CALIFORNIA STATE UNIVERSITY, NORTHRIDGE
The Zariski Topology on the Prime Spectrum of a Commutative Ring
A thesis submitted in partial ful llment of the requirements for the degree of Master of Science in Mathematics
by
Choomee Kim
December 2018 The thesis of Choomee Kim is approved:
Jason Lo, Ph.D. Date
Katherine Stevenson, Ph.D. Date
Jerry Rosen, Ph.D., Chair Date
California State University, Northridge
ii Dedication
I dedicate this to my beloved family.
iii Acknowledgments With deep gratitude and appreciation, I would like to rst acknowledge and thank my thesis advisor, Professor Jerry Rosen, for all of the support and guidance he provided during the past two years of my graduate study. I would like to extend my sincere appreciation to Professor Jason Lo and Professor Katherine Stevenson for serving on my thesis committee and taking time to read and advise. I would also like to thank the CSUN faculty and cohort, especially Professor Mary Rosen, Yen Doung, and all of my o ce colleagues, for their genuine friendship, encouragement, and emotional support throughout the course of my graduate study. All this has been truly enriching experience, and I would like to give back to our students and colleagues as to become a professor who cares for others around us. Last, but not least, I would like to thank my husband and my daughter, Darlene, for their emotional and moral support, love, and encouragement, and without which this would not have been possible.
iv Table of Contents
Signature Page...... ii
Dedication...... iii
Acknowledgements...... iv
Table of Contents...... v
Abstract...... vii
1 Preliminaries...... 1 1.1 Commutative Rings...... 1 1.1.1 Basic Notions...... 1 1.1.2 Noetherian Rings...... 6 1.1.3 Localization...... 8 1.2 Topology...... 10
2 An Introduction to the Zariski Topology on the Prime Spectrum of R . . . 14 2.1 Introduction...... 14 2.2 Mappings...... 20 2.3 Noetherian Topological Spaces...... 23
3 Connectedness and Dimension in Spec(R) ...... 25 3.1 Connectedness in Spec(R) ...... 25 3.2 Krull Dimension...... 28 3.3 Irreducibility...... 34
4 Worked Examples...... 36 4.1 ...... 36 Spec(Zn)
v 4.2 Spec(Z[x]) ...... 37 4.3 Spec of C(X) ...... 40 4.4 Spec of a Boolean Ring...... 45
Bibliography...... 50
vi Abstract
The Zariski Topology on the Prime Spectrum of a Commutative Ring
by
Choomee Kim
Master of Science in Mathematics
The Zariski Topology is an interesting topic in algebraic geometry that combines commutative algebra and topology to deal with questions that are algebraic and ge- ometrical in nature. It was introduced by Oscar Zariski in the mid twentieth century and has had great success helping to answer several long standing, famous questions, such as the solution(by Andrew Wiles in the 1990s) to Fermat's famous last theorem. Initially, algebraic geometry dealt with polynomial rings in nitely many variables over a eld. A great deal of research went to understanding the qualitative and quan- titative nature of solutions to polynomial equations in several variables. As is the case
in studying rings, the understanding the ideal structure of F [x1, x2, ··· , xn] where F is a eld of characteristic zero was of paramount importance. Initially, the maximal ideal structure of such rings played a vital role in algebraic geometry, and Hilbert's
Nullstellensats completely characterized the maximal ideal of F [x1, x2, ··· , xn] when F has characteristic zero. In particular, the maximal ideals of this ring are of the form
(x1 − a1, ··· , xn − an) where ai ∈ F . The signi cance of this result is that it provides
a natural one to one correspondence between the maximal ideals of F [x1, x1, ··· , xn] and the points in a ne n-space F n. However, one drawback to de ning topologies in
vii terms of spaces of maximal ideals is that under a homomorphism, from a commutative ring to itself, the inverse image of a maximal ideal may not be maximal. The problem with this is that ring homomorphisms, when extended to the maximal spectrum of the ring, may fail to be continuous. Hence, the maximal spectrum, while important, may not be the best topology to consider. Eventually, Zariski and others realized that the ideal topology is one that is de ned using the prime ideals of a commutative ring. This topology is called the prime spectrum of the commutative ring, and it has the maximal spectrum as a topological subspace. The primary purpose of this thesis is to investigate the prime spectrum of commutative ring. We will prove some general results about this spectrum and then specialize to speci c examples. For instance, we
will investigate the prime spectrum of Z[x], Z the ring of integers, and C(X), the ring of continuous functions on a compact Hausdor space X. One interesting feature of the Zariski topology on a commutative ring is that it often produces surprising prop- erties. For example, it is always compact, but rarely Hausdor. We will, however, give necessary and su cient conditions on a ring, which result in this topology being Hausdor. We will also discuss connectedness and see that properties on the ring will lead to the topology being connected. Hence, the aim of our thesis is to simply
investigate the connection between a prime ring R and the topological properties of Spec(R).
viii Chapter 1 Preliminaries 1.1 Commutative Rings
1.1.1 Basic Notions
In this thesis, all rings are assumed to be commutative.
De nition 1.1. A subset I of a ring R is called an ideal if it is closed under addition, contains 0 and satis es the ideal property. That is for any a ∈ I and r ∈ R, ar ∈ I.
De nition 1.2. An ideal P 6= R is called a prime ideal if ab ∈ P implies a ∈ P or b ∈ P . An ideal M 6= R is called a maximal ideal if M ⊆ I E R, then either M = I or I = R.
Proposition 1.1.
1. An ideal P 6= R is a prime ideal if and only if R/P is an integral domain.
2. An ideal M 6= R is a maximal ideal if and only if R/M is a eld.
Proof.
1. Assume P 6= R is a prime ideal and let (a + P )(b + P ) = ab + P = 0 + P . Then ab ∈ P and so a ∈ P or b ∈ P implies a + P = 0 + P or b + P = 0 + P . Thus R/P is an integral domain. Now assume R/P is an integral domain and let (a + P )(b + P ) = ab + P = 0 + P . Then ab ∈ P and either a + P = 0 + P or b + P = 0 + P which implies a ∈ P or b ∈ P . Thus P is a prime ideal.
2. First we assume M is a maximal ideal of R. If r + M 6= 0 + M, then r∈ / M and by maximality we have M + (r) = R. Then 1 = m + ar for some m ∈ M and a ∈ R which gives 1 + M = m + ar + M = ar + M = (a + M)(r + M)
1 proving r + M is invertible. Now we assume R/M is eld and let M ( I E R. If a ∈ I − M, then a + M 6= 0 + M and there exists b + M ∈ R/M such that
1 + M = ab + M, since R/M is eld. Hence 1 − ab ∈ M ( I and since ab ∈ I, we obtain 1 ∈ I. This implies I = R so M is maximal ideal.
De nition 1.3. An integral domain R is called a Principal Ideal Domain (PID) if
every ideal is principal of the form (a) = aR
Proposition 1.2. In a PID, every nonzero prime ideal is maximal.
Proof. Let P 6= (0) be prime ideal and P ( I E R. Since it is PID, we let P = (a) ( I = (b) E R. Then a ∈ (a) ( (b) gives a = bc ∈ (a), so we have b ∈ (a) or c ∈ (a) but b∈ / (a) by our assumption. Thus c = ar = bcr and 1 = br, showing (b) = R, so I = R.
Theorem 1.3. (First Isomorphism Theorem) If f : R → S is a ring epimorphism,
then R/Ker(f) ∼= S.
Proof. De ne f˜ : R/ker(f) → S by f˜(r + Ker(f)) = f(r). It is easy to check f˜ is well de ned ring homomorphism. If r + Ker(f) ∈ Ker(f˜) if and only if 0 + Ker(f) = f˜(r + Ker(f)) = f(r) if and only if r ∈ Ker(f) if and only if r + Ker(f) = 0. Hence f˜ is also one-to-one proving it is isomorphism.
Theorem 1.4. (Correspondence Theorem) For given onto homomorphism f : R → S, f induces a bijection between the ideal of R containing Ker(f) and all the ideal of S.
−1 −1 Proof. Let J/S and we have {0s} ⊆ J. Then Ker(f) = f ({0s}) ⊂ f (J) /R and f(f −1(J)) = J since f is onto proving it is onto. To prove one-to-one, we suppose A, B are ideals of R containing Ker(f) and f(A) = f(B). For any a ∈ A, f(a) ∈ f(B),
2 so there exists b ∈ B such that f(a) = f(b). Then f(a) − f(b) = f(a − b) = 0 by homomorphism, so a − b ∈ Ker(f) ⊆ B showing a ∈ B, proving A ⊆ B. Analogously B ⊆ A and so we have a one-to-one correspondence.
It is important to investigate what happens to prime and maximal ideals under this correspondence theorem.
Corollary 1.5. Let f : R → R0 be a ring epimorphism.
1. If P ⊇ Ker(f) is a prime ideal of R, then f(P ) is a prime ideal of R0.
2. If P 0 is a prime ideal of R0, then f −1(P 0) is a prime ideal of R.
3. If M ⊇ Ker(f) is a maximal ideal of R, then f(M) is a maximal ideal of R0.
4. If M 0 is a maximal ideal of R0, then f −1(M 0) is a maximal ideal of R.
Proof.
1. If f(1) = 1 ∈ f(P ), then f(1) = f(p) for some p ∈ P . This implies 1 − p ∈ Ker(f) giving 1 ∈ P which is contradiction. Thus 1 ∈/ f(P ). Now let f(a)f(b) = f(ab) ∈ f(P ). This implies ab ∈ P hence a ∈ P or b ∈ P since P is prime ideal of R. Thus f(a) ∈ P or f(b) ∈ P proving f(P ) is prime ideal of R0.
2. If 1 ∈ f −1(P 0), then f(1) = 1 ∈ P 0 a contradiction. Thus 1 ∈/ f −1(P 0). Now let a, b ∈ R and ab ∈ f −1(P 0). Then f(ab) = f(a)f(b) ∈ P 0. Since P 0 is prime ideal of R0, either f(a) ∈ P 0 or f(b) ∈ P 0. This implies either a ∈ f −1(P 0) or b ∈ f −1(P 0) proving f −1(P 0) is prime ideal of R.
3. If f(M) = R0 = f(R) since f is onto, then M = R, a contradiction. Now
0 −1 −1 suppose f(M) ( J E R . This implies M ( f (J) so f (J) = R since M is a maximal ideal. Thus J = f(R) = R0 proving f(M) is a maximal ideal of S.
3 0 −1 4. Let N be a maximal ideal of R and suppose f (N) ( I E R. This implies N ( f(I) so f(I) = R0 since N is a maximal ideal. Then R0 = f(R) and R = I by onto and Correspondence Theorem above proving f −1(N) is maximal ideal of R.
We are now able to characterize the prime and maximal ideals of a quotient ring.
Corollary 1.6. The prime and maximal ideals of R/I are of the form P/I and M/I respectively, where P is a prime ideal and M is a maximal ideal of R.
As we proved, in a PID, every nonzero prime ideal is maximal ideal. This leads to an characterization of the prime and maximal ideals of Z and F [x] where F is a eld. In these rings, the only prime ideal which is not maximal is (0). The maximal ideals are generated by prime(=irreducible) elements. Hence, maximal ideals of Z are of the form pZ, p a prime. In F [x], the maximal ideals are of the form (f(x)), f(x) is irreducible. However, they are not simple in Z[x]. First note that Z[x] is not a PID as (2, x) is not principal. Hence, the determination of the prime and maximal ideals of Z[x] is not as easy.
De nition 1.4. De ne by where if P i, then φ : Z[x] → Zp[x] φ(f) = f f = aix P i f = aix . The map φ is a ring homomorphism with Ker(φ) = (p)(x) and is called reduction of coe cient mod p.
Theorem 1.7. The prime ideals of Z[x] are either (0), (f) where f is irreducible in and hence in , or where is prime in and is irreducible in . Z[x] Q[x] (p, g) p Z g Zp[x] The last one is maximal ideal of Z[x].
Proof. Let P 6= 0 be a prime ideal of Z[x] and we rst assume P = (f) is principal. Claim 1. f is irreducible. Otherwise, f = gh and neither g nor h is ±1. So gh ∈ P
4 and one of them, say g is in P . Thus g = ff1 and f = ff1h so 1 = f1h showing h = ±1, a contradiction.
