Note Am Ica 0.1.2

Shengtian Yang

Notes

on Atiyah and MacDonald'sIntroduction to Commutative Algebra

Version: 0.1.2Date: 2013-06-08

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Title: Notes on Atiyah and MacDonald's Introduction to Commutative Algebra.

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Contents

Preface v

1 Rings and Ideals 11.1 Notes on Text . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Solutions to Exercises . . . . . . . . . . . . . . . . . . . . . . . . . 4

Bibliography 23

iii

iv Contents

Preface

This document includes my solutions to the exercises of the book Introductionto Commutative Algebra (Atiyah and MacDonald, 1969), as well as my notes onsome interesting facts in the book. The exercises of the book are quite good. Someof them in fact introduce more advanced material related to algebraic geometry.For me, a beginner, most of these exercises are not easy, but I enjoy doing themand writing down the solutions and ideas got in this process. Because of copyrightreasons, the original text of the exercises is not included in the public release ofthis document.

I am very glad for bug reports as well as any comments on all aspects of thisdocument.

Shengtian YangHangzhou, [email protected] 16, 2013

(Last revised on June 8, 2013)

v

vi Contents

Chapter 1

Rings and Ideals

1.1 Notes on Text

Note 1.1 (p. 8, Exercise 1.12) b

Proof. i) It is clear that xb a for any x 2 a, which implies that a (a : b).ii) By denition, every x 2 (a : b) satises xb a, so that (a : b)b a.iii) Since bc = cb, it suces to prove the second equality. Using property (ii),

we have

((a : c) : b)(bc) = (((a : c) : b)b)c (a : c)c aso that ((a : c) : b) (a : bc). On the other hand, for x 2 (a : bc), we havexbc a, and in particular, xbc 2 a for all b 2 b and c 2 c, so that xb 2 (a : c) forb 2 b, hence x 2 ((a : c) : b), and therefore (a : bc) ((a : c) : b).

iv) x 2 (Tt at : b) , xb 2 Tt at , xb 2 at for all t , x 2 (at : b) for all t ,x 2 Tt(at : b).

v) x 2 (a :Pt bt) , xPt bt 2 a , xbt 2 a for all t , x 2 (a : bt) for all t ,x 2 Tt(a : bt).Note 1.2 (p. 9, Exercise 1.13) b

Proof. i) It is clear that x1 a for all x 2 a.ii) By (i) we have r(r(a)) r(a), and for x 2 r(r(a)) we have xmn = (xm)n 2 a

for some m;n > 0, so that x 2 r(a), and therefore r(r(a)) r(a).iii) Since ab a \ b, it is clear that r(ab) r(a \ b). On the other hand,

x 2 r(a \ b) implies xn 2 a \ b, so that x2n 2 ab, hence x 2 r(ab), and thereforer(a \ b) r(ab).

Analogously, it is clear that r(a \ b) r(a) \ r(b). On the other hand, x 2r(a) \ r(b) implies xm 2 a for some m > 0 and xn 2 b for some n > 0, so thatxm+n 2 a \ b, hence x 2 r(a \ b), and therefore r(a) \ r(b) r(a \ b).

iv) r(a) = (1) , 1 2 r(a) , 1 2 a , a = (1).v) Since a+b r(a)+r(b), it is clear that r(a+b) r(r(a)+r(b)). On the other

hand, x 2 r(r(a) + r(b)) implies that xm = y + z for some y 2 r(a) and z 2 r(b).Then for suciently large n, we have xmn = (y + z)n =

Pi

ni

yizni 2 a+ b, so

r(r(a) + r(b)) r(a+ b).

1

2 1. Rings and Ideals

vi) It is trivially true for n = 1 by the denition of prime ideal. By inductionon n and (iii), we have r(pn) = r(pn1 \ p) = r(pn1) = p.Note 1.3 (p. 10, the example before Proposition 1.17) This example containssome basic facts:

Fact 1.1. Z[i] is a principal ideal domain.

This can be proved by showing that it has a Euclidean algorithm (see e.g.,Hungerford, 1974, Section 3.3). Dene

(x) := xx = a2 + b2 for x = a+ bi 2 Z[i]:It has the following simple properties:

1. (xy) = (x)(y);

2. (x) = 0, x = 0;3. (x) 1 for x 6= 0.

Then for xy 6= 0, we have (y) 1, so that(x) (x)(y) = (xy):

The dicult part is to show that if x; y 2 Z[i] and x 6= 0, then there exist q; r 2 Z[i]such that y = qx+ r with r = 0 or (r) < (x).

