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AFFINE VARIETIES
A Project Report Submitted
in Partial Fulfilment of the Requirements
for the Degree of
MASTER OF SCIENCE
in
Mathematics and Computing
by
USHA SINGH
(Roll No. 132123045)
to the
DEPARTMENT OF MATHEMATICS
INDIAN INSTITUTE OF TECHNOLOGY GUWAHATI
GUWAHATI - 781039, INDIA
April 2015
CERTIFICATE
This is to certify that the work contained in this report entitled “ AFFINE
VARIETIES” submitted by USHA SINGH (Roll No: 132123045.)
to Department of Mathematics, Indian Institute of Technology Guwahati
towards the requirement of the course MA699 Project has been carried
out by her under my supervision.
Guwahati - 781 039
(Dr.SHYAMASHREE UPADHYAY)
April 2015
Project Supervisor
ii
ABSTRACT
Algebraic geometry is a branch of mathematics that is concerned with
the geometric structure of solution set to a system of polynomial equations.
The zero locus of a set of polynomials is called an affine algebraic set and
an irreducible affine algebraic set is called an affine variety. Classically, this
theory is developed over algebraically closed fields. In this thesis, we study
affine varieties and provide a natural duality between their geometry and
algebra.
iii
Contents
1
Affine varieties, definition and properties
1
1.1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
1.2
Algebraic sets . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
1.3
Zariski Topology on An
. . . . . . . . . . . . . . . . . . . . .
4
1.4
Affine varieties . . . . . . . . . . . . . . . . . . . . . . . . . .
5
2
Hilbert’s Nullstellensatz
10
2.1
Hilbert’s Nullstellensatz . . . . . . . . . . . . . . . . . . . . . 10
2.2
Affine Coordinate Ring . . . . . . . . . . . . . . . . . . . . . . 15
3
Dimension of affine algebraic set
16
3.1
Noetherian topological space . . . . . . . . . . . . . . . . . . . 16
3.2
Dimension of a Topological Space . . . . . . . . . . . . . . . . 17
Bibliography
22
iv
Chapter 1
Affine varieties, definition and properties
1.1
Introduction
We begin with the proceed to introduce our main objects of study, affine
algebraic sets. After examining a number of examples, we study their algebraic counterpart, the vanishing ideals. The thesis continues by describing
relationships between the two, necessarily providing an introduction to the
Zariski topology.
In this thesis, K will always denote an algebraically closed field.
1.2
Algebraic sets
Definition 1.2.1. We define affine n-space over K, denoted Ank or simply
An , to be the set of all n-tuples of elements of K. An element p ∈ An will be
called a point, and if p = (a1 , ..., an ) with ai ∈ K, then the ai will be called
the coordinates of p.
1
Let A = K[x1 , ..., xn ] be the polynomial ring in n variables over K. We will
interpret the elements of A as functions from the affine n-space to K, by
defining f (p) = f (a1 , ..., an ), where f ∈ A and p ∈ An . Thus if f ∈ A is a
polynomial, we can talk about the set of zeros, namely Z(f ) = {p ∈ An |
f (p) = 0}. More generally,
Definition 1.2.2. If T is any subset of A, we define the zero set of T to be
the common zeros of all the elements of T , namely
Z(T ) = {p ∈ An | f (p) = 0 ∀f ∈ T }
Proposition 1.2.3. If a is the ideal of A generated by T , then prove that
Z(T ) = Z(a).
Proof. T is a subset of A = k[x1 , ..., xn ]. a is the ideal generated by T , then
P
a = { ni=1 fi gi | gi ∈ A, fi ∈ T }.
Claim: Z(T ) = Z(a).
Let p ∈ Z(T ) ⇒ Z(T ) = {p ∈ An | f (p) = 0 ∀f ∈ T }. Since p ∈ Z(T ) we
have f (p) = 0 ∀f ∈ T .
P
Take any f ∈ a, since f = ni=1 gi fi , for some fi ∈ T, gi ∈ A.