Hence, we can assume there exists polynomial f1, f2 ∈ P with no common factors in
Z[x]. Claim 2. have no common factors in . Suppose not and let be a common f1, f2 Q[x] h factor. Then and for and degree of is greater or f1 = hg1 f2 = hg2 h, g1, g2 ∈ Q[x] h equal to 1. There exists elements such that , and α, β, γ ∈ Q h = αh0 g1 = βq1 g2 = γq2 where are primitive polynomials in . By Gauss' Lemma, are h0, q1, q2 Z[x] h0q1, h0q2 also primitive polynomials. Now , and f1 = hg1 = (αh0)(βq1) = (αβ)(h0q1) ∈ Z[x] this gives . Similarly we have . Hence divides , and divides , αβ ∈ Z αγ ∈ Z h0 f1 h0 f2 a contradiction to claim 1 above. Hence, in . gcd(f1, f2) = 1 Q[x] Now there exists such that since is a PID. If is u(x), v(x) ∈ Q[x] 1 = uf1 + vf2 Q[x] c common denominator for all rational coe cient of and , u v c = cuf1+cvf2 ∈ (f1f2)∩Z and . Hence if is prime ideal that is not principal ideal, then c ∈ (f1f2) ∩ Z 6= (0) P Z∩P 6= (0). It is easy to check that Z∩P is prime ideal of Z and hence Z∩P = (p) for some prime integer . Consider reduction of coe cient mod map, . p p φ : Z[x] → Zp[x] This is onto homomorphism with Kerφ = (p)[x] ( P . By Correspondence Theorem 1.3, is prime ideal of and hence maximal ideal. This is also principal φ(P ) Zp[x] ideal since is PID. Hence and must be monic irreducible in Zp[x] φ(P ) =< g > g . Also by Theorem 1.3 again, is maximal ideal in . Now for any , Zp[x] P Z[x] f ∈ P φ(f) = f = gh so p divides f − gh in Z[x]. Then f − gh = pt for some t ∈ Z[x], and so f = pt + gh ∈< p, g >. We have P ⊆< p, g >6= Z[x], so P =< p, g > since P is maximal ideal.
De nition 1.5. An element a ∈ R is called nilpotent if there exists natural number n such that an = 0. Let N(R) be the set of all nilpotent element.
De nition 1.6. The set of all prime ideals of R will be denoted Spec(R).
5 Theorem 1.8. N(R) = ∩P where P ∈ Spec(R).
Proof. If x ∈ N(R), then xn = 0 for some natural number n. Thus for any P in Spec(R), 0 = xn ∈ P so x ∈ P and x ∈ ∩P . Now we suppose x ∈ ∩P where P ∈ Spec(R) and assume x is not nilpotent element. Then xn ∈/ (0) for any natural
n number n. Let J = {I E R : x ∈/ I for any n ≥ 1}. Note that J is not empty since (0) ∈ J. Then J has a maximal element P , by Zorn's Lemma. We claim P is prime ideal. If not, there exists ab ∈ P but a, b∈ / P . Hence, P ( P + (a) and P ( P + (b). Since P is maximal element, there exists natural number m, n such that xm ∈ P +(a)
n m n m+n and x ∈ P + (b). Thus x = p1 + ra and x = p2 + sb and x ∈ P + (ab) = P , a contradiction to our construction of set J. Thus P is prime ideal and x ∈ ∩P so we get x ∈ P , a contradiction. Hence x must be nilpotent element.
Remark. Theorem 1.8 shows that N(R) is ideal of R. √ De nition 1.7. If I is a ideal of R, the radical of I is the set rad(I) = I = {a ∈ √ R : an ∈ I for some n ≥ 1}. I is called a radical ideal if I = I. √ Remark. a ∈ I if and only if an ∈ I for some natural number n if and only if (a + I)n = 0 + I if and only if π(a) = a + I ∈ N(R/I) if and only if a ∈ π−1(N(R/I)). Thus we have rad(I) = π−1(N(R/I)) where π is canonical homomorphism. Hence
rad(I) E R.
1.1.2 Noetherian Rings
De nition 1.8. R is called a Noetherian ring if every ideal is nitely generated.
De nition 1.9. A ring satis es the Ascending Chain Condition on ideals if given
I1 ⊆ I2 ⊆ · · · , there exists some n such that In = In+1 = ··· .
Theorem 1.9. The followings are equivalent where R is a ring.
1. R is Noetherian ring
6 2. R satis es Ascending Chain Condition on ideals
3. every nonempty collection of ideal contains a maximal element.
Proof. 1 2: Suppose we have and let ∞ . Since is an ⇒ I1 ⊆ I2 ⊆ · · · I = ∪i=1Ii I ideal of R, then I = (a1, a2, ··· , an) and there exists su ciently large m such that
ai ∈ Im for any i = 1, 2, ··· , n. Thus we get (a1, a2, ··· , an) = I ⊆ Im and hence ∞ showing it satis es ascending chain condition. ∪i=1 = I = Im = Im+1 = ··· 2 ⇒ 3: Let S be nonempty collection of ideals. We suppose S does not have a maximal
element. We choose I1 ∈ S and I1 not maximal implies there exists I2 ∈ S such that since is not maximal ideal. In a similar manner, we get I1 ( I2 I1 I1 ( I2 ( I3 ( ··· contradicting to ascending chain condition.
3 ⇒ 1: Let I be ideal of R and S = {J : J ⊆ I and J is nitely generated}. S is not
empty since (0) is in there. Thus S has maximal element K satisfying K ( I and
K is nitely generated. Let K = (a1, ··· , an). We claim I = K. If it is not, there exists and . This contradicts that is maximal b ∈ I − K K ( (a1, ··· , an, b) ⊆ I K and nitely generated ideal. Thus ideal I of R is nitely generated and hence R is Noetherian.
Theorem 1.10. (Hilbert Basis Theorem) If R is a Noetherian ring, then R[x] is also a Noetherian ring.
Proof. Let and de ne n . Then I E R[x] In = {a ∈ R : f = ax +· · · ∈ I}∪{0} I0 = {a ∈
R : a ∈ I} and I1 = {a ∈ R : ax+b ∈ I}. It is easy to check that In is ideal of R. Thus there are nitely many generators such that and such In = (bn1 , bn2 , ··· bnk ) fnj ∈ I that n . We have and , and there fnj = bnj x +··· fn1 , fn2 , ··· , fnk ∈ I I0 ⊆ I1 ⊆ I2 ⊆ · · ·
exists m such that Im = Im+1 = ··· since R is Noetherian by Theorem 1.7. We claim
I = {fij : 0 ≤ i ≤ m, 1 ≤ j ≤ k}. We check this by induction on degree of f ∈ I. If
degree of f is 0, f ∈ R so f ∈ (b01, b02, ··· , b0k). Now assume degree of f is bigger that
0 and all polynomial of degree less than n are generated by {fij} as above. If degree of
7 n f is less than or equal to m, f = cx ··· , so c ∈ In and c = r1bn1 + r2bn2 + ··· + rkbkn. Let Pkn n n . Then degree of g = f − i=1 rifni = f −(r1(bn1x +··· )+···+rkn(bnkx +··· )) g is less than n, it works by induction hypothesis. Now assume degree of f is greater than m. Then Pkm n−m of degree less than n. Thus as g = f − ( i=1 rifmi)x ∈ I g ∈ (fij) above and we get f ∈ (fij). Hence R[x] is Noetherian if R is Noetherian ring.
Remark. By Hilbert basis Theorem and induction, R[x1, ··· , xn] where R is Noethrian ring and F [x, y] where F is eld are also Noetherian rings.
De nition 1.10. R is called a Artinian ring if it satis es descending chain condition
on ideals. That is, for given I1 ⊇ I2 ⊇ · · · , there exists n such that In = In+1 = ··· .
Proposition 1.11. If R is an Artinian ring, every prime ideal is maximal ideal.
Proof. Suppose P is prime ideal of R and let x∈ / P . If we consider (xm) as ideal of R, there exists n such that (xn) = (xn+1) since it is Artinian ring. Hence there exists r ∈ R such that xn = rxn+1. Now we have xn − rxn+1 = xn(1 − rx) = 0 showing 1 − rx ∈ P . Thus we get R = P + Rx proving P is maximal ideal.
1.1.3 Localization
We recall that if R is an integral domain then we can embed it in a eld F where elements are fractions of elements from R. Instead of reviewing this construction, we will show how it is possible to embed a commutative ring R into a localization of R which can be a better behaved commutative ring. Suppose R is a commutative ring
and S is a multiplicative subset. That is, 1 ∈ S and for any s1, s2 ∈ S, s1s2 ∈ S. Let R × S = {(r, s) : r ∈ R and s ∈ S} and de ne a relation ∼ on R × S by
(r1, s1) ∼ (r2, s2) if there exists t ∈ S such that t(r1s2 − r2s1) = 0.
Proposition 1.12. The relation ∼ above is an equivalence relation on R × S.
8 Proof. It is re exive since there exists t = 1 ∈ S such that t(r1s1 − r1s1) = 0. That is,
(r1, s1) ∼ (r1, s1). Also it is symmetric since if (r1, s1) ∼ (r2, s2) with t(r1s2 −r2s1) = 0
for t ∈ S, then (r2, s2) ∼ (r1, s1) with −t(r2s1 − r1s2) = 0. Now let (r1, s1) ∼ (r2, s2)
and (r2, s2) ∼ (r3, s3). Then there exists t1, t2 ∈ S such that t1(r1s2 − r2s1) = 0 and
t2(r2s3 − r3s2) = 0. Multiplying these two equation by t2s3 and t1s1 respectively gives
t1t2s2(r1s3 − r3s1) = 0 showing (r1, s1) ∼ (r3, s3). Thus it is transitive.
We write r for . We introduce arithmetic on the set of equivalence classes. Let s [(r, s)] RS−1 denote the set of equivalence class. Addition is de ned as r1 + r2 = r1s2+r2s1 , s1 s2 s1s2 and multiplication is de ned as r1 r2 = r1r2 . It is easy to check that addition and s1 s2 s1s2 multiplication are well de ned and the other ring axioms are easily checked. We note
that zero in −1 is 0 and in −1 is 1 . −1 r is a ring called RS 1 1 RS 1 RS = { s : r ∈ R, s ∈ S} the ring of fraction, and elements in becomes invertible in −1 since s −1 1 S RS ( 1 ) = s and s 1 1 . 1 s = 1
Example.
1. Let P be a prime ideal and S = R − P . This S is a multiplicatively closed set.
−1 Then RS is denoted as Rp, and it is a local ring having unique maximal ideal p . PRp = { s : p ∈ P, s ∈ S}
n −1 2. Let R be any commutative ring and S = {s : n ≥ 0}. We denote RS as Rs.
Proposition 1.13. Let R be a commutative ring and S be subset of R that is multiply closed. Then,
−1 −1 1. If I E R, then IS E RS .
−1 −1 2. If J E RS , then J = IS for some I E R.
3. There exists one-to-one correspondence between {P : P ∈ Spec(R),P ∩ S = ∅} and Spec(RS−1) given by P → PS−1.
9 Proof.
1. It is easy to check that IS−1 is closed under addition and the ideal property, so it is an ideal.
2. Let a . Then a a 1 if and only if a s a and I = {a ∈ R : s ∈ J} s = 1 s ∈ J s 1 = 1 ∈ J can be written as a . Considering the map −1, I {a ∈ R : 1 ∈ J} l : R → RS −1 . Now we want to show −1. If a −1, and so I = l (J) E R J = IS λ = s ∈ S a ∈ I a . Thus a a 1 . If a −1, a and so . Thus −1. 1 ∈ J s = 1 s ∈ J µ = s ∈ S 1 ∈ J a ∈ J µ ∈ IS
3. Let Q be prime ideal of RS−1 and l : R → RS−1. Then l−1(Q) is prime ideal of R and Q = PS−1 where P = l−1(Q) ∈ spec(R) by part 2. Now suppose p ∈ P ∩S. Then , so p and p 1 1 , a contradiction. To show one-to-one, p ∈ P 1 ∈ Q 1 p = 1 ∈ Q −1 −1 suppose P1S = P2S . Then Pi ∩ S = ∅ and there exists p1 ∈ P1, p2 ∈ P2
such that p1 p2 . Thus there exists such that and so 1 = s t ∈ S t(sp1 − p2) = 0
tsp1 = tp2 showing tsp1 ∈ P2. Since S ∩ P2 = ∅, p1 ∈ P2 showing P1 ⊆ P2. We
also have P2 ⊆ P1 analogously, so it is one-to-one.
1.2 Topology
De nition 1.11. Let X be a topological space. X is called
1. T0 if for any two distinct points x, y ∈ X, there exists open set containing one but not the other.
2. T1 if for any two distinct points x, y ∈ X, there exists open sets U and V containing x and y each such that y∈ / U and x∈ / V .
3. T2 or Hausdor if for any two distinct points x, y ∈ X, there exists disjoint open sets U and V such that x ∈ U and y ∈ V .