We rst assume that y = a + bi and x 2 Z n f0g. Then there exist q1, r1, q2,r2 such that

a = q1x+ r1 with jr1j x=2;b = q2x+ r2 with jr2j x=2;

so thaty = (q1 + q2i)x+ (r1 + r2i);

where (r1 + r2i) = r21 + r

22 x2=2 < (x).

Generally, for any x 6= 0, we apply the above result to the number x0 = xxand y0 = yx, and we have y0 = qx0 + r0 with r0 = 0 or (r0) < (x0). Then

r := y qxsatises (r) = 0, or (rx) = (r0) < (xx), that is, (r) < (x), since (x) > 0.VTODO: More comments to be added. b

Note 1.4 (p. 10, Exercise 1.18) b

1.1. Notes on Text 3

Proof. i) Since a1 + a2 ai and b1 + b2 bi, it is clear that (a1 + a2)e ae1 + ae2and (b1 + b2)

c bc1 + bc2. On the other hand,

(a1 + a2)e (aec1 + aec2 )e (ae1 + ae2)ce ae1 + ae2:

ii) Since a1 \ a2 ai and b1 \ b2 bi, it is clear that (a1 \ a2)e ae1 \ ae2 and(b1 \ b2)c bc1 \ bc2. On the other hand,

(b1 \ b2)c (bce1 \ bce2 )c (bc1 \ bc2)ec bc1 \ bc2:

i This proof is also right for arbitrary intersections, and hence we have(Ti2I ai)e Ti2I aei and (Ti2I bi)c = Ti2I bci (cf. Note 1.6).iii) On one hand, it is clear that f(a1a2) ae1ae2 and hence (a1a2)e ae1ae2, so

that

(b1b2)c (bce1 bce2 )c (bc1bc2)ec bc1bc2:

On the other hand, any y 2 ae1ae2 is a nite sum of the following termsXi

b1;if(a1;i)Xj

b2;jf(a2;j) =Xi;j

b1;ib2;jf(a1;ia2;j) 2 (a1a2)e;

where a1;i 2 a1, a2;j 2 a2, and b1;i; b2;j 2 B. Therefore (a1a2)e ae1ae2.iv) If y 2 (a1 : a2)e, then y =

Pi bif(xi) where xia2 a1 and bi 2 B.

Then for any y0 =P

j b0jf(x

0j) 2 ae2 with x0j 2 a2 and b0j 2 B, it is clear that

yy0 =P

i;j bib0jf(xix

0j) 2 ae1, so that (a1 : a2)e (ae1 : ae2).

If x 2 (b1 : b2)c, then f(x)b2 b1, so that f(xf1(b2)) f(x)b2 b1, hencexf1(b2) f1(b1), or x 2 (bc1 : bc2), and therefore (b1 : b2)c (bc1 : bc2).

v) Since every x 2 r(a) satises (f(x))m = f(xm) 2 f(a) for some m > 0, it isclear that f(r(a)) r(ae), so that r(a)e r(ae).

Since every x 2 r(b)c satises f(xm) = (f(x))m 2 b for some m > 0, it is clearthat r(b)c r(bc). On the other hand, r(b)c r(bce)c r(bc)ec r(bc).

vi) The last statement is obviously true by the properties already proved. Forexample, for a1; a2 2 C, we have

(a1 : a2) = (aec1 : a

ec2 ) (ae1 : ae2)c (a1 : a2)ec;

so that (a1 : a2) = (a1 : a2)ec and therefore (a1 : a2) 2 C.


1.2 Solutions to Exercises

1.

Proof. If 1+ x were not a unit, then 1+x is contained in a maximal ideal a of A,so that 1 = (1+x)x 2 a (Proposition 1.8), which is absurd. A similar argumentshows that the sum of a nilpotent element and a unit is a unit.

2.

Proof. i) If f is a unit, then there is a polynomial g = b0+ b1x+ + bmxm suchthat fg = 1, and hence it follows that a0b0 = 1 and

rXi=0

anibmr+i = 0 for 0 r m:

Here, we assume that ani = 0 for i > n. Multiplying each equation with arn, wend that ar+1n annihilates bmr. In particular, a

m+1n b0 = 0. Since b0 is a unit, an

is nilpotent. Then f 0 = fanxn is a unit (Ex. 1). Repeating the above argument,we deduce that a0 is a unit in A and a1, . . . , an are nilpotent.