P
Then f (p) = ni=1 gi (p)fi (p) = 0, since f (p) = 0 ∀f ∈ T
such that p ∈ Z(a) and hence
Z(T ) ⊆ Z(a)
(1.1)
Conversely, let q ∈ Z(a), f (q) = 0 ∀f ∈ a.
but T ⊆ a ⇒ f (q) = 0 ∀f ∈ T ⇒ q ∈ Z(T ), therefore
Z(a) ⊆ Z(T )
2
(1.2)
By both equation, hence Z(a) = Z(T ).
• By Hilbert Basis theorem, we know that A is Noetherian, so any ideal
of A has finite set of generators f1 , f2 , . . . , fr . Thus Z(T ) can be expressed
as the common zeros of a finite set of polynomials f1 , f2 , . . . , fr .
Definition 1.2.4. A subset Y of An is said to be an algebraic set if there
exist a subset T ⊆ A such that Y = Z(T ).
In this case Y = Z(T ) is called the locus of T ∈ A = K[x1 , . . . , xn ].
Proposition 1.2.5. The union of two algebraic sets is an algebraic set.
Proof. If Y1 = Z(T1 ) and Y2 = Z(T2 ), prove that Y1 ∪ Y2 = Z(T1 T2 ), where
T1 T2 = {f1 f2 | f1 ∈ T1 , f2 ∈ T2 }.
Let p ∈ Y1 ∪ Y2
⇒ either p ∈ Y1 or p ∈ Y2
⇒ either p ∈ Z(T1 ) or p ∈ Z(T2 )
⇒ either p is zero of all polynomials in T1 or p is zero of all polynomials in
T2 .
Take any f ∈ T1 T2 , then f = f1 f2 for some f1 ∈ T1 and f2 ∈ T2 .
f (p) = f1 (p)f2 (p) = 0 ⇒ p is a zero of every f ∈ T1 T2 .
⇒ Y1 ∪ Y2 ⊆ Z(T1 T2 )
Conversely, let p ∈ Z(T1 T2 )
⇒ f (p) = 0 ∀ f ∈ T1 T2 .
⇒ f1 (p)f2 (p) = 0 ∀f1 ∈ T1 and f2 ∈ T2 .
⇒ either f1 (p) = 0 ∀ f1 ∈ T1 or f2 (p) = 0 ∀ f2 ∈ T2
⇒ either p ∈ Y1 or p ∈ Y2 ⇒ p ∈ Y1 ∪ Y2 .
3
(1.3)
⇒ Z(T1 T2 ) ⊆ Y1 ∪ Y2
(1.4)
Hence by (1.3) and (1.4), Y1 ∪ Y2 = Z(T1 T2 )
Proposition 1.2.6. The intersection of an arbitrary family of algebraic sets
is an algebraic set.
Proof. If Y∝ = Z(T∝ ) is any family of algebraic sets, then
T
so Y∝ is also an algebraic set.
T
T
Y∝ = Z( T∝ ),
Proposition 1.2.7. The empty set and whole space are algebraic sets.
Proof. The empty set φ = Z(1), and whole space An = Z(0). So they are
affine algebraic sets.
1.3
Zariski Topology on An
Definition 1.3.1. We define the Zariski topology on An by taking the open
subsets to be the complements of the algebraic sets. This is a topology,
because according to the above proposition (1.2.5) the intersection of two
open sets is open and proposition (1.2.6) the union of any family of open sets
is open.
Example 1.3.2. Let us consider the Zariski topology on the affine line A1 .
Every ideal in A = K[x] is principal, so every algebraic set is the set of zeros
of a single polynomial. Since K is algebraically closed, every nonzero polynomial f (x) can be written f (x) = c(x − a1 ) . . . (x − an ) with c, a1 , . . . , an ∈ K.
Then Z(f ) = {a1 , . . . , an }. Thus the algebraic sets in A1 are just the finite
4
subsets.
Thus the open sets are empty set, the whole space and complement of finite
Sets.
1.4
Affine varieties
Definition 1.4.1. A non empty subset Y of a topological space X is irreducible if it can not be expressed as the union Y = Y1 ∪ Y2 of two proper
subsets, each one of which is closed in Y . The empty set is not considered
to be irreducible.
Example 1.4.2. A1 is irreducible, because its only proper closed subsets
are finite, yet it is infinite because K is algebraically closed, hence infinite.