10 4. T3 or regular if for any x ∈ X and closed subset B of X such that x∈ / B, there exists disjoint open sets U and V such that x ∈ U and B ⊆ V .
5. T4 or normal if for any disjoint closed set A and B, there exists disjoint open sets U and V such that A ⊆ U and B ⊆ V .
Remark. By De nition above, T(n+1) implies T(n).
Proposition 1.14. The topological space X is T1 if and only if one point sets are closed.
Proof. Suppose X is T1 and let x ∈ X. Then for any y ∈ X − {x}, there exists open set Uy with y ∈ Uy and x∈ / Uy. Then U = ∪Uy is open and complement of {x}. This means {x} is closed. Now suppose one point set is closed and let x, y ∈ X with x 6= y.
Then X − {x} is open with y ∈ X − {x}. Thus X is T1.
De nition 1.12. A space X is compact if for given any collection of open set that covers X, there exists nite subcover.
Theorem 1.15. Every closed subset of a compact space is compact.
Proof. Let X be a compact space and Y be closed set in X. Also let Oy = {Y ∩Uα} be
open cover of Y . Note that X−Y is open since Y is closed. Thus Y ⊆ ∪(Y ∩Uα) implies
X ⊆ (∪(Y ∩Uα)∪(X −Y )). Since X is compact, there is a nite subcover. If X −Y/∈ nite subcover of , then so . X X ⊆ Uα1 ∪ · · · ∪ Uαn Y ⊆ (Y ∩ Uα1 ) ∪ · · · ∪ (Y ∩ Uαn ) Now if nite subcover of , , so X − Y ∈ X X ⊆ Uα1 ∪ · · · ∪ Uαn ∪ (X − Y ) Y ⊆ . In either case, we get nite subcover for . (Y ∩ Uα1 ) ∪ · · · ∪ (Y ∩ Uαn ) Y
Theorem 1.16. Every compact subset of Hausdor space X is closed.
Proof. Let Y be compact subset of X. If x0 ∈ X − Y , then for any y ∈ Y there exists disjoint open set Uy,Vy of x0, y respectively. Since {Vy} is an open cover of Y , there
11 exists a nite subcover . Then and Vy1 , ··· ,Vyn Y ⊆ V = Vy1 ∪ · · · ∪ Vyn x0 ∈ Uyi for so n . We have and . Hence i = 1, ··· , n x0 ∈ ∩i=1Uyi = U U ∩ V = ∅ U ∩ Y = ∅ x0 ∈ U ⊆ X − Y proving X − Y is open so Y is closed.
Theorem 1.17. Every compact Hausdor space is normal.
Proof. Let X be a compact Hausdor space. We will rst show it is regular space. We suppose x ∈ X and x∈ / B, for closed set B in X. Then B is compact by Theorem 1.15. Thus there exists disjoint open set containing x and B proving it is regular space (This follows from the part of Theorem 1.16 proof). Now suppose A, B are disjoint closed set in X (hence they are both compact). Now for any a ∈ A there exists a
disjoint open sets Ua and Va containing a and B respectively. Since A is compact, there are nitely many . Then and Ua1 , ··· ,Uan U = Ua1 ∪· · ·∪Uan V = Va1 ∩· · ·∩Van provide disjoint open set containing A, B respectively proving it is normal space.
De nition 1.13. The space X is connected if it can not be written as the disjoint union of two nonempty open subsets. That is, the only clopen subsets are X and ∅.
Proposition 1.18. A product of nitely many connected spaces is connected.
Proof. It is enough to show X × Y is connected if X and Y are connected spaces. Let
y be xed element of Y and x be arbitrary element of X. Considering the set Ux = ({x} × Y ) ∪ (X × {y}), {x} × Y is homeomorphic to Y and X × {y} is homeomorphic
to X. Also, ({x} × Y ) ∩ (X × {y}) = (x, y). Thus Ux is connected since it is union of connected set with nonempty intersection. Hence we have ∩x∈X Ux = X × {y} and
∪x∈X Ux = X × Y proving X × Y is connected.
De nition 1.14. The space X is path-connected if for any x, y ∈ X, there exists a continuous function f : [0, 1] → X such that f(0) = x and f(1) = y. We remark that path-connected implies connected.
12 Theorem 1.19. (Urysohn's Lemma) Suppose X is a normal space and C,D are
disjoint closed subsets of X. Then there exists a continuous function f : X → R such that f(x) ⊂ [0, 1] and f(C) = 0 and f(D) = 1.
Urysohn's lemma states that if a topological space is normal, then any two disjoint closed subsets can be separated by a continuous function. We skip the proof and we will use this lemma later in this thesis to construct and use a continuous function on normal space.
13 Chapter 2
An Introduction to the Zariski Topology on the Prime Spectrum of R The principal topic in this thesis is the study of a certain topology de ned on the collection of all prime ideals of a commutative ring R.
2.1 Introduction
De nition 2.1. The set of all prime ideal of R is called Prime Spectrum of R and will be denoted Spec(R). The subset of spec(R) consisting of the maximal ideals will be denoted Maxspec(R) and is called Maximal Spectrum.
Example.
1. Spec(Z) = {(0), (2), (3), (5), (7)...} and
Maxspec(Z) = {(2), (3), (5), (7)...}
2. For any eld F, Spec(F ) = MaxSpec(F ) = {(0)}
3. By the Fundamental Theorem of Algebra, Spec(C[x]) = {(0), (x − a) : a ∈ C} and Maxspec(C[x]) = {(x − a) : a ∈ C}
4. More generally, if F is a eld, we know F [x] is a PID and in this case every nonzero prime ideal is maximal and generated by an irreducible polynomials.
Hence, SpecF [x] = {(0), (f(x)) : f(x)irreducible}. So, for example, over the real eld, the nonzero prime ideals are generated by linear polynomials and irreducible quadratic polynomials. As we know, over the rational eld we can have irreducible polynomials of any positive degree.
5. By Theorem 1.7, we have that Spec(Z[x]) = {(0), (g(x)), (p, f(x))}, where g(x) is irreducible in Z[x] (and hence in Q[x]), p is a prime integer and f(x) is ir-
14 reducible mod p. Note that Maxspec(Z[x]) = {(p, f(x))}, for p and f(x) as above.
Later in this thesis, we investigate Spec(R) where R is over a wider variety of rings. The topology de ned on Spec(R) will be given in terms of closed sets, whose de nition is given by the following.
De nition 2.2. If S is subset of a ring R, then V (S) = {P ∈ Spec(R) : S ⊆ P }
Remark. It is easy to see that V (S) = V ((S)), where (S) is ideal of R generated by S. Hence, we will usually only have to deal with sets of the form V (I) where I is an ideal of R.
Theorem 2.1. The collection of all V (I), that is where I runs over all ideals of R, are the closed sets for a topology on SpecR.
1. ∅ and Spec(R) are closed.
2. The collection V (I) is closed under nite unions.
3. The collection V (I) is closed under arbitrary intersections.
Proof. We will show the V (I)'s satisfy the axioms for closed sets:
1. V (R) = {P ∈ Spec(R) : R ⊆ P } = ∅ and V ({0}) = {P ∈ Spec(R) : (0) ⊆ P } = Spec(R).
2. Claim: n Qn . ∪j=1V (Ij) = V ( j=1 Ij) If for some then , which implies Qn , proving P ∈ V (Ij) j Ij ⊆ P j=1 Ij ⊆ P n Qn . The reverse inclusion follows from the property that ∪j=1V (Ij) ⊆ V ( j=1 Ij) if a prime ideal contains the product of nitely many ideals, then it must contain one of the ideals.
15 3. Claim: P , where P represent the ideal consisting of all ∩jV (Ij) = V ( j Ij) j Ij
nite sums of elements from the Ij's.
If for all , then for all , which immediately gives P , P ∈ V (Ij) j Ij ⊆ P j j Ij ⊆ P proving P . Now, if P , then for all , P ∩jV (Ij) ⊆ V ( j Ij) P ∈ V ( j Ij) j Ij ⊆ j Ij ⊆ giving . Thus, P . P P ∈ ∩jV (Ij) V ( j Ij) ⊆ ∩jV (Ij)
The preceding theorem implies that V (I)s are the closed sets for a certain topology on Spec(R), called the Zariski topology, which will be denoted Z(R). We will now take up the problem of determining the open sets for the Zariski topology.
Remark. Naturally, Spec(R) − V (I) is open in Z(R) for any I, a ideal of R. Also, X − V (I) = X − ∩V ({a}) = ∪(X − V ({a})) for a ∈ I by DeMorgan's law. This is union of open set, so it is open.
De nition 2.3. For all a ∈ R, let D(a) = {P ∈ SpecR : a∈ / P }
We will prove that the collection of all D(a) forms a basis for the Zariski topology on Spec(R). Before doing so, we gather up some elementary properties of these sets.
Proposition 2.2. For a commutative ring R, the following holds:
1. D(a)= ∅ if and only if a ∈ N(R), the nilradical of R
2. D(a) = SpecR if and only if a ∈ U(R), the unit of R
3. For all a, b ∈ R, D(ab) = D(a) ∩ D(b)
Proof.
1. D(a) = ∅ implies a ∈ P for all prime ideals P of R, thus a ∈ N(R). On the other hand, a ∈ N(R) means that a is nilpotent and so an = 0, for some natural
16 number n. Thus, for any prime ideal P we have an ∈ P , which yields a ∈ P and consequently we must have D(a) = ∅.
2. If D(a) = Spec(R), then a is not in any prime ideal and so it is not in any maximal ideal. Since every non-unit is contained in some maximal ideal, a must be a unit. Since a prime ideal cannot contain an invertible element, it follows
that a is not in any prime ideal and so D(a) = Spec(R).
3. This is an immediate consequence of the de nition of prime ideal.
√ Proposition 2.3. For any ideal I of a commutative ring R, V (I) = V ( I). Further- more, there is a one-to-one correspondence between radical ideals of R (that is, ideals √ satisfying I = I) and the closed sets in Spec(R). √ √ Proof. Since I ⊆ I always holds, we have V ( I) ⊆ V (I). Now Q ∈ V (I) implies √ √ I ⊆ Q and hence I = ∩{P : P ∈ Spec(R),I ⊆ P } ⊆ Q implies Q ∈ V ( I). Now, let φ : {radical ideals of R} → {closed sets in Spec(R)} by φ(I) = V (I). To show one-to-one, assume I,J are radical ideals such that V (I) = V (J). Now √ √ √ I = ∩{P : P ∈ Spec(R),I ⊆ P } and J = ∩{P : O ∈ Spec(R),J ⊆ P }. If x ∈ I and let Q ∈ Spec(R) such that Q ∈ V (J) = V (I), implying I ⊆ Q. So, x ∈ Q and √ √ √ hence is in every prime ideal containing J, giving x ∈ J. Thus, I ⊆ J and the reverse inclusion is analogous. Hence the correspondence is one-to-one. For onto, √ √ √ recall that for any ideal I, I is radical ideal. Thus φ( I) = V ( I) = V (I), giving onto.
Theorem 2.4. The sets D(a) are open and form a basis for the Zariski topology.
Proof. To establish the rst statement, if su ces to show that Spec(R) − D(a) = V ((a)). Now, P ∈ Spec(R) − D(a) if and only if P/∈ D(a) if and only if a ∈
17 P if and only if P ∈ V ((a)). To prove the second statement, it su ces to show
Spec(R) − V (I) = ∪a∈I D(a), where I is and ideal of R. We have P ∈ Spec(R) − V (I) if and only if I * P if and only if there exists some a ∈ I − P if and only if
P ∈ ∪a∈I D(a).
Example. We look at the closed and basic open sets in Spec(Z): For a prime p, D(p) = {(0), (q) : q a prime 6= p}. It is easy to see that for any natural number , m . Hence, for any integer m1 m2 mk , m ≥ 1 D(p ) = D(p) a > 1, a = p1 p2 ...pk so we have m1 m2 mk m1 m2 mk D(a) = D(p1 p2 ...pk ) = D(p1 ) ∩ D(p2 )... ∩ D(pk ) = D(p1) ∩
... ∩ D(pk) = {(0), (q): q a prime 6= pi, i = 1, 2, ..k}. For a prime p, since (p) is maximal, m . For any integer m1 m2 mk , we V ((p)) = {(p)} = V ((p )) a > 1, a = p1 p2 ...pk have V ((a)) = {(p1), .., (pk)}.
The following result gives a characterization of the closure of a subset of Spec(R).
De nition 2.4. If Y ⊆ Spec(R), let 4(Y ) = ∩P ∈Y P , where the intersection runs over all the elements of Y .
Proposition 2.5. With the notation as above, we have Y = V (4(Y )).
