W implies I(W

) ⊇ I(W ) for i = 1, 2. By Proposition 1.6 (ii), V(I(W )) = W and V(I(W

)) = W

for i = 1, 2, so W

6= W implies I(W

) % I(W ) for i = 1, 2. By the assumption on W , this implies that W

and W

are both good, they are unions of finitely many irreducible affine algebraic varieties. Then W = W

∪ W

is also the union of finitely many irreducible affine algebraic varieties, contradicting the assumption that W is bad.

This shows that every affine algebraic variety V can be written as a finite union of irreducible affine algebraic varieties, then discarding the varieties which are contained in one of the terms, we can write V = V

∪ V

. . . ∪ V

with V

irreducible for each i, 1 ≤ i ≤ k and V

6⊆ V

if i 6= j.

Let V = V

∪ V

∪ . . . ∪ V

be another decomposition of V into irreducible varieties with V

6⊆ V

for i 6= j. Then

V

= V

∩ V = (V

∩ V

) ∪ (V

∩ V

) ∪ . . . ∪ (V

∩ V

).

As V

is irreducible, there exists i such that V

= V

∩ V

, i. e., V

⊆ V

. By applying this argument to V

, we obtain that there exists j such that V

⊆ V

, therefore V

⊆ V

. This implies j = 1 by our assumption on the V

, V

, . . . , V

, and therefore V

= V

.

By applying this argument to each V

(1 ≤ r ≤ k) and V

(1 ≤ s ≤ l), we obtain a bijection between {V

, V

, . . . , V

} and {V

, V

, . . . , V

}, so the decompositions are the same apart from possibly the order of the components.

Definition. Let R be a commutative ring. An ideal I / R is called a prime

ideal if and only if ab ∈ I implies a ∈ I or b ∈ I.

Proposition 1.9 An affine algebraic variety W is irreducible if and only if I(W ) is prime.

Proof. We shall prove the contrapositive, namely that W is reducible if and only if I(W ) is not prime.

Assume that W is reducible, then W can be written as W = W

∪ W

, where W

and W

are affine algebraic varieties and W 6= W

, W 6= W

. W

⊂ W implies I(W

) ⊇ I(W ), but V(I(W )) = W and V(I(W

)) = W

by Proposition 1.6 (ii), therefore I(W

) 6= I(W ), so I(W

) % I(W ). Let f

∈ I(W

) \ I(W ). Similarly, let f

∈ I(W

) \ I(W ). If P ∈ W

, then (f

f

)(P ) = 0 because f

(P ) = 0, if P ∈ W

, then (f

f

)(P ) = 0 because f

(P ) = 0, therefore (f

f

)(P ) = 0 for every P ∈ W , so f

f

∈ I(W ).

Neither f

, nor f

is an element of I(W ), therefore I(W ) is not prime. This shows that if I(W ) is prime, then W is irreducible.

Assume now that I(W ) is not prime. Let f

, f

be polynomials such that f

, f

∈ I(W ), but f /

f

∈ I(W ). Let

W

= {P ∈ W | f

(P ) = 0} = W ∩ V(hf

i) = V(hI(W ), f

i) and similarly

W

= {P ∈ W | f

(P ) = 0} = W ∩ V(hf

i) = V(hI(W ), f

i).

W

, W

are affine algebraic varieties. Clearly W

⊆ W , W

⊆ W , but W

6= W since f

∈ I(W ) and similarly W /

6= W since f

∈ I(W ). For any / P ∈ W , 0 = (f

f

)(P ) = f

(P )(f

)(P ), so f

(P ) = 0 or f

(P ) = 0. In the first case, P ∈ W

, in the second P ∈ W

, therefore W = W

∪ W

and W is reducible. This shows that if W is irreducible, then I(W ) is prime.

Remark. The criterion involves I(W ), so if W = V(J ) for some ideal J , the primality or otherwise of J is not sufficient in general to tell whether W is irreducible or not. If the field is algebraically closed, then I(V(J )) = √

J by the Nullstellensatz. If J is prime, then it is automatically radical, so I(V(J)) = J and V(J) is irreducible. If J is radical but not prime, then V(J) is reducible.

The functions V and I create a bijection between prime ideals of K[x

, x

, . . . , x

] and irreducible affine algebraic varieties in A

.

Properties of irreducible varieties and examples

1. Irreducibility is preserved under affine equivalence. If X and Y are affine

equivalent affine algebraic varieties, then X is irreducible if and only if Y

is. Let Φ : X → Y be an affine equivalence and assume that X can be

written as X = X

∪ X

, where X

, X

are also affine algebraic varieties and X

6= X 6= X

, then Y = Φ(X

) ∪ Φ(X

), Φ(X

), Φ(X

) are also affine algebraic varieties and Φ(X

) 6= Y 6= Φ(X

).