Proposition 1.4.3. Any nonempty open subsets of an irreducible space is
irreducible and dense.
Proof. Let X be irreducible topological space. Let U be a nonempty open
subset of the irreducible space X.
To prove: U is dense in X, that is to prove that for any nonempty open
subset V of X, U ∩ V 6= φ.
Suppose not, that is suppose U ∩ V = φ ⇒ (U ∩ V )c = (φ)c ⇒ U c ∪ V c = X.
U and V both are open sets ⇒ U c and V c both are closed sets.
Note that U c 6= X because U 6= φ, similarly V c 6= X because V 6= φ
⇒ U c and V c are proper closed subsets of X therefore X = U c ∪ V c , which
is a contradiction because X is given to be irreducible.
Hence U is dense in X.
5
To prove: U is irreducible.
Suppose not,
Then U = U1 ∪ U2 for some U1 , U2 closed in U in the induced topology,
which are proper subsets of U .
U1 is closed in U ⇒ U1 = V1 ∩ U , where V1 is closed in X.
U2 is closed in U ⇒ U2 = V2 ∩ U where V2 is closed in X.
Consider the equation,
X = U ∪ Uc
= (U1 ∪ U2 ) ∪ U c
= (V1 ∩ U ) ∪ (V2 ∩ U ) ∪U c
= [(V1 ∪ V2 ) ∩ U ] ∪ U c
= [(V1 ∪ V2 ) ∪ U c ] ∩ [U ∪ U c ]
= (V1 ∪ V2 ) ∪ U c
U is open in X ⇒ U c is closed in X.
U 6= φ ⇒ U c 6= X ⇒ U c is a proper subset of X.
V1 and V2 both are closed in X ⇒ V1 ∪ V2 is closed in X.
V1 ∪ V2 6= X because if V1 ∪ V2 = X, then (V1 ∪ V2 )c = φ ⇒ V1c ∩ V2c = φ ⇒ V1c
and V2c are both open sets in X whose intersection is empty. But since X is
irreducible, every open subset in X is dense ⇒ V1c ∩ V2c can not be empty.
So that V1 ∪ V2 is a proper closed subset of X.
X = [V1 ∪ V2 ] ∪ U c which is a contradiction, since given X is irreducible.
Proposition 1.4.4. If Y is an irreducible subset of X, then its closure Y in
X is also irreducible.
Proof. Suppose Y is not irreducible. Then Y = Y1 ∪Y2 for two proper subsets
of Y , each one of which is closed in Y . Then Y1 = V1 ∩ Y and Y2 = V2 ∩ Y
6
for some V1 ,V2 proper and closed in X, Then
Y = Y1 ∪ Y2
=(V1 ∩ Y )∪ (V2 ∩ Y )
=(V1 ∪ V2 )∩Y
Also we have, Y = Y ∪ Y
= (V1 ∪ V2 ) ∩ Y ∩ Y
= ((V1 ∪ V2 ) ∩ Y
= ((V1 ∩Y )∪(V2 ∩Y ), where (V1 ∩Y ) and (V2 ∩Y ) are proper
closed subsets in Y . This is contradiction the fact that Y is irreducible. So
Y is irreducible.
Definition 1.4.5. An affine algebraic variety is an irreducible closed subset
of An .
In other words, irreducible affine algebraic sets are called affine algebraic
variety.
These affine varieties are our first objects of study. But before we can go
further, in fact before we can even give any interesting examples, we need to
explore the relationship between subsets of An and ideals in A more deeply.
So for any subset Y ⊆ An , let us define the ideal of Y in A by
I(Y ) = {f ∈ A | f (p) = 0 ∀ p ∈ Y }
Now we have a function Z which maps subsets of A to algebraic sets, and a
function I which maps subsets of An to ideals. Their properties are summarized in the following proposition.
Proposition 1.4.6. If T1 ⊆ T2 are subsets of A, then Z(T1 ) ⊇ Z(T2 )
Proof. Let T1 ⊆ T2 .