Proof. For a prime ideal Q ∈ Y , we have 4(Y ) = ∩P ∈Y P ⊆ Q, and thus Q ∈ V (4(Y )). Consequently Y ⊆ V (4(Y )) and thus Y ⊆ V (4(Y )). For the reverse inclusion we will show that any closed set containing Y must contain V (4(Y )). If Y ⊆ V (A), then if P ∈ Y we must have A ⊆ P and this yields A ⊆ 4(Y ). Consequently, V (4(Y )) ⊆ V (A), proving V (4(Y )) is the smallest closed set containing Y .
Corollary 2.6. Spec(Z) is not T 1. However, Maxspec(Z) is T 1, but not T 2.
Proof. In Z, 4({0}) = the intersection of all the prime ideals of Z = (0). By the preceding theorem, we have {0} = V (4({0})) = V ({0}) = Spec(Z). Thus, {0}
18 is not closed and hence Spec(Z) is not T 1. Due to the fact that Z is a PID, it follows that the singletons in Maxspec(Z) are {(p)} = V ((p)) and are hence closed. It follows that Maxspec(Z) is T 1. Finally, the basic open sets in Maxspec(Z) (using the subspace topology inherited from Spec(Z)) are of the form Maxspec(Z)∩D(a) = {Q ∈ Maxspec(Z) : a∈ / Q}. If a 6= b are integers greater than 1, choose a prime p, which does not divide either a or b. Then (p) ∈ D(a) ∩ D(b), proving Maxspec(Z) is not T 2.
Remark. Actually, Spec(R) is always T 0 , and Maxspec(R) is always T 1 for any
commutative ring. For any two distinct prime ideal P and Q, P * Q or Q * P . Then P ∈ Spec(R) − V (Q) or Q ∈ Spec(R) − V (P ) showing it is T 0. Now, for any two
distinct maximal ideal M1 and M2, we can choose a1 ∈ M1 − M2 and a2 ∈ M2 − M1.
Hence M2 ∈ D(a1) − D(a2) and M1 ∈ D(a2) − D(a1) showing it is T 1.
Theorem 2.7. For any commutative ring R, Spec(R) is compact.
Proof. It su ces to show any cover of Spec(R) by basic open sets, has a nite sub-
cover. Assume Spec(R) ⊆ ∪α∈AD(aα) and let I =< {aα : α ∈ A} >, be the ideal of
R generated by the aα. Hence, for any P ∈ Spec(R),P ∈ D(aα) for some α. Thus, I cannot be contained in any prime ideal of R, which implies V (I) = ∅. Since ev- ery proper ideal is contained in a maximal, and hence prime ideal, we must have that . However, this means that Pn for . Now, for any I = R 1 = i=1 riai i = 1, 2, .., n
P ∈ Spec(R), 1 ∈/ P , which means aαi ∈/ P , for some i = 1, ..n. Hence, P ∈ D(aαi), for some , proving n , showing is compact. i = 1, ...n Spec(R) ⊆ ∪i=1D(aαi) Spec(R)
Remark. One unusual feature of the Zariski topology on Spec(R) is that the basic open sets D(a) are compact. This can be proven directly, in a manner similar to the preceding proof. However, we will give an alternate proof using mapping and localization later in this chapter.
19 2.2 Mappings
Given that the Spec of ring is a topological space, it is natural and useful to nd continuous functions on Specs. It turns out that homomorphisms between rings induce continuous functions between their spectrums.
Theorem 2.8. Suppose φ : R → S is ring homomorphism. Then φ induces a contin- uous function φ∗ : Spec(S) → Spec(R) given by φ∗(Q) = φ−1(Q). If φ is onto, φ∗ is one-to-one, and if φ is one-to-one, then φ∗ is onto.
Proof. We show φ∗ is continuous by proving the pre-image of a closed set is closed. In fact, we claim (φ∗)−1(V (I)) = V (φ(I)) (note: in general, the image of an ideal under a homomorphism may not be an ideal, but recall that the closed sets are de ned for any subset of a ring.) If Q ∈ (φ∗)−1(V (I)), Then φ∗(Q) = φ−1(Q) ∈ V (I). Hence I ⊆ φ−1(Q) giving φ(I) ⊆ φ(φ−1(Q)) ⊆ Q, showing Q ∈ V (φ(I)). Now if Q ∈ V (φ(I)), then φ(I) ⊆ Q and thus I ⊆ φ−1 = φ∗(Q), giving Q ∈ (φ∗)−1(V (I)). If φ is onto, assume φ∗(P ) = φ∗(Q), implying φ−1(P ) = φ−1(Q) and φ onto yields P = Q Finally, if φ is one-to-one, then for any P ∈ Spec(R) we have φ∗(φ(P )) = φ−1(φ(P )) = P , proving φ∗ is onto.
Example.
1. If φ : R → S is onto, then φ∗ : Spec(S) → Spec(R) may not be onto.
Consider the onto substitution homomorphism φ : Z[x] → Z given by x → 0. Also consider φ∗ : Spec(Z) → Spec(Z[x]). The prime ideal of Z are (0) and (p), where p is prime number. Then, φ∗((0)) = φ−1(0) = (x) and φ∗((p)) = (p, x).
These prime ideals do not cover all the prime ideal of Z[x], that is, they do not include the prime ideals of the form pZ[x]. Hence, φ∗ can not be onto.
20 2. If φ is one-to-one, then φ∗ may not be one-to-one.
Consider the one-to-one natural embedding φ : Z → Z[x] given by φ(a) = a. Also, consider φ∗ : SpecZ[x] → Spec(Z). Then, φ∗((x)) = φ−1((x)) and a ∈ φ−1((x)) if and only if φ(a) = a ∈ (x) if and only if a = 0. Hence, φ∗((x)) = (0). Similarly, φ∗((x + 1)) = (0), but (x) 6= (x + 1) proving φ∗ is not one-to-one.
The next result reduces the study of the spec of an arbitrary commutative ring to the spec of a ring with no nonzero nilpotent elements.
Corollary 2.9. For any commutative ring R, Spec(R) is homeomorphic to Spec(R/N(R)), where N(R) denotes the nilradical of R.
Proof. By the preceding result, the canonical surjection π : R → R/N(R) induces the one-to-one, continuous map π∗ : Spec(R/N(R)) → Spec(R), given by π∗(P/N(R)) = P , where P is an arbitrary prime ideal of R. By the correspondence theorem, P/N(R) ∈ Spec(R/N(R)) and hence π∗ is onto. To show it is a homeomorphism, it su ces to show that it is a closed map. Let V (I/N(R)) be an arbitrary closed set in Spec(R/N(R)). It su ces to show that π∗(V (I/N(R))) = V (I). Now, P/N(R) ⊇ I/N(R) if and only if π−1(P/N(R)) ⊇ π−1(I/N(R)) if and only if P ⊇ I. Hence π∗ is a homeomor- phism.
We now consider the relationship between Spec(R) and Spec(S−1R) where S−1R is localizataion. By Proposition 1.13, there is one-to-one correspondence between prime ideal of S−1R and prime ideals of R disjoint from S. That is, an arbitrary prime ideal of S−1R is of the form S−1P where P is prime ideal of R with P ∩ S = ∅. We let
Specs(R) = {P ∈ Spec(R) : P ∩ S = ∅} and view this as subspace of Spec(R) using subspace topology. Hence, the closed set in Specs(R) looks like Specs(R) ∩ V (I).
21 −1 −1 Corollary 2.10. The map f : Specs(R) → Spec(S R) given by f(P ) = S P is a homeomorphism.
Proof. We show the continuity of f by proving the pre-image of closed set is closed
−1 −1 in subspace topology. In fact, we claim f (V (S I)) = Specs(R) ∩ V (I). Let P ∈ f −1(V (S−1I)) if and only if f(P ) = S−1P ∈ V (S−1I) if and only if S−1I ⊂ S−1P if and only if for any , a p for if and only if there exists such that a ∈ I 1 = s p ∈ P t ∈ S t(sa − p) = 0 if and only if sta ∈ P if and only if a ∈ P if and only if I ⊂ P if and
only if P ∈ Speccs(R) ∩ V (I). Hence, f is continuous. To show one-to-one, supposed S−1P = S−1Q for P,Q ∈ Spec(R), so for any p ∈ P , p q for . Then there exists such that if and only if 1 = s q ∈ Q t ∈ S t(sp − q) = 0 tsp ∈ Q if and only if p ∈ Q if and only if P ⊆ Q. Similarly Q ⊆ P . Hence P = Q. By construction of f, onto is automatic.
Finally, we need to show f is closed to show it is homeomorphism. Let V (I)∩Specs(R)
−1 be closed in Specs(R) and claim f(V (I))∩Specs(R) = V (S I). Let f(P ) ∈ f(V (I))∩
Specs(R), then P ∈ V (I) ∩ Specs(R), since f is onto and one-to-one, which implies I ⊂ P and S−1I ⊂ S−1P . Hence, S−1P ∈ V (S−1P ) and f(P ) ∈ V (S−1I). For the
other inclusion, let −1 −1 , so −1 −1 . Thus, for any , a p for S P ∈ V (S I) S I ⊂ S P a ∈ I 1 = s p ∈ P if and only if there exists t ∈ S such that t(as − p) = 0 if and only if tas ∈ P
−1 if and only if a ∈ P and I ⊂ P . Therefore, S P = f(P ) ∈ f(f(V (I)) ∩ Specs(R)), showing f is a closed map.
Theorem 2.11. For any a ∈ R, D(a) is compact with respect to the subspace topology on Spec(R).
Proof. We will prove this using Corollary 2.10. We consider the special case where
n S = {a : n ≥ 0}, a multiplicatively closed set. In this case, we claim Specs(R) =
n D(a). So P ∈ Specs(R) if and only if P ∩ S = ∅ if and only if a ∈/ P for any n ≥ 0 if and only if a∈ / P , since P is prime, if and only if P ∈ D(a). We already know that
22 Spec(R) is always compact by Theorem 2.7, so Spec(S−1R) is compact by Corollary
2.10. Hence, D(a) = Specs(R), which is homeomorphic to Spec(R), and the result follows.
Corollary 2.12. Suppose φ : R → S is an onto homomorphism between commutative rings. Then, Spec(S) is homeomorphic to subspace of Spec(R).
Proof. Recall the continuous map φ∗ : Spec(S) → Spec(R), which is one-to-one by Theorem 2.8. As we have mentioned, the map φ∗ is, not in general, onto. However, if we consider φ∗ : Spec(S) → φ∗(Spec(S)). Then φ∗ is a continuous bijection. To show it id homeomorphism, we only need to prove that this induced map is closed.
To show this, we claim φ∗(V (I)) = V (φ−1(I)) for any I of S. Take φ∗(P ) ∈ φ∗(V (I)), then P ⊂ V (I). That is, I ⊂ P where P is prime ideal S. Hence φ−1(I) ⊂ φ−1(P ) = φ∗(P ) ∈ V (φ−1(I)). For the other inclusion, take P ∈ V (φ−1(I)) so φ−1(I) ⊂ P . Then Kerφ = φ−1({0}) ⊂ P . Also, φ−1(I) ⊂ P implies I ⊂ φ(P ). By the correspondence theorem, we have φ(P ) ∈ Spec(S) and we can conclude φ(P ) ∈ V (I). Finally, again using the fact that P ⊃ Kerφ, we have P = φ−1(φ(P )) = φ∗(φ(P )) ∈ φ∗(V (I)), so the reverse inclusion is established. Thus, φ∗ : Spec(S) → φ∗(Spec(S)) ⊂ Spec(R) gives the desired homeomorphic between Spec(S) and subspace of R.
2.3 Noetherian Topological Spaces
De nition 2.5. A topological space is Noetherian if it satis es the descending chain condition of closed sets or equivalently, the ascending chain condition on open sets.
Theorem 2.13. If R is a Noetherian ring, then Spec(R) is a Noetherian topological space.
23 √ √ Proof. First note if I and J are ideals such that V (I) ⊆ V (J), then J ⊆ I by Proposition 2.3. So the correspondence given above is order reversing. Now, assume
V (I1) ⊇ V (I2) ⊇ · · · is a descending chain of closed sets in Spec(R). Then we obtain √ √ the ascending chain I1 ⊆ I2 ⊆ · · · of ideals in R. Hence there exists a natural √ √ √ number n such that In = In+1 = ··· . By the above we get V (In) = V ( In) = √ V ( In+1) = V (In+1) = ··· , showing Spec(R) is Noetherian.