2. Affine subspaces over an infinite field K are irreducible. (Note that al- gebraically closed fields are infinite.) Every affine subspace is equivalent to one of the form x

= x

= . . . = x

= 0 for some k. It can be proved that I(V(hx

, x

, . . . , x

i)) = hx

, x

, . . . , x

i if K is infinite and that hx

, x

, . . . , x

i/K[x

, x

, . . . , x

] is a prime ideal directly from the definition.

3. Let K be an algebraically closed field. Let H ⊂ A

(K) be a hypersur- face, that is, an algebraic variety defined by an ideal generated by a sin- gle polynomial f ∈ K[x

, x

, . . . , x

], H = V(hf i). f can be factorised as f = f

f

· · · f

, where the f

, 1 ≤ i ≤ r, are irreducible polynomials such that f

is not a scalar multiple of f

if i 6= j, and α

, 1 ≤ i ≤ r, are positive integers. This factorisation is unique up to scalar factors.

We shall now show that phfi = hf

f

· · · f

i. Let g ∈ phfi. g can be factorised as g = g

g

· · · g

, where the g

, 1 ≤ i ≤ s, are irreducible polynomial such that g

is not a scalar multiple of g

if i 6= j, and β

, 1 ≤ i ≤ s, are positive integers. g ∈ phfi means that there exists a positive integer m such that g

= g

g

· · · g

∈ hf i, therefore for each i, 1 ≤ i ≤ r, f

|g

g

· · · g

. As f

is irreducible, it must divide g

for some j, 1 ≤ j ≤ s, so f

|g. The f

are pairwise coprime, therefore it follows that f

f

. . . f

|g, g ∈ hf

f

· · · f

i. Conversely, if g ∈ hf

f

· · · f

i, then g

∈ hf i. This shows that I(H) = phfi = hf

f

· · · f

i.

By Proposition 1.3 (i), S

i=1

V(hf

i) = Vhf

f

· · · f

i = H. By the argument used previously, hf

i / K[x

, x

, . . . , x

] is a prime ideal, therefore it is radical, so I(V(hf

i)) = hf

i by the Nullstellensatz, by using the fact that hf

i is prime again, it follows that V(hf

i) is irreducible. V(hf

i) does not contain V(hf

i) if i 6= j, because f

is not a multiple of f

. This shows that the irreducible components are V(hf

i), 1 ≤ i ≤ r.

In particular, H is irreducible if and only if r = 1, i. e., f is the power of a single irreducible polynomial.

Lemma 1.10 Let n ≥ 2 be an integer, let K be a field and let p ∈ K[x

, x

, . . . , x

] be a polynomial which does not involve the variable x

for some i, 1 ≤ i ≤ n, and which is not a square of a polynomial in K[x

, x

, . . . , x

]. Then x

− p is irreducible in K[x

, x

, . . . , x

].

Proof. We can assume for simplicity that i = n. Let’s assume that for a

contradiction that x

−p can be factorised. By considering it as a polynomial

in x

with coefficients in K[x

, x

, . . . , x

], there are two possibilities: either

one factor has degree 2 in x

and the other degree 0, or both factors have degree 1.

In the first case the factors are ax

+ bx

+ c and d, where a, b, c, d ∈ K[x

, x

, . . . , x

]. From the coefficient of x

we get ad = 1, therefore a and d must be constants, so we do not get a proper factorisation.

In the second case the factors are ax

+ b and cx

+ d with a, b, c, d ∈ K[x

, x

, . . . , x

]. Similarly to the previous case, from the coefficient of x

we get ac = 1, therefore a and c must be constants. By dividing the first factor by a and multiplying the second by a, we can assume that a = c = 1.

Then x

− p = (x

+ b)(x

+ d) = x

+ (b + d)x

+ bd, so by comparing the coefficient of x

and the degree 0 term in x

we get b + d = 0 and −p = bd.

Hence d = −b, so −p = −b

, p = b

, but p is not a square, this is a contra- diction, therefore x

− p is irreducible as claimed. .

This simple lemma has lots of useful consequences.

4. Ellipses, hyperbolas, parabolas are irreducible over C and R.

x

+ y

− 1 = x

− (1 − y

) and 1 − y

is not a square, therefore x

+ y

− 1 is irreducible in C[x, y] by Lemma 1.10 and hx

+ y

− 1i / C[x, y] is a prime ideal. Prime ideals are radical, so by the Nullstellensatz (Theorem 1.7), I(V(hx

+ y

− 1i)) = hx

+ y

− 1i. By Proposition 1.9, V(hx

+ y

− 1i) is irreducible over C as hx

+ y

− 1i / C[x, y] is a prime ideal.

It can also be proved that I(V(hx

+ y

− 1i)) = hx

+ y

− 1i over R, too. The main idea is that if x

+ y

− 1 does not divide f ∈ R[x, y], then f (x, y) = x

+ y

− 1 = 0 only has finitely many solutions, but the unit circle x

+ y

− 1 = 0 has infinitely many points over R, therefore f / ∈ I(V(hx

+ y

− 1i)). The rest of the proof is the same as over C.