7
Prove that Z(T1 ) ⊇ Z(T2 ).
suppose any p ∈ Z(T2 ) ⇒ f (p) = 0 ∀ f ∈ T2 . Since T1 ⊆ T2 , therefore
f (P ) = 0 ∀ f ∈ T1 ⇒ p ∈ Z(T1 ).
Hence Z(T1 ) ⊇ Z(T2 ).
Proposition 1.4.7. If Y1 ⊆ Y2 are subsets of An , then I(Y1 ) ⊇ I(Y2 ).
Proof. Let Y1 ⊆ Y2 .
Prove that I(Y1 ) ⊇ I(Y2 ).
suppose any f ∈ I(Y2 )
⇒ f (p) = 0 ∀ p ∈ Y2 . Since Y1 ⊆ Y2 , therefore f (P ) = 0 ∀ p ∈ Y1
⇒ p ∈ I(Y1 ).
Hence I(Y1 ) ⊇ I(Y2 ).
Proposition 1.4.8. I(Y1 ∪ Y2 ) = I(Y1 ) ∩ I(Y2 ).
Proof. Let, f ∈ I(Y1 ∪ Y2 )
⇔ f (P ) = 0 ∀ p ∈ Y1 ∪ Y2
⇔ f (P ) = 0 ∀ P ∈ Y 1 and f (P ) = 0 ∀P ∈ Y2
⇔ f ∈ I(Y1 ) and f ∈ I(Y2 )
⇔ f ∈ I(Y1 ) ∩ I(Y2 ).
Proposition 1.4.9. For any ideal a ⊆ A, I(Z(a)) =
Proof. The fact that I(Z(a)) ⊆
√
√
a, the radical of a.
a is a direct consequence of Hilbert’s Null-
stellensatz, stated below, since the radical of a is defined as
√
a = {f ∈ A|f r ∈ a f or some r > 0}
√
The other way inclusion is easy: Let f ∈ a. Then f r ∈ a. We claim that
f (p) = 0 for all p ∈ Z(a).
8
proof of claim: Take any p ∈ Z(a), then since f r ∈ a, we must have f r (p) = 0.
That is [f (p) · · · f (p)]r = 0. Since K is an integral domain, we must have
f (p) = 0.
Proposition 1.4.10. For any subset Y ⊆ An , Z(I(Y )) = Y , the closure of
Y.
Proof. First, We note that Y ⊆ Z(I(Y )), which is a closed set, so clearly
Y ⊆ Z(I(Y )). On the other hand, let W be any closed set containing Y .
Then W = Z(a) for some ideal a. So Z(a) ⊇ Y , and by proposition (1.4.7),
I(Z(a)) ⊆ I(Y ). Be certainly, a ⊆ I(Z(a)), so by proposition (1.4.6) we have
W = Z(a) ⊇ Z(I(Y )). Thus Z(I(Y )) = Y .
9
Chapter 2
Hilbert’s Nullstellensatz
2.1
Hilbert’s Nullstellensatz
Given a field extension K ⊆ L and and I an ideal in the polynomial ring P =
K[x1 , . . . , xn ], let IL[x1 , . . . , xn ] denote the ideal of L[x1 , . . . , xn ] generated
by I. We state the following result without proof:
Proposition 2.1.1. Let K ⊆ L be a field extension and let I] ⊆ A =
K[x1 , . . . , xn ] an ideal. Then
IL[x1 , . . . , xn ] ∩ K[x1 , . . . , xn ] = I.
In particular, we have IL[x1 , . . . , xn ] = L[x1 , . . . xn ] if and only if I =
K[x1 , . . . , xn ].
This result is the key to the weak form of Nullstellensatz which tells
us that algebraic closure is enough to guarantee that the only ideal which
represents the empty variety is the entire polynomial ring.
10
Theorem 2.1.2. (Weak Nullstellensatz) Let K be a field and let a be a proper
ideal of A. Then Z(a) 6= 0.
Proof. Let K be the algebraic closure of K, Ā=K[x1 , . . . , xn ]. By proposition
(2.1.1), a Ā is a proper ideal of Ā which is a Noetherian ring. It follows that
a Ā is contained in a maximal ideal m of Ā. But the corollary 1.4 says that
n
there exists (a1 , . . . , an ) ∈ K such that m = (x1 − a1 , . . . , xn − an ) and thus
(a1 , . . . , an ) is a zero of the ideal m.