Remark. The converse of Theorem 2.13 does not hold. For example, consider a ring
2 2 which is not a Noethrian ring. Note that A = F (x1, x2, ··· )/(x1, x2, ··· ) rad(A) = ∼ ∼ (x1, x2, ··· ) and A/rad(A) = F (x1, x2, ··· )/(x1, x2, ··· ) = F so Spec(A) is Noethe- rian.
Theorem 2.14. Every Noetherian topological space is compact.
Proof. Let {Uα} be open cover of a Noetherian space X and let J be collection of all
nite union of sets in {Uα}. Since X is Noetherian, J must have a maximal element, say . We claim . If this is not true, there exists Uα1 ∪ · · · ∪ Uαn X = Uα1 ∪ · · · ∪ Uαn and we must have for some , since is covered by x ∈ X − (Uα1 ∪ · · · ∪ Uαn ) x ∈ Uβ β X
0 . But this gives contradicting the maximality. Uα s Uα1 ∪· · ·∪Uαn ( Uα1 ∪· · ·∪Uαn ∪Uβ Thus X is compact.
24 Chapter 3
Connectedness and Dimension in Spec(R) In this chapter we will consider the problem of when Spec(R) is connected and when it is Hausdorf.
3.1 Connectedness in Spec(R)
The connectedness of Spec(R) is directly related to the presence of certain elements in a ring called idempotents.
De nition 3.1. An element e ∈ R is called idempotent if e2 = e. Note that the identity and zero in a ring are always idempotent element.
Theorem 3.1. If R is a commutative ring, then Spec(R) is connected if and only if the only idempotent elements in R are 0 and 1.
The proof of this will be broken up into the next two propositions.
Proposition 3.2. If R is a commutative ring having an idempotent dierent from 0 and 1, then Spec(R) is not connected.
Proof. Suppose e ∈ R be an idempotent element that is not 0 or 1. Then e2 = e and (1−e)2 = 1−2e+e2 = 1−2e+e = 1−e, so (1−e) is also idempotent element. Then for any r ∈ R, we have r = re + r(1 − e), hence R = Re + R(1 − e). Furthermore, if x ∈ Re ∩ R(1 − e), then x = re = s(1 − e) and we have x = re = re2 = s(1 − e)e = 0. This shows that R = R(e) L R(1 − e), that is R is direct sum of the ideals Re and R(1 − e). We now claim that the set V (Re) is clopen (open and closed) in Spec(R), which is not equal to the empty set or all of Spec(R) (and hence Spec(R) is not connected). We know V (Re) is closed by de nition, so we need to show it is open. To this end, we claim that V (Re) = Spec(R)−V (R(1−e)). Let P ∈ V (Re), then Re ⊆ P .
25 If P ∈ V (R(1 − e)), then we would have Re ⊆ P , R(1 − e) ⊆ P and hence R ⊆ P , a contradiction. For the other inclusion, assume R(1−e) * P , which means (1−e) ∈/ P . However, 0 = e2 − e = (1 − e)e ∈ P and this gives e ∈ P , proving P ∈ V (Re). Hence V (Re) is clopen. If V (Re) = ∅, then Re is not contained in any prime ideal which implies Re = R. Thus 1 − e = re, for some r ∈ R, and 0 = (1 − e)e = re, giving e = 1, a contradiction. Now assume V (Re) = Spec(R). If V (R(1 − e)) 6= ∅, then choose P ∈ V (R(1 − e)) and thus R(1 − e) ⊆ P . But V (Re) = Spec(R) implies Re ⊆ P and so R ⊆ P , a contradiction. Hence V (R(1 − e)) = ∅ and by an argument analogous to the one above, we would obtain e = 0, a contradiction. Thus Spec(R) is not connected if it contains an idempotent other than 0 and 1.
Proposition 3.3. If R is a commutative ring such that Spec(R) is not connected, then there exists an idempotent e ∈ R dierent from 0 and 1.
Proof. Suppose Spec(R) is not connected. Then there exists ideal A and B of ring R such that V (A) ∪ V (B) = Spec(R) and V (A) ∩ V (B) = ∅ with V (A),V (B) 6= ∅ or Spec(R). It is easy to check the following: 1. V (A) ∪ V (B) = V (A ∩ B) = Spec(R) 2. V (A) ∩ V (B) = V (A + B) = ∅ 3. A + B = R (This follows from 2) 4. AB = A ∩ B (This follows from 3) 5. V (AB) = V (A ∩ B) = Spec(R) Note that 5 implies that AB is contained in every prime ideal of R, that is AB ⊆ N(R) and hence every element of AB is nilpotent. Hence, for any a ∈ A and for any b ∈ B, there exists a natural number n such that (ab)n = 0. Thus 1 = (a+b)n = an +bn +abc, for some c ∈ R. This gives an+bn = 1−abc and hence is the sum of unit and a nilpotent element and, as such an + bn must be a unit. Hence there exists some u ∈ R such that u(an + bn) = 1, which yields uan = uan(u(an + bn)) = u2a2n + u2(ab)n = u2a2n,
26 proving uan is an idempotent element. In an analogous manner, we obtain ubn is also idempotent. Finally, we must show uan 6= 0 or 1. If uan = 1, then a is invertible which implies A = R and thus V (A) = V (R) = ∅, a contradiction. Now if uan = 0, an = 0 so 1 = u(an + bn) = ubn. Therefore b is invertible and so V (B) = ∅. This is a contradiction using the same argument as in case of A. Thus, e = uan is our desired nontirivial idempotent if Spec(R) is not connected.
By the preceding two results, we obtain Theorem 3.1.
Example.
1. SpecZ is connected since the only idempotents in the integers (or in any integral domain) are 0 and 1. However, Spec(Z × Z) has nontrivial idempotents, (0, 1) and (1, 0). This shows that Spec(Z×Z) is not homeomorphic to SpecZ×SpecZ with the product topology, since the product of connected space is connected by Proposition 1.18.
2. We already have seen that has the indiscrete topology in Corollary Spec(Zpn ) 3.2, so it must be connected. We will show directly that has only trivial Zpn idempotents, which will give another proof that is connected. Suppose Spec(Zpn ) is a nontrivial idempotent. This implies 2 n and so x ∈ Spec(Zpn ) x ≡ x mod(p ) pn | (x2 − x), and x nontrivial implies pm | x and pn−m | (x − 1) for some 0 < m < n. Thus x = pmqand x − 1 = pn−mt. Thus 1 = pmq − pn−mt, a contradiction. If has more than one prime divisor, then we know m Spec(Zm) is discrete and thus is not connected (in fact it is totally disconnected).
3. C[0, 1] is connected ; if f ∈ C[0, 1] is idempotent such that f 6= 0, we can choose a ∈ [0, 1] such that f(a) 6= 0. Then f(a)2 = f(a) so f(a) = 1. Hence, if f(x) 6= 0, then f(x) = 1. By the Intermediate Value Theorem, f have to take all values between 0 and 1 since f is continuous. Thus if f(b) 6= 1 for some b ∈ [0, 1], f would have to take an every value between f(b) and 1 which is a contradiction.
27 4. A Boolean ring is not connected if it has more than two elements 0 and 1 since a Boolean ring consists of only idempotent elements.
3.2 Krull Dimension
De nition 3.2. Krull Dimension of commutative ring R is the dimension of its spectrum as a topological space. It is the supremum of the length of all chains of prime ideals.
Example.
1. Dim(Z) = 1 since (0) ⊂ (p) in Z.
2. Dim(F ) = 0 where F is a eld since only prime ideal is (0) which is maximal in eld.
3. Dim(Z[x]) = 2 since (0) ⊂ (f(x)) ⊂ (p, f(x)) or (0) ⊂ (p) ⊂ (p, f(x)) where p is prime number in Z and f(x) is irreducible polynomial in Z[x] by Theorem 1.7.
Q Q 4. Dim( Fk) = 0 ( nite or in nite) where F is a eld; If Dim( Fk) ≥ 1, there
exists prime ideal P such that P ( M where M is maximal ideal. Without loss of generality, since F is a eld, if x ∈ M −P , and we can say all the components of x are 0s and 1s. Then 0 = x2 − x = x(x − 1) ∈ P since x2 = x. But x∈ / P by assumption, so x − 1 ∈ P . Then x − 1 ∈ M and x ∈ M give 1 ∈ M which is a contradiction.
5. Dim(R) = 0 if R is Artinian ring; If x is not a zero divisor, then (x) ⊃ (x2) ⊃ · · · and there exists n such that (xn) = (xn+1). Thus xn = xn+1y which gives 1 = xy showing every nonzero divisor is a unit. If R is Artinian and P is prime ideal
28 of R, R/P is also Artinian and R/P is integral domain. Hence, every nonzero element in R/P is invertible, so it is eld implying P is maximal.
Proposition 3.4. For a commutative ring R, Spec(R) is T1 if and only if DimR = 0.
Proof. Suppose Spec(R) is T 1. Then {P } for P ∈ Spec(R) is closed by Proposition 1.14, so {P } = V (I). P is the only prime ideal that contains I, so P is maximal since I has to be contained in maximal ideal. This gives DimR = 0 if Spec(R) is T 1. Now suppose DimR = 0. Then every prime ideal is maximal ideal. We claim {P } = V (P )
since if Q ∈ V (P ), we have P ⊂ Q and Q = P . This gives Spec(R) is T1 since one point sets are closed.
Proposition 3.5. For a commutative ring R, if Spec(R) is T1, then is is T2.
Proof. If Spec(R) is T1, then DimR = 0 by Proposition 3.4. If P 6= Q ∈ Spec(R), we can choose x ∈ P − Q and let S = R − P . We consider the local ring having
unique maximal ideal. −1 a . Now, −1 , so S P = { b : a ∈ P, b∈ / P } N(PRp) = S P every element in −1 is nilpotent. If x −1 , then x n xn . Thus there S P 1 ∈ S P ( 1 ) = 1 = 0 exists s ∈ S = R − P such that sxn = 0. Now Q ∈ D(xn) because if xn ∈ Q, then x ∈ Q which contradicts our choice of x. Also, P ∈ D(s) because s∈ / P . D(xn) ∩ D(s) = D(xns) = D(0) = ∅. Thus, D(xn) forms an open set containing Q, and D(s) forms an open set containing P and their intersection is empty. Hence
Spec(R) is T2.
De nition 3.3. A commutative ring R is called V an Neumann Regular (denoted VNR) if for any a ∈ R, there exists x ∈ R such that a = a2x.
Example.
29 1. Any eld F is VNR since we can just take x = a−1 if a 6= 0 and it is immediate when a = 0.
2. Any product of VNR ring is VNR. In particular, Q F must be VNR.
Lemma 3.6. In a VNR ring R, every principal ideal is generated by an idempotent in the ideal.
Proof. Consider principal ideal (a). There exists x ∈ R such that a = a2x since R is VNR. If we let e = ax, then e2 = a2x2 = (a2x)x = ax = e. Since e = ax, (e) ⊂ (a). Now for any ra ∈ (a), we have ra = ra2x = rae ∈ (e) proving (a) = (e).
Lemma 3.7. If DimR = 0, then R/N(R) is VNR.
Proof. Let R = R/N(R). The arbitrary prime ideal of R is of the form P where P ∈ Spec(R). Since DimR = 0, P must be a maximal ideal and so P is maximal ideal and DimP = 0 by Correspondence Theorem. Choose an arbitrary a ∈ R and let S = {an(1 + ab) : n ≥ 0, b ∈ R}. It is easy to check that S is mutiplicatively closed. We claim 0 ∈ S. If not, we can nd a prime ideal P such that P ∩ S = ∅. Since we assumed 0 ∩ S = ∅, we enlarge (0) to an ideal maximal with respect to the condition of being disjoint from S. Such an ideal must be a prime ideal. We claim either a ∈ P or 1 − ab ∈ P for some b ∈ R. If a∈ / P , 1 = p + ab for some p ∈ P and b ∈ R, since P is maximal in R. Then we get 1 − ab ∈ P ∩ S, a contradiction. Thus 0 ∈ S and 0 = an(1 − ab) for some n ≥ 0. Then a(1 − ab) ∈ N(R) = N(R/N(R)) = 0, so a(1 − ab) = 0. Finally we have a = a2b proving R/N(R) is VNR.
Remark. Since Spec(R) ' Spec(R/N(R)) by Corollary 2.9, R is VNR if R/N(R) is VNR.
Theorem 3.8. If DimR = 0, then Spec(R) is T2 and it is totally disconnected. That is, the only connected subsets are singletons.