Any ellipse is affine equivalent to V(hx

+y

−1i) over either C or R, therefore all ellipses are irreducible.

The proof for hyperbolas and parabolas is similar.

5. If p(x) ∈ C[x] is a polynomial of odd degree, then V(hy

− p(x)i) is irreducible over C, as p(x) cannot be a square.

This includes as special cases the nodal cubic curve, V(hx

+ x

− y

i), the

cuspidal cubic curve V(hx

− y

i) and elliptic curves (to be studied later in

the course).

-1 1 2

-4 -2 2 4

-0.5 0.5 1.0 1.5 2.0 2.5

-4 -2 2 4

Nodal cubic Cuspidal cubic

6. Let C = V(hx

y

+ y

− x

− 2y

+ 1i) ⊂ A

(C). This polynomial factorises as

x

y

+ y

− x

− 2y

+ 1 = (y

− 1)

+ x

(y

− 1) = (x

+ y

− 1)(y

− 1)

= (x

+ y

− 1)(y − 1)(y + 1)

y − 1 and y + 1 have degree 1, so they are irreducible. We have already showed that x

+ y

− 1 is irreducible. Therefore the irreducible components of C are the lines V(hy − 1i), V(hy + 1i) and V(hx

+ y

− 1i).

-2 -1 1 2

-1.5 -1.0 -0.5 0.5 1.0 1.5

Warning. The correspondence between the irreducible factors of the gener- ator and the irreducible components described above only applies to hyper- surfaces. An ideal generated by irreducible polynomials need not be prime.

Factorisation is still a useful tool in decomposition of varieties into irreducible

componenent, but all elements of the ideal have to be considered, not just

the generators.

7. The unit sphere V(hx

+ y

+ z

− 1i) ⊂ A

(C) is irreducible, since x

+ y

+ z

− 1 = x

− (1 − y

− z

) and 1 − y

− z

is not a square, therefore x

+ y

+ z

− 1 is irreducible in C[x, y, z].

It can be proved that I(V(hx

+ y

+ z

− 1i)) = hx

+ y

+ z

− 1i over R similarly to the case of the circle, therefore the unit sphere is also irreducible over R.

Any ellipsoid is affine equivalent to the unit sphere, therefore ellipsoids are irreducible over R or C.

8. Let I = hx

+ y

+ z

− 1, 3x

+ y

− z

− 1i / C[x, y, z] and let W = V(I) ⊂ A

be the variety defined by I. The graph below shows this variety, the two surfaces are the sphere defined the equation x

+ y

+ z

− 1 = 0 and the hyperboloid of one sheet defined by 3x

+ y

− z

− 1 = 0, W is their intersection, the black curve.

The generators themselves are irreducible, but

x

− z

= (3x

+ y

− z

− 1) − (x

+ y

+ z

− 1)

2 ∈ I,

too, and it factorises as x

− z

= (x + z)(x − z). We can try to follow the argument of the proof Proposition 1.9, without proving that x + z, x − z / ∈ I.

W can be written as the union of W

= V(hx

+ y

+ z

− 1, 3x

+ y

− z

−

1, x + zi) and W

= V(hx

+ y

+ z

− 1, 3x

+ y

− z

− 1, x − zi). We shall

prove that these varieties are irreducible.

Let’s consider W

= V(hx

+y

+z

−1, 3x

+y

−z

−1, x+zi). The difference of x

+ y

+ z

− 1 and 3x

+ y

− z

− 1 is divisible by x + z so we can omit one of them from the set of generators, therefore W

= V(hx

+y

+z

−1, x+zi).

The graph below show the intersection of the plane x + z = 0 with the sphere and the hyperboloid.

The plane x + z = 0 intersects both in the same curve, shown in black, this is W

. The intersection of a sphere with a plane is a circle. By rotating W

through π/4 about the y-axis, we can transform it to the unit circle x

+ y

− 1 = 0 in the xy-plane. We have already proved that the unit circle is irreducible, the rotation is a Euclidean transformation, which is a special case of an affine equivalence, therefore W

is irreducible.

The irreducibility of W

= V(hx

+ y

+ z

− 1, 3x

+ y

− z

− 1, x − zi) = V(hx

+ y

+ z

− 1, x − zi) can be proved similarly.

It is also clear that neither of W

and W

contains the other, for exam- ple, because (1/ √

2, 0, −1/ √

2) ∈ W

, but (1/ √

2, 0, −1/ √

2) / ∈ W

, while (1/ √

2, 0, 1/ √

2) ∈ W

, but (1/ √

2, 0, 1/ √

2) / ∈ W

. Therefore they are the

irreducible components of W .