Because a ⊆ a Ā ⊆ m, we obtain that Z(a) ⊇ Z(m) and finally that Z(a) 6=
∅.
Theorem 2.1.3. (Hilbert’s Nullstellensatz). Let K be an algebraically closed
field, let a be an proper ideal in A = K[x1 . . . xn ], and let f ∈ A be a polynomial which vanishes at all points of Z(a). Then f r ∈ a for some integer
√
r > 0. i.e., I(Z(a)) = a.
Proof. For ‘ ⊇’ Let f ∈ A such that f i ∈ a for some i ≥ 0.
Then
f i (a1 , . . . , an ) = 0 for all (a1 , . . . , an ) ∈ Z(a). It follows that f (a1 , . . . , an ) = 0
for every (a1 , . . . , an ) ∈ Z(a), which means that f ∈ I(Z(a)).
Now for ‘ ⊆’ We may assume that a 6= 0. Let f ∈ I(Z(a)) − {0} and
let {g1 , . . . , gs } be a system of generators of a. Let xn+1 be a new indeterminate and let a0 = aA[xn+1 ] + (xn+1 f − 1) be an ideal in A[xn+1 ]. Then
for every point (a1 , . . . , an+1 ) ∈ Z(a0 ) we have an+1 f (a1 , . . . , an ) = 1 and
gi (a1 , . . . , an ) = 0 for i = 1, . . . , s. We obtain that (a1 , . . . , an ) ∈ Z(a) and
f (a1 , . . . , an ) 6= 0 which contradict the choice of f . It follows that Z(a0 ) = ∅
and then by 2.1.2, 1 ∈ a0 .This means that there exit h, h1 . . . hs ∈ A[xn+1 ]
such that
11
1=
Ps
i=1
hi gi + h(xn+1 f − 1)
In the field K(x1 , . . . , xn , xn+1 ) we substitute xn+1 with 1/f and we obtain
P
1 = si=1 hi (x1 , . . . , xn , 1/f )gi
Ps
By clearing the denominators, we find f m =
i=1 h̃i gi for some m > 0
and suitable polynomials h̃1 , . . . ,h̃s in A.
√
i.e.,f ∈ a.
Finally we get that f m ∈ a,
The most important consequence of the Nullstellensatz is that we obtain
an ideal-variety correspondence given by the following.
Definition 2.1.4. An ideal a of the polynomial ring A is called a radical
√
√
ideal if a = a where a is as defined in Proposition (1.4.9).
Corollary 2.1.5. If K is an algebraically closed field, and if we restrict to
radical ideals, then the maps
I
affine varieties →
− radical ideals
and
Z
radical ideals −
→ affine varieties
are inclusion-reversing bijections which are inverses of each other. So, any
question about varieties can be rephrased as an algebraic question and conversely, provided that we are working over an algebraically closed field.
Corollary 2.1.6. An algebraic set is irreducible if and only if its ideal is a
prime ideal.
Proof. If Y is irreducible, we show that I(Y ) is prime. Indeed, if f g ∈ I(Y ),
then Y ⊆ Z(f g) = Z(f )∪Z(g). Thus Y = (Y ∩Z(f ))∪(Y ∩Z(g)), both being
closed subsets of Y . Since Y is irreducible, we have either Y = Y ∩ Z(f ), in
which case Y ⊆ Z(f ), or Y ⊆ Z(g). Hence either f ∈ I(Y ) or g ∈ I(Y ).
12
Conversely, let p be a prime ideal, and suppose that Z(p) = Y1 ∪ Y2 . Then
p = I(Y1 ) ∩ I(Y2 ), so either p = I(Y1 ) or p = I(Y2 ). Thus Z(p) = Y1 or Y2 ,
hence it is irreducible.
Example 2.1.7. An is irreducible, since it correspondence to the Zero ideal
in A, which is prime.
An = Z(0), since we know that 0 is a prime ideal, then Z(0) is irreducible.
Example 2.1.8. Let f be an irreducible polynomial in A = K[x, y]. Then
f generates a prime ideal in A, since A is a unique factorization domain, so
the zero set Y = Z(f ) is irreducible. we call it affine curve defined by the
equation f (x, y) = 0.