30 Proof. The fact that Spec(R) is T2 follows from Proposition 3.5. We must show the only connected subsets of Spec(R) are singletons. Suppose Y ⊆ Spec(R) and choose M 6= N ∈ Y and take a ∈ M − N. Since R is VNR by Lemma 3.7 and Remark above, there exists idempotent e ∈ R such that (a) = (e) by Lemma 3.6. a 6= 0, 1 by our choice of a , so e 6= 0, 1. Furthermore, e ∈ M − N gives M ∈ V (Re) and N ∈ V (R(1 − e)) since 0 = e2 − e = e(1 − e) ∈ N and e∈ / N. By Proposition 3.2, V (Re) and V (R(1 − e)) are clopen and they are not empty or equivalent to Spec(R). Therefore we get Y = (Y ∩ V (Re))∪˙ (Y ∩ (V (R(1 − e))). Hence Spec(R) is totally disconnected.
Example.
1. A Boolean ring R has dimension 0, so Spec(R) is both Hausdor and totally disconnected by Theorem 3.8.
2.A Q F has dimension 0 where F is a eld, so Spec(Q F ) is both Hausdor and totally disconnected by Theorem 3.8.
Corollary 3.9. If DimR = 0, then R must be a normal space.
Proof. By Theorem 2.7, Spec(R) is always compact for any commutative ring R,
and Dim(R) = 0 implies T2 by Proposition 3.4 and Proposition 3.5. Finally, every
compact T2 space is normal by Theorem 1.17.
We can summarize what we have proved as below.
Theorem 3.10. If DimR = 0, then Spec(R) must be T2, in fact T4. Hence if Spec(R)
is T1, it must be T4.
Proof. Proposition 3.4, Proposition 3.5 and Corollary 3.9 provide proof.
31 Proposition 3.11. If R is a commutative local ring with unique maximal ideal M, then Spec(R) with the Zariski topology is path-connected.
Proof. We must show that for any prime ideal P ∈ Spec(R), there exists a path connecting P to M. Then since path-connectedness is an equivalence relation, it follows that Spec(R) is path connected. De ne γ : [0, 1] → Spec(R) by
P t ∈ [0, 1) γ(t) = M t = 1
We must show γ is continuous. To show this, we show γ−1(D(a)) is open for any basic open set D(a) for any a ∈ R. Case 1: a∈ / M. Since P ⊆ M as M is unique maximal ideal, a∈ / M. Hence P,M ∈ D(a). We claim γ−1(D(a)) = [0, 1]. Now t ∈ γ−1(D(a)) if and only if γ(t) ∈ D(a). But γ(t) = P orM for any t ∈ [0, 1] and P,M ∈ D(a). Thus γ−1(D(a)) = [0, 1] which is open in [0, 1]. Case 2: a ∈ M so M/∈ D(a). If a ∈ P then P/∈ D(a). We claim γ−1(D(a)) = ∅. If there exists t ∈ γ−1(D(a)), then γ(t) = MorP ∈ D(a), a contradiction. Hence γ−1(D(a)) = ∅ which is open. If a∈ / P then P ∈ D(a). We claim γ−1(D(a)) = [0, 1). t ∈ γ−1(D(a)) if and only if γ(t) ∈ D(a) if and only if γ(t) = P if and only if t ∈ [0, 1). Thus in this case, we have γ−1(D(a)) = [0, 1) and [0, 1) is open in [0, 1] using subspace topology.
Therefore, γ is continuous since γ−1(basic open sets) is open proving Spec(R) is path- connected.
Proposition 3.12. If R is an integral domain, then R is path-connected.
Proof. We must show there exists path between any prime ideal P ∈ Spec(R) and
32 (0). De ne γ : [0, 1] → Spec(R) by
P t ∈ [0, 1) γ(t) = (0) t = 1
To establish continuity, we consider γ−1(D(a)). We assume a 6= 0 since D(0) = ∅ and γ−1(D(0)) = γ−1(∅) = ∅ which is open. This implies 1 ∈/ γ−1(D(a)) otherwise γ(1) = (0) ∈ D(a) and a = 0 contradicting our assumption. Case 1: P ∈ D(a) so a∈ / P . For any 0 ≤ t < 1, γ(t) ∈ D(a) if and only if t ∈ γ−1(D(a)) if and only if P ∈ D(a), thus γ−1(D(a)) = [0, 1) which is open. Case 2: P/∈ D(a) so a ∈ P . If t ∈ [0, 1) such that γ(t) ∈ D(a), then P ∈ D(a), a contradiction. Thus γ−1(D(a)) = ∅ which is open. Therefore, γ is continuous proving any integral domain is path-connected.
Proposition 3.13. MaxspecC[0, 1] is path-connected.
Proof. We de ne π : [0, 1] → MaxspecC[0, 1] by π(x) = Mx where Mx = {f ∈ C[0, 1] : f(x) = 0 for some x ∈ X}. The map π is onto by its construction. We must show π is continuous. Then MaxspecC[0, 1] is continuous image of path connected space and hence path-connected. To show continuity, if su ces to show that π−1(MaxspecC[0, 1] ∩ D(f)) is open for any basic open set D(f). We let a ∈
−1 π (MaxspecC[0, 1] ∩ D(f)). Then π(a) = Ma ∈ D(f) so f∈ / Ma and f(a) 6= 0. We assume f(a) > 0. There exist δ > 0 such that f(x) > 0 for any x ∈ (a − δ, a + δ). We claim (a − δ, a + δ) ⊂ π−1(MaxspecC[0, 1] ∩ D(f)) and hence π−1(MaxspecC[0, 1] ∩
D(f)) is open. If b ∈ (a − δ, a + δ), then f(b) > 0 and so f∈ / Mb. Therefore,
Mb ∈ MaxspecC[0, 1] ∩ D(f) that is, π(b) = Mb ∈ MaxspecC[0, 1] ∩ D(f) and b ∈ π−1(MaxspecC[0, 1] ∩ D(f)).
33 3.3 Irreducibility
Now we will investigate the concept of irreducibility in a topological space.
De nition 3.4. A topological space X is irreducible if X can not be written as the union of two proper closed sets.
Proposition 3.14. A topological space X is irreducible if and only if the intersection of any two non empty open sets is not empty.
Proof. First we suppose X is irreducible. Let X = U ∪ V where U and V are two non empty open sets with U ∩V = ∅. Then X = X −∅ = X −(U ∩V ) = (X −U)∪(X −V ). Since X is irreducible, either U or V is empty which is a contradiction. Now we suppose intersection of any two non empty open sets is not empty and let X = C ∪ D where C and D are closed. Then ∅ = X − (C ∪ D) = (X − C) ∩ (X − D) and X − C and X − D are open which is a contradiction. Thus, X is irreducible.
The following result gives the connection of irreducibility with the Zariski toplogy.
Proposition 3.15. If P is prime ideal of R, then V(P) is an irreducible set in Spec(R).
Proof. Suppose V (P ) = V (I) ∪ V (J) for ideals I and J of R. Since P ∈ V (P ), P ∈ V (I) or P ∈ V (J). Assume P ∈ V (I), then I ⊂ P . Now for Q ∈ V (P ), we have P ⊂ Q and so I ⊂ Q implying Q ∈ V (I). This shows V (P ) ⊂ V (I) and clearly V (I) ⊂ V (P ). So we have equality showing V (P ) is irreducible.
√ There is an interesting converse to Proposition 3.15. Recall that V (I) = V ( I). √ Proposition 3.16. If V (I) is irreducible, then I is a prime ideal. √ √ √ Proof. Suppose ab ∈ I but a∈ / I and b∈ / I. Hence there exists prime ideal P and Q such that I ⊆ P and I ⊆ Q with a∈ / P and b∈ / Q. This means P ∈
34 D(a) and Q ∈ D(b), so P ∈ V (I) ∩ D(a) and Q ∈ V (I) ∩ D(b). V (I) ∩ D(a) and V (I)∩D(b) are open set in V (I) and they are non-empty. By Proposition 3.14, we have √ (V (I)∩D(a))∩(V (I)∩D(b)) 6= ∅. Let T be in the intersection, so T ∈ V (I) = V ( I) √ by Proposition 2.3, and it gives I ⊂ T and I ⊂ T . Also, R ∈ D(a) ∩ D(b) = D(ab) √ √ implies ab∈ / T which contradicts the fact that ab ∈ I ⊂ T . Hence, I must be prime ideal.
35 Chapter 4 Worked Examples 4.1 Spec(Zn)
Lemma 4.1. For any prime and , , that is, the principal ideal p n > 1 N(Zpn ) = ([p]) generated by the equivalence class in . More generally, for any integer , [p] Zpn m > 1 , where the are the prime divisors of . N(Zm) = ([p1p2 ··· pt]) pi (i = 1, 2, ··· , t) m
Proof. First note that is nilpotent in implies k n and hence [a] N(Zpn ) a ≡ 0 mod p if n k, which implies . Thus and the reverse inclusion is p | a p | a N(Zpn ) ⊆ ([p])
clear. Now, suppose s1 s2 st and the Chinese Remainder Theorem gives m = p1 p2 ··· pt
that the map f : m → s1 × s2 × · · · st given by f([x]) = ([x], [x], ··· , [x]) is Z Zp1 Zp2 Zpt a ring isomorphism. Hence [x] ∈ N( m) if and only if [x] ∈ N( si ) if and only if Z Zpi
[x] ∈ ([pi]) if and only if pi | x for i = 1, 2, ··· , t if and only if p1p2 ··· pt | x, proving , and so . [x] ∈ ([p1p2 ··· pt]) N(Zm) = ([p1p2 ··· pt])
Remark. In , the ideal consist of the non-units and hence is a local Zpn ([p]) Zpn ring having ([p]) as the unique maximal ideal. We claim that this ideal is, in fact, the only prime ideal of . We may as well assume and let be a prime ideal Zpn n > 1 I of . Then would be a nite integral domain and hence a eld. This implies Zpn Zpn /I is maximal ideal and thus . Thus for all positive integers , I I = ([p]) n Spec(Zpn ) = {([p])}. This is not surprising for the following reason: we have the well-de ned ring epimorphism given by with . By Corollary f : Zpn → Zp f([x]) = [x] Kerf = ([p]) 2.9, and the First Isomorphism Theorem, we have ∼ ∼ Spec(Zpn ) = Spec(Zpn /([p])) = . In the next result, we use the preceding results to determine Spec(Zp) Spec(Zm) when m has more than one prime divisor.
36 Corollary 4.2. If n, then is indiscrete. If s1 s2 st , then m = p Spec(Zm) m = p1 p2 ··· pt ∼ and hence is discrete. Spec(Zm) = Spec(Zp1 × Zp2 × · · · × Zpt ) Spec(Zm)
Proof. By Lemma 3.1 and the First Isomorphism Theorem, we have ∼ Spec(Zpn ) = ∼ . Since is a eld, must have the indiscrete Spec(Zpn /([p])) = Spec(Zp) Zp Spec(Zp)
topology. Now let s1 s2 st . By the Chinese Remainder Theorem, the map m = p1 p2 ··· pt given by is a ring epimorphism f : Zm → Zp1 ×Zp2 ×· · ·×Zpt f([x]) = ([x], [x], ··· , [x]) and . Thus ∼ ∼ Kerf = ([p1p2 ··· pt]) = N(Zm) Spec(Zm) = Spec(Zm/([p1p2 ··· pt])) = . Finally , we know that the prime ideals of Spec(Zp1 ×Zp2 ×· · ·×Zpt ) Zp1 ×Zp2 ×· · ·×Zpt are of the form . It is easy to see that Zp1 × · · · × (0)i × · · · × Zpt D(0, ··· , 1i, ··· , 0) = and consequently, in , the singletons {Zp1 ×· · ·×(0)i×· · ·×Zpt } Spec(Zp1 ×Zp2 ×· · · Zpt ) are open. Thus and hence are discrete. Spec(Zp1 × Zp2 × · · · × Zpt ) Spec(Zm)
Example. Consider the map that maps to . It is an φ : Z36 → Z4 × Z9 x (x, x) isomorphism by the Chinese Remainder Theorem. It is easy to see that N(Z4) =< , and . By the First Isomorphism Theorem (2) > N(Z9) =< (3) > N(Z36) =< (6) > and Chines Remainder Theorem, we have ∼ ∼ . Now consider Z36/ < (6) >= Z6 = Z2 ×Z3 the map such that maps to . Hence ψ : Z4 ×Z9 → Z2 ×Z3 (x, x) (x, x) (ψ ◦φ)((x)36) =
ψ((x)4, (x)9) = ((x)2, (x)3). It is easy to see that kernel of ψ ◦ φ is < (6) >. Finally, we have by the First Isomorphism Spec(Z36) ' Spec(Z36/ < (6) >) ' Spec(Z2 × Z3) Theorem and Corollary 2.9. Since is discrete, is discrete by Spec(Z2 × Z3) Spec(Z36) using a homeomorphism.