Lemma 2.1.9. The maximal ideals of the polynomial ring K[x1 , . . . , xn ] are
in bijective correspondence with points of An . A point a = (a1 , . . . , an ) of An
corresponds to the kernel Ma of the substitution map sa : K[x1 , . . . , xn ] → K
that sends xi to ai . The kernel Ma is generated by the n linear polynomials
x i − ai .
Proof. Let a be a point of An and let Ma be the kernel of sa . Since sa is
surjective and since K is a field, Ma is a maximal ideal. To verify that Ma
is generated by the linear polynomials as asserted, we first consider the case
that the point a is the origin (0, . . . , 0). We must show that the kernel of the
map s0 that evaluates a polynomial at the origin is generated by the variables
x1 , . . . , xn . Well, f (0, . . . , 0) = 0 if and only if the constant term of f is zero.
If so, then every monomial that occurs in f is divisible by at least one of
the variables, so f can be written as a linear combination of the variables,
with polynomial coefficients. The proof for an arbitrary point a can be made
13
using the change of variable xi = x0i + ai to move a to the origin.
It is harder to prove that every maximal ideal has the form Ma . Let M be
a maximal ideal, and let F denote the field K[x1 , . . . , xn ]/M . We restrict
the canonical projection map π : K[x1 , . . . , xn ] → F to the subring K[x1 ] of
polynomials in the first variable, obtaining a homomorphism φ1 : K[x1 ] → F.
We will now use the following result: If F is a field and φ : F [x] → R is a
homomorphism to an integral domain R, and P is the kernel of φ, then
either P is a maximal ideal or P = (0). Using this result, we get that the
kernel of φ1 is either the zero ideal or one of the maximal ideals (x1 − a1 )
of K[x1 ]. We will now show that it cannot be the zero ideal. Suppose that
Ker(φ1 ) = 0. Then φ1 maps K[x1 ] isomorphically to its image, a subring of
F. The mapping property of fraction fields shows that this map extends to
an injective map : K(x1 ) → F. So F contains a field isomorphic to K(x1 ).
But this is impossible, hence ker(φ1 ) 6= 0.
Hence the kernel of φ1 is a maximal ideal of the form < x1 − a1 >. The
same will be true if the index 1 is replaced by any other index. So M will
contain linear polynomials of the form xi − ai for each i. This will show that
M contains one of the ideals Ma , and since Ma is maximal, M will be equal
to that ideal.
Lemma 2.1.10. A maximal ideal m of A = K[x1 , . . . , xn ] corresponds to
a minimal irreducible closed subset of An , which must be a point, say p =
(a1 , . . . , an ).
Proof. It follows from the previous lemma that every maximal ideal of A is of
the form < x1 −a1 , . . . , xn −an >. Now a maximal ideal is a prime ideal, so it
corresponds to an irreducible closed subset of An . Since the ideal is maximal,
14
the corresponding irreducible closed subset of An is minimal. And since the
closed subset of An corresponding to the ideal < x1 − a1 , . . . , xn − an > is
the point (a1 , . . . , an ) the result follows.
2.2
Affine Coordinate Ring
Definition 2.2.1. If Y ⊆ An is an affine algebraic set, we define the affine
coordinate ring A(Y ) of Y , to be A/I(Y ).
Example 2.2.2. If Y = An , then A(Y ) = A = K[x1 , . . . , xn ].
Remark 2.2.3. If Y is an affine variety, then A(Y ) is an integral domain. Furthermore, A(Y ) is a finitely generated K− algebra. Conversely, any finitely
generated K− algebra B which is a integral domain is the affine coordinate
ring of some affine variety. Indeed, write B as the quotient of a polynomial
ring A = K[x1 , . . . , xn ] by an ideal a, and let Y = Z((a)).
Fact : Any finitely generated K− algebra B is a quotient of a polynomial
ring in finitely many variables.
Next we will study the topology of our varieties. To do so we introduce
an important class of topological spaces which includes all varieties.
15
Chapter 3
Dimension of affine algebraic set
3.1
Noetherian topological space
Definition 3.1.1. A topological space X is called noetherian if it is satisfies
the descending chain condition for closed subsets: for any sequence Y1 ⊇
Y2 ⊇ . . . of closed subsets, there is an integer r such that Yr = Yr+1 = . . ..
Example 3.1.2. An is a noetherian topological space.Indeed, if Y1 ⊇ Y2 ⊇
. . . is a descending chain of closed subsets, then I(Y1 ) ⊆ I(Y2 ) ⊆ . . . is an
ascending chain of ideals in A = K[x1 , . . . , xn ]. Since A is a noetherian ring,
this chain of ideals is eventually stationary. But for each i, Yi = Z(I(Yi )), so
the chain Yi is also stationary.
Proposition 3.1.3. In a noetherian topological space X, every nonempty
closed subset Y can be expressed as a finite union Y = Y1 ∪ . . . ∪ Yr of
irreducible closed subsets Yi . If we require that Yi + Yj for i 6= j, then the Yi
are uniquely determined. They are called the irreducible components of Y .
16
Proof. First we show the existence of such a representation of Y . Let s be the
set of nonempty closed subset of X which cannot be written as a finite union
of irreducible closed subsets. If s is nonempty, then since X is noetherian,
it must contain a minimal element, say Y . Then Y is not irreducible, by
construction of s. Thus we can write Y = Y 0 ∪ Y 00 , where Y 0 and Y 00 are
proper closed subsets of Y . By minimality of Y , each of Y 0 and Y 00 can be
expressed as a finite union of closed irreducible subsets, hence Y also, which
is a contradiction. We conclude that every closed set Y can be written as
a union Y = Y1 ∪ . . . ∪ Yr of irreducible subsets By throwing away a few if
necessary, we may assume Yi + Yj for i 6= j.
Now suppose Y = Y10 ∪ . . . ∪ Ys0 is another such representation. Then Y10 ⊆
Y = Y1 ∪ . . . ∪ Yr , so Y10 = ∪(Y10 ∩ Yi ). But Y10 is irreducible, so Y10 ⊆ Yi for
some i, say i = 1. Similarly, Y1 ⊆ Yj0 for some j. Then Y10 ⊆ Yj0 , so j = 1,
and we find that Y1 = Y10 . Now let Z = (Y − Y1 ). Then Z = Y2 ∪ . . . ∪ Yr
and also Z = Y20 ∪ . . . ∪ Ys0 . So proceeding by induction on r, we obtain the
uniqueness of the Yi .
Corollary 3.1.4. Every algebraic set in An can be expressed uniquely as a
union of varieties, no one containing another.
Proof. Since An is noetherian, therefore it follows from the definition of
noetherian topological space that any algebraic set is also noetherian.
3.2
Dimension of a Topological Space
Definition 3.2.1. If X is a topological space, we define the dimension of X
(denoted dim X) to be the supremum of all integers n such that there exists
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a chain Z0 ⊂ Z1 ⊂ . . . ⊂ Zn of distinct irreducible closed subsets of X. We
define the dimension of an affine variety to be its dimension as a topological
space.
Example 3.2.2. The dimension of A1 is 1. Indeed, the only irreducible
closed subsets of A1 are the whole space and single points.
Definition 3.2.3. In a ring A, the height of a prime ideal p is the supremum
of all integers n such that there exist a chain p0 ⊂ p1 ⊂ . . . ⊂ pn = p of
distinct prime ideals. We define the dimension (or Krull dimension) of A to
be the supremum of heights of all prime ideals.
Example 3.2.4. The Krull dimension of a field is zero.
Proposition 3.2.5. If Y is an affine algebraic set, then the dimension of Y
is equal to the dimension of its affine coordinate ring A(Y ).
Proof. If Y is an affine algebraic set in An , then the closed irreducible subsets
of Y correspond to prime ideals of A = K[x1 , . . . , xn ] containing I(Y ). These
in turn correspond to prime ideals of A(Y ). Hence dim Y is the length of
the longest chain of prime ideals in A(Y ), which is its dimension.