4.2 Spec(Z[x])
Recall that Spec(Z[x]) = {(0), (g(x)), (p, f(x)) : g(x) is irreducible in Z[x], p is prime, and f(x) is irreducible mod p}. In the next two results, we will explicitly prove that
37 contains homeomorphic copies of and , for some Spec(Z[x]) Spec(Q[x]) Spec(Zp[x]) prime p. In fact, Spec(Z[x]) is made up of these two copies.
Proposition 4.3. There is a copy of homeomorphically embedded in Spec(Zp[x]) Spec(Z[x]).
Proof. De ne be the reduction of coe cient . This is a φ : Z[x] → Zp[x] mod p ring epimorphism, and by Theorem 2.8 it induces one-to-one continuous map φ∗ :
given by ∗ −1 . This also gives that Spec(Zp[x]) → Spec(Z[x]) φ (f(x)) = φ ((f(x))) is homeomorphic to ∗ . We claim that ∗ . Spec(Zp[x]) φ (Spec(Zp[x])) φ (f(x)) = (p, f(x)) To see this; g(x) ∈ φ∗(f(x)) if and only if φ(g(x)) = g(x) = f(x)h(x) if and only if p | (g(x) − f(x)h(x)) if and only if g(x) ∈ (p, f(x))
Remark. For any , we have , where and f(x) ∈ Q[x] f(x) = rf1(x) r ∈ Q f1 ∈ Z[x] is primitive. If any is irreducible in , then is irreducible in (and f(x) Q[x] f1(x) Z[x] hence in Q[x]). We will make use of this fact and notation in the following result.
Proposition 4.4. There is a copy of Spec(Q[x]) homeomorphically embedded in Spec(Z[x]).
Proof. Before starting the proof, we remark that the mapping from Spec(Q[x]) into Spec(Z[x]), de ned in the proof below, does not arise from a homomorphism as in the previous result.
De ne ψ : Spec(Q[x]) → Spec(Z[x]) as follows: if (0) 6= (f(x)) ∈ Spec(Q[x]), then is irreducible and as in the remark, , where and f(x) f(x) = rf1(x) r ∈ Q f1(x) ∈ Z[x] is primitive and irreducible in . Thus . De ne Z[x] (f1(x)) ∈ Spec(Z[x]) ψ((f(x))) =
(f1(x)). We will show ψ is well de ned homeomorphism onto its image. For well
de ned, assume (f(x)) = (g(x)), which implies g(x) = af(x), for a ∈ Q. Now, as in the
38 Remark, we have , , where and f(x) = rf1(x) g(x) = sg1(x) r, s ∈ Q f1(x), g1(x) ∈ Z[x] are primitive and irreducible. Hence, we have g(x) = arf1(x) = sg1(x), and clearing denominators gives . However, are irreducible αf1(x) = βg1(x), α, β ∈ Q f1(x), g1(x) in , a UFD, implies , proving . Z[x] f1(x) = ±g1(x) (f1(x)) = (g1(x)) We next prove ψ is continuous by showing ψ−1(V (I)) = V ((I)). Note that on the left side of this equality we view I as an ideal of Z[x] and on the right side we consider (I) to be the ideal of Q[x], generated by I. If (f(x)) ∈ ψ−1(V (I)), then
ψ((f(x))) = (f1(x)) ⊇ I, f(x) = rf1(x) are as above. Now h(x) ∈ I implies h(x) = (where ) = −1 where . Hence f1(x)g(x) g(x) ∈ Z[x] rf1(x)r g(x) = f(x)p(x) p(x) ∈ Q[x] (I) ⊆ (f(x)) ⊆ Q[x], and so (f(x)) ∈ V ((I)). For the other inclusion, take (p(x)) ∈ V ((I)) (note: since (I) is a non-zero ideal of Q[x], we can assume an element is a principal ideal generated by an irreducible polynomial.) As above, we have a primitive, irreducible polynomial and some such that . Thus p1(x) t ∈ Q p(x) = tp1(x) (p(x)) =
−1 (p1(x)) = ψ((p(x))) and so (p(x)) ∈ ψ (V (I)).
Finally, to complete the proof, it su ces to show that ψ(D(f(x))) = ψ(Spec(Q[x])) ∩
D((f1(x))). Consider ψ((p(x))), for (p(x)) ∈ D(f(x)) where p(x) = tp1(x). We must show . If not, then ψ((p(x))) = (p1(x)) ∈ D((f1(x))) f1(x) = p1(x)g(x), g(x) ∈ Q[x] and so −1 ( ) , a f(x) = rf1(x) = (sp1(x))(rs g(x)) = p(x)h(x) h(x) ∈ Q[x] ∈ (p(x)) contradiction. For the other inclusion, take and we p1(x) ∈ ψ(Spec(Q[x])) ∩ D(f1(x)) must prove f(x) ∈/ (p(x)). Otherwise, f(x) = p(x)g(x) implies rf1(x) = tp1(x)g(x) and so in , say . Then there exists a such p1(x) | f1(x) Q[x] f1(x) = p1(x)q(x) c ∈ Z that , . However, is irreducible in , so this cf1(x) = p1(x)(cq(x)) cq(x) ∈ Z[x] p1(x) Z[x] last equality implies in , contrary to . p1(x) | f1(x) Z[x] p1(x) ∈ D(f1(x))
The preceding two propositions imply that Spec(Z[x]) contains homeomorphic copies of and . Furthermore, by Theorem 1.7, every element of Spec(Q[x]) Spec(Zp[x]) Spec(Z[x]) is of the form (g(x)), where we can take g(x) ∈ Z[x] irreducible in Z[x] or (p, f(x)), p
39 a prime and f(x) ∈ Z[x] irreducible mod p. Thus Spec(Z[x]) is the set theoretic union of the copies of and . Spec(Q[x]) Spec(Zp[x])
4.3 Spec of C(X)
De nition 4.1. C(X) = {f(x) : X → R : X is compact Hausdor, and f(x) is continuous}
C(X) has a ring structure with usual addition and multiplication. We will study C(X) where X is a compact Hausdor space hence normal.
Theorem 4.5. Suppose X is a compact Hausdor space and M a maximal ideal of
C(X). Then M = Mx = {f ∈ C(X) : f(x) = 0} for some x ∈ X. Conversely, Mx is a maximal ideal of C(X).
Proof. Consider by with . Clearly is a φx : C(X) → R φx(f) = f(x) Kerφx = Mx φx
ring epimorphism, and this shows that Mx is a maximal ideal. Now we prove every
maximal ideal of C(X) is of the form Mx for some x ∈ X. Let M be an arbitrary maximal ideal of C(X) and set V = V (M) = {x ∈ X : f(x) = 0 for any f(x) ∈ M}.
We claim V 6= ∅. We can nd fx ∈ M such that fx(x) 6= 0. Since fx is continuous,
there exist open set Ux of x ∈ X such that fx 6= 0 on Ux. Now X ⊂ ∪Ux where
Ux is an open cover of X and since X is compact, there exists nite subcover. Let and for any , on . If we set X ⊆ Ux1 ∪ Ux2 ∪ · · · ∪ Uxn i = 1, ··· , n fxi 6= 0 Uxi 2 2 , it follows that for any since each is in some f = fx1 + ··· + fxn f(x) 6= 0 x ∈ X x and each so . Finally, for any implies that is Uxi fxi ∈ M f ∈ M f(x) 6= 0 x ∈ X f(x) invertible which is contradiction since M is maximal ideal. Thus V 6= ∅ and for x ∈ V ,
f(x) = 0 for any f ∈ M. We now have M ⊆ Mx, so M = Mx by maximality.
40 Remark.
1. The intersection of all maximal ideal of C(X) is the zero function since any function in the intersection vanishes for all x ∈ X.
2. For any ideal I of C(X), let V (I) = {x ∈ X : f(x) = 0 for any f ∈ I}. Then
−1 V (I) = ∩f∈I f ({0}) which is the intersection of closed sets and hence closed. Every closed subset of compact space is compact by Theorem 1.15. Thus, V (I) is compact.
Proposition 4.6. Every prime ideal in C(X) is contained in a unique maximal ideal.
Proof. Let P be a prime ideal and consider V (P ). We claim that V (P ) 6= ∅.A
prime ideal is always contained in some maximal ideal which is of the form Mx (refer Theorem 4.5). Hence for any f ∈ P, f(x) = 0 implying x ∈ V (P ). We claim | V (P ) |= 1, that is, it has exactly one element. To prove this, we suppose | V (P ) |≥ 2
and choose x 6= y in V (P ). X is T2 space, so there exists open set U and V such that x ∈ U, y ∈ V and U ∩V = ∅. Hence X = X −∅ = X −(U ∩V ) = (X −U)∪(X −V ). By Urysohn's Lemma in Theorem 1.19, there exists f ∈ C(X) such that f(X − U) = 0 giving f(y) = 0 and f(x) = 1. Similarly there exists g ∈ C(X) such that g(X − V ) = 0 giving g(x) = 0 and g(y) = 1. Now if f ∈ P , then x ∈ V (P ) and f(x) = 0 which is a contradiction. Thus f∈ / P and similarly g∈ / P . But f(X − U) = 0 and g(X − V ) = 0, so fg = 0 ∈ P which is contradiction. Hence, | V (P ) |= 1. Finally
suppose P is contained in two dierent maximal ideals Mx and My for x 6=y. This implies x, y ∈ V (P ) which is a contradiction. Hence every prime ideal is contained in a unique maximal ideal.
This result raises the question of whether every prime ideal is maximal ideal. We will show this is not true in general.
41 Example. Let S be the set of all nonzero polynomial for functions in C[0, 1] and let A be the collection of all ideal in C[0, 1] disjoint from S. Since (0) ∈ A, A is not empty. Then there exist prime ideal P maximal with respect to P ∩ S = ∅ by Zorn's Lemma.
If P were maximal, then P = Ma for some a ∈ [0, 1] by Theorem 4.5. But then the polynomial function g(x) = x − a can be in P ∩ S which contradicts that P ∩ S = ∅. Thus this example gives the existence of prime ideal that is not a maximal.
Theorem 4.7. If X is a compact T 2 space, then X and MaxspecC(X) are homeo- morphic with respect to the subspace topology on C(X).
Proof. De ne µ : X → maxspecC(X) by µ(x) = Mx. By Theorem 4.5, µ must be onto. To show one-to-one, suppose x 6= y. Then {x}and{y} are disjoint closed subset of the normal space X. By Urysohn's Lemma in Theorem 1.19, there exists a continuous function such that and . Hence, but . f : X → R f(x) = 0 f(y) = 1 f ∈ Mx f∈ / My
Thus x 6= y implies Mx 6= My proving µ is one-to-one. We now need to show that µ and µ−1 are continuous. We know the set D(f) = {µ ∈ maxspecC(X) : f∈ / µ} forms basis for maxspecC(X) in the subspace topology inherited from Zariski topology on SpecC(X). To come up with basis for X, we de ne U(f) = {x ∈ X : f(x) 6= 0} for any f ∈ C(X) and claim {U(f)} is a basis of X. Let W be an open set in X and choose x ∈ W . By Urysohn's Lemma again, there exist f ∈ C(X) such that f(X − W ) = 0 and f(x) = 1 taking disjoint closed sets to be X − W and {x}. Now if a ∈ U(f) ⊂ W , W = ∪U(f) implying U(f) form a basis of X. Finally, x ∈ µ−1(D(f)) if and only if µ(f) = Mx ∈ D(f) if and only if f∈ / Mx if and only if f(x) 6= 0 if and only if x ∈ U(f). Hence µ−1(D(f)) = U(f) and so µ(U(f)) = D(f). This proves µ and µ−1 are continuous.
1 n Remark. C(X) is not Noetherian in general. For example, let fn(x) = x 2 for n = 0, 1, 2, ··· . We claim C[0, 1] has a properly ascending chain of ideals that does not
42 1 1 2 1 1 1 terminate. Since x 2n = (x 2n+1 ) ∈ (x 2n+1 ), (x 2n ) ⊆ (x 2n+1 ). Now we show each
1 1 1 1 containment is proper. That is (x 2n+1 ) ∈/ (x 2n ). Otherwise x 2n+1 = x 2n g(x) for g ∈ 1 −1 x 2n+1 . But then 2n+1 which is not continuous at . Hence, C[0, 1] g(x) = 1 = x 0 (x) ( x 2n 1 1 (x 2 ) ( (x 4 ) ( ··· does not terminate.
Theorem 4.8. X is a nite set with the discrete topology (hence X is compact T 2 space) if and only if C(X) is a Noetherian ring.
The proof of this will be broken up into the next two propositions.
Proposition 4.9. If X is nite set with the discrete topology, then C(X) is Noethe- rian ring.
Proof. We will show every ideal of C(X) is nitely generated. Let X = {x1, x2, ··· , xn} and assume 0 6= I is ideal of C(X). We choose 0 6= f ∈ I and say f(xi) 6= 0 for some i = 1, 2, ··· n. Now de ne f 0 : X → R by
1 k = i 0 f(xk) f (xk) = 0 k 6= i
Then it gives 1 k = i 0 (ff )(xk) = 0 k 6= i
0 Also, ff ∈ (f) ⊂ I. We can assume there exists f1 ∈ I such that
1 i = 1 f1(xi) = 0 i 6= 1
Note that there is no loss in generality in assuming i = 1 here. Now if for any
43 P ∈ I, we have P (x2) = ··· = P (xn) = 0. We claim I = (f1) for if P ∈ I is arbitrary. Now we de ne h by h(x1) = P (x1) and h(xi) = 0 for any i ≥ 2. Then,
(hf1)(x1) = h(x1)f(x1) = P (x1)1 = P (x1) and (hf1)(xi) = h(xi)f1(xi) = 0 = P (xi)
for any i ≥ 2. In this case, P = hf1 ∈ (f1), so I = (f1) is principal. Now assume
there exists some P ∈ I such that P (xi) 6= 0 for some i ≥ 2 and as above we can also
assume there exists f2 ∈ I where
1 i = 2 f2(xi) = 0 i 6= 2
Now if P (x3) = ··· = P (xn) = 0 for any P ∈ I, we claim I = (f1, f2). To see this, let
P ∈ I be arbitrary and de ne h1, h2 as follows.
P (x1) k = 1 h1(xk) = 0 k 6= 1
P (x2) k = 2 h2(xk) = 0 k 6= 2
Then we get (f1h1 + f2h2)(x1) = f1(x1)h1(x1) + f2(x1)h1(x1) = 1P (x1) + 0 − P (x1)
(f1h1 + f2h2)(x2) = P (x2) and (f1h1 + f2h2)(xi) = 0 = P (xi) for any i ≥ 3. Then
P = f1h1 + f1h2 ∈ (f1, f2) and I = (f1, f2) is nitely generated. On the other hand,
if there exists some P ∈ I such that P (xi) 6= 0 for some i ≥ 3, then as above there exists f ∈ I such that 3 1 i = 3 f3(xi) = 0 i 6= 3
and if P (x4) = ··· = P (xn) = 0 for any P ∈ I, we will get P ∈ (f1, f2, f3) a contradiction. Continuing this process, we see that I is nitely generated. Hence
44 C(X) is Noetherian.
Proposition 4.10. If C(X) is a Noetherian ring, Then X is a nite set with the discrete topology.
Proof. By Corollary 2.13, R = C(X) is Noetherian implies Spec(R) is Noetherian topological space. Hence every subset of R is compact. In particular maxspec(R) and every subset of maxspec(R) is compact. By Theorem 4.7, maxspec(R) = maxspecC(X) ' X (homeomorphic), so maxspec(R) is Hausdor. This gives every subset of X is com- pact by the above homeomorphism. Thus X is a compact Hausdor space in which every subset is compact. Also, a compact subset of Hausdor space must be closed by
Theorem 1.16. Therefore, every subset of X is closed and hence open which implies X is a discrete space. However, a compact discrete space can not be in nite because singletons provide open cover with no nite subcover. Thus X must be nite set with the discrete topology.
4.4 Spec of a Boolean Ring
De nition 4.2. A ring R is called a Boolean ring if a2 = a for all a ∈ R.
Example.
1. Any product ( nite or in nite) of is Boolean ring Z2
2. is Boolean ring. C(X, Z2)
3. The power set of any set X is Boolean ring where A + B = (A − B) ∪ (B − A) (symmetric dierence) and AB = A ∩ B
45 Proposition 4.11. If R is a Boolean ring, then followings hold.
1. The characteristic of R is 2.
2. R is a commutative ring.
3. Every prime ideal of R is maximal.
4. For any prime ideal P of R, R/P is eld with 2 elements.
5. Every nitely generated ideal of R is principal.
Proof.
1. For all a ∈ R, 2a = (a + a)2 = a2 + 2a + a2 = 4a, hence 2a = 0 proving characteristic 2.
2. For all a, b ∈ R, (a + b) = (a + b)2 = a2 + ab + ba + b2 = a + ab + ba + b. Thus ab = −ba and we get ab = ba by part 1.
3. Suppose P is prime ideal of R and assume P ( I for some ideal of R. If a ∈ I−P , then 0 = a − a2 = a(1 − a) ∈ P . Thus we should have (1 − a) ∈ P ( I showing 1 ∈ I so I = R and P is maximal ideal.
4. Suppose x + P 6= 0 + P in R/P . Then since R/P is eld by part 3, there exists y + P such that (x + P )(y + P ) = 1 + P . Now x + P = x2 + P = (x + P )2 and so (x + P )2(y + P ) = 1 + P = x + P . Hence R/P has exactly two elements.
5. Suppose a, b ∈ P . We claim (a, b) = (a + b + ab). The right side is contained in the left side clearly. For the reverse inclusion, note that a = (a + b + ab)a and b = (a + b + ab)b. Then it follows by simple induction.
The next result summarizes properties of prime spectrum of Boolean ring.
46 Theorem 4.12. If R is Boolean ring, then followings hold.
1. The basic open sets D(a) in Spec(R) are also closed.
2. For a1, a2, ··· , an, D(a1) ∪ D(a2) ∪ · · · ∪ D(an) = D(b) for some b ∈ R.
3. The sets D(a) are the only sets in Spec(R) that are both open and closed.
4. Spec(R) is compact Hausdor space.
Proof.
1. It su ces to show D(a) = V ((a − 1)) for all a ∈ R. Note that if a = 0, then a − 1 = 1 and so D(a) = V ((a − 1)) = ∅. Otherwise, P ∈ D(a) if and only if a∈ / P if and only if a − 1 ∈ P since 0 = a2 − a = a(a − 1) ∈ P if and only if P ∈ V ((a − 1)).
2. If su ces to show that for all a, b ∈ R, D(a) ∪ D(b) = D(c) for some c ∈ R. By Proposition 4.11 part 5, (a, b) = (c) and we claim D(a) ∪ D(b) = D(c). Since c = ra + sb for r, s ∈ R, if c∈ / P , then either a∈ / P or b∈ / P . Now we suppose P ∈ D(a) ∪ D(b), say P ∈ D(a). If c ∈ P , then since (a) ⊆ (c), we would have a ∈ P , a contradiction. An easy induction gives the result as stated.
3. Suppose Y is clopen. Since it is open, we know that Y must be a union of
basic open sets, that is Y = ∪a∈RD(a). By part 1, Y must be closed, and since Spec(R) is compact Y must be compact by Theorem 1.15 and hence it is a nite union of the D(a)0s. By part 2, Y = D(b) for some b ∈ R.
4. We only need to check that Spec(R) is Hausdor space. To this end, choose P 6= Q ∈ Spec(R). If e ∈ P − Q, then P ∈ V ((e)) and Q ∈ D(e). But e is an idempotent and hence Spec(R) = V ((e)) ∪ V ((1 − e)) (disjoint union) by Theorem 3.1. It follows that V ((e)) is also open. Finally, it is clear that V ((e)) ∩ D(e) = ∅ proving Spec(R) is Hausdor space.
47 Lemma 4.13. If X is totally disconnected compact Hausdor space, then it has a basis of clopen sets.
Proof. Let C(x) be the connected component of x ∈ X and Q(x) be intersection of clopen sets containing x ∈ X. It is clear that C(x) ⊆ Q(x), and we claim C(x) = Q(x) for all x ∈ X, when X is a compact Hausdor space. It su ces to show that Q(x) is connected. If not, Q(x) = A ∪ B with disjoint clopen sets A and B. Without loss of generality, we assume x ∈ A. Since A and B are closed in X, A and B are disjoint
compact subset of T2 which are normal by Theorem 1.15 and Theorem 1.17. Hence, there exists disjoint open sets U and V containing A and B respectively. By de nition
of Q(x) and compactness, there exists clopen sets C1, ··· ,Cn containing x such that
C1 ∩ · · · ∩ Cn ⊂ U ∪ V . Consider the set D = C ∩ U. It contains x and is clopen. Now it follows that Q(x) ⊂ D and B = ∅ since B = Q(x) ∩ V , and D does not
intersect with V . Thus, Q(x) = C(x) = {x} for x ∈ Ux ∈ X, by the above and totally connectedness. Finally, there exists clopen set C such that x ∈ C ∈ U showing there is a basis of clopen sets.
Lemma 4.14. Suppose is a set and where are functions X fi : X → Z2 i = 1, 2 ··· , n such that at least one f (x) = 1. If we let f = Pn P f ··· f , then i k=1 1≤i1 Proof. To illustrate, for n = 2, we get f = f1 +f2 +f1f2. If both fi = 1, f = 1+1+1 = 1. If one of fi = 1, f = 1 + 0 + 0 = 1. Thus f = 1 in any cases. For any x ∈ X, let k be the number of functions where fi(x) = 1 for 1 ≤ k ≤ n. By the binomial theorem, k k values of fi(x) = 1 produce 2 − 1 nonzero terms in the de nition of f, and hence f(x) is odd for all x ∈ X and hence f(x) = 1 mod 2. The next result serves as a converse to Theorem 4.12 and uses preceding two Lemmas. 48 Theorem 4.15. If X is compact T 2 space which is totally disconnected, then X is homeomorphic to where we are taking topology on X to be the one Spec(C(X, Z2)) generated by the basis of clopen sets. Proof. By Lemma 4.13, there is a basis B of X consisting of clopen sets. We will show that with topology generated by is homeomorphic to . X B Spec(C(X, Z2)) For any , let −1 , and hence it is clopen f ∈ C(X, Z2) Xf = {x ∈ X : f(x) = 1} = f {1} in . For any , we de ne by if , and 0 if . X B ∈ B f : X → Z2 f(x) = 1 x ∈ B x ∈ X −B Since B is clopen, f is continuous and x ∈ B if and only if f(x) = 1 if and only if . Hence for some x ∈ Xf B = Xf f ∈ C(X, Z2) Now for any , let . x ∈ X Px = {f ∈ C(X, Z2) : f(x) = 0} ∈ Spec(C(X, Z2)) Note that since is Boolean, is also maximal by Proposition 4.10. De ne C(X, Z2) Px by . We claim is homeomorphism. Recall φ : X → Spec(C(X, Z2)) φ(x) = Px φ that is a basic open set of . Now D(f) = {P ∈ Spec(C(X, Z2)) : f∈ / P } C(X, Z2) −1 x ∈ φ (D(f)) if and only if φ(x) = Px ∈ D(f) if and only if f∈ / Px if and only if −1 f(x) = 1 if and only if x ∈ Xf . Thus φ (D(f)) = Xf showing φ is continuous. If x 6= y ∈ X, then we have y ∈ X − {x} which is open since X is T 2. Then there exists some Xf such that y ∈ Xf ⊆ X −{x}. This means that f(y) = 1 and f(x) = 0. Thus proving is one-to-one. Now for any , we want Px 6= Py φ P ∈ Spec(C(X, Z2)) to show P = Px for some x ∈ X to show φ is onto. If not, then for any x ∈ X, there exists some such that . Thus we have −1 and fx ∈ P fx(x) 6= 0 X ⊆ ∪x∈X fx (1) X compact gives −1 −1 , where each . For simplicity, we X ⊆ fx1 ({1}) ∪ · · · ∪ fxn ({1}) fxi ∈ P denote these n functions by fi. Note that for any x ∈ X, at least one fi(x) = 1. Now, de ne f = Pn P f f ··· f and note that f ∈ P . However f(x) = 1 k=1 1≤i1<··· 49 Bibliography [1] Katz, D. Abstract Algebra Revisited. Lecture notes from CSUN courses Math 460 and Math 560. [2] Munkres, J. Topology. Pearson, 2000. [3] Ueno, Kenji Algebraic Geometry 1. American Mathematical Society, 1997. [4] Atiyah, M, and Macdonald, I. Introduction to Commutative Algebra. Westview Press, 1994. [5] Pandy, Vaibhav. On the Prime Ideals of C[0,1]. 2016. [6] Reid, Miles Undergraduate Commutative Algebra. Cambridge University Press, 1995 [7] Runde, Volker A Taste of Topology. Springer, 2005 [8] Jacobson, Nathan Basic Algebra. Dover Publications, 2009 [9] Walker, Elbert Introduction to Abstract Algebra. Randon House, 1998 [10] Hungerford, Thomas Algebra. Springer, 1980 50