If A is a ring and I is an ideal of A, by I[x] we shall mean the set of all
polynomials in x with coefficients in I. It is easy to see that I[x] is an ideal
of A[x] such that I[x] ∩ A = I. Observe that p is prime ideal of A[x] if and
only if p ∩ A is a prime ideal of A, and that q is a prime ideal of A if and
only if q[x] is a prime ideal of A[x].
Theorem 3.2.6. Let A be a noetherian ring and I ⊆ A[x] be an ideal, then
ht(I) ≤ ht(I ∩ A) + 1.
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Proof. omitted.
Theorem 3.2.7. If A is a noetherian ring, then dim(A[x]) =dim(A) + 1.
Proof. observe that every chain
q0 $ . . . $ qs
of prime ideals of A gives a chain
q0 [x] $ . . . $ qs [x]
prime ideals of A[x]. Moreover, for any maximal ideal m of A, since the
quotient A[x]/m[x] ∼
/ Max(A[x]). Therefore,
= (A/m)[x] is not a field, m[x] ∈
it follows that
dim(A[x]) ≥ dim(A) + 1
If dim(A) = ∞, there is nothing to prove, so assume that dim(A) = d < ∞.
Suppose, if possible that dim(A[x]) >dim(A)+1. Then we can find a maximal
ideal m of A[x] such that ht(m) >dim(A) = d + 1. But then theorem (3.1.10)
implies that ht(m ∩ A) > d, contradicting that dim(A) = d. Therefore, we
must have
dim(A[x]) = d + 1 = dim(A) + 1
.
The following two corollaries are immediate:
Corollary 3.2.8. If A is a noetherian ring, then we have dim(A[x1 , . . . , xn ]) =dim(A)+
n.
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Corollary 3.2.9. The dimension of An is n.
The following 3 results are stated without proof:
Theorem 3.2.10. Let K be a field, and let B be an integral domain which
is a finitely generated K− lgebra.Then:
For a prime ideal p in B, we have
heightp + dimB/p = dimB
.
Theorem 3.2.11. Let A be a noetherian ring, and let f ∈ A be an element
which is neither a zero divisor nor a unit. Then every minimal prime ideal
p containing f has height 1.
Proposition 3.2.12. A noetherian integral domain A is a unique factorization domain if and only if every prime ideal of height 1 is principal.
Proposition 3.2.13. A variety Y in An has dimension n − 1 if and only
if it is the zero set Z(f ) of a single non constant irreducible polynomials in
A = K[x1 , . . . , xn ].
Proof. Suppose f is an irreducible polynomial, then Z(f ) is clearly a variety
and its ideal is the prime ideal p = hf i. Now, since f is irreducible, it is
neither a zero divisor nor a unit.
Moreover, p = hf i is a maximal prime ideal containing f . Hence by theorem
(3.2.11), p has height 1.
Now, by theorem (3.2.10), we get dimA/p =dim(A)−heightp, where A =
K[x1 , . . . , xn ].
Therefore A/p = n − 1.
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But A/p = A/hf i is nothing
but the affine coordinate ring of the variety Z(f ). By proposition (3.2.5),
dimA/p =dim(Z(f )).
Hence dim(Z(f )) = n − 1.
Conversely, suppose Y ⊆ An is an affine variety of dimension n − 1. Then
again by theorem (3.2.10), the ideal I(Y ) of Y will be a prime ideal of
height 1. Now the polynomial ring A = K[x1 , . . . , xn ] is a unique factorization domain. So by proposition (3.2.12), I(Y ) is a principal ideal, therefore I(Y ) = hf i for some f ∈ A. Since I(Y ) is a prime ideal, therefore
the polynomial f must be a non constant irreducible polynomial. Hence
Y = Y = Z(I(Y )) = Z(hf i) = Z(f ).
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Bibliography
[1] Algebraic Geometry, Robin Hartshorne, Graduate texts in Mathematics,
Springer, 1977.
[2] Algebra, second edition, Michael Artin, PHI Learning Private Limited,
Pearson Education Inc., 2011.
[3] Commutative Algebra Lecture Notes, Anand Prabhakar Sawant,
T.I.F.R Mumbai.
[4] Hilberts Nullstellensatz, Alina Ostafe, Seminar Computational Algebra.
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