Conversely, a polynomial with a0 a unit in A and a0, . . . , an nilpotent mustbe a unit (Ex. 1).

ii) If f is nilpotent, then 1 + xf is a unit, and hence it follows from (i) thata0, . . . , an are all nilpotent. The converse is trivially true.

iii) Let g = b0+ b1x+ bmxm be the polynomial of least degree m such thatfg = 0. Then anbm = 0, hence ang = 0 because angf = 0 and ang has degree< m. Let f 0 = f anxn. We have f 0g = fg = 0, and then a similar argumentshows that an1g = 0. Continuing this process, we have arg = 0 for all 0 r n.In particular arbm = 0 for all 0 r n, and therefore bmf = 0.

The converse is clearly true.iv) For convenience, we denote by C(f) the ideal (a0; : : : ; an) of A and we

adopt the convention that ai = 0 for i < 0 or i > n when calculating a sum.Let g = b0 + b1x+ bmxm, and then

fg =m+nXr=0

crxr =

m+nXr=0

rX

i=0

aibri

!xr:

If fg is primitive, then (1) = C(fg) C(f) and (1) = C(fg) C(g), so thatf and g must be primitive.

Conversely, suppose that f and g are primitive. Let I be a maximal idealcontaining C(fg) or be A if C(fg) = A. Let s and t be the least integers suchthat as =2 I and bt =2 I. If s (or t) does not exist, then we are done by observing

1.2. Solutions to Exercises 5

that I = A is necessary to contain all the coecients of f because f is primitive.If however both s and t exist, note that

asbt = cs+t Xi and 2 Zn0, then + > + .(3) > is a well-ordering on Zn0.iv) The proof is similar to Ex 2.(iv) and also requires a monomial ordering

in the proof of (iii). For illustration, we consider A[x1; x2] by a lexicographicmonomial ordering (with indeterminate order x1 > x2).

Let

f =X

(i;j)2Z20

ai;jxi1x

j2; g =

X(i;j)2Z20

bi;jxi1x

j2:


Note that almost all of the ai;j (i.e., all but a nite set) are zero. We denote byC(f) the ideal generated by all coecients ai;j of f . It is clear that

fg =X

(i;j)2Z20

ci;jxi1x

j2 =

X(i;j)2Z20

0BB@ X0i0i0j0j

ai0;j0bii0;jj0

1CCAxi1xj2:The statement that fg is primitive ) f and g are primitive is obviously true.Conversely, let I be a maximal ideal containing C(fg) or be A if C(fg) = A.

Let ; 2 Z20 be the smallest pairs of integers (in lexical ordering) such thata 2 I and b 2 I. If = (0; 0) (or = (0; 0)), then we are done by observing thatI = A is necessary to contain all ai;j because f is primitive. If however > (0; 0)and > (0; 0), we let and be the predecessors of and , respectively, andwe note that

ab = c+ X

(0;0) 6=(;);0+0=+a0b0 2 I;

which implies a 2 I or b 2 I, a contradiction.4.

Proof. It suces to show that the Jacobson radical is a subset of the nilradical.Let f = a0 + a1x + + anxn be an element of the Jacobson radical of A[x].Then 1 + xf is a unit (Proposition 1.9) and hence a0, a1, . . . , an are nilpotent(Ex. 2.(i)), so that f is nilpotent.

5.

Proof. i) If f is a unit, then there is a power series g =P1

n=0 bnxn such that

fg = 1. In particular, we have a0b0 = 1 and therefore a0 is a unit.If a0 is a unit in A, then there is an element b0 of A such that a0b0 = 1. If

choosing bn in the following way:

b1 = b0(b0a1);b2 = b0(b0a2 + b1a1);

:::

bn = b0n1Xk=0

bkank;

:::


we have fg = 1 for g =P1

n=0 bnxn.

ii) If f is nilpotent, then fm = 0 for some m > 0. In particular, am0 = 0, sothat a0 is nilpotent. Then f1 = f a0 is also nilpotent and hence a1 is nilpotent.Repeat this kind of argument and by induction on n, we prove that an is nilpotentfor all n 0.

The converse is false. VTODO: Need a counterexample.iii) If f belongs to the Jacobson radical of A[[x]], then for all g 2 A[[x]], 1fg

is a unit in A[[x]] (Proposition 1.9). From (i) it follows that 1 a0b0 is a unit inA for all b0 2 A, so that a0 belongs to the Jacobson radical of A.

Conversely, if a0 belongs to the Jacobson radical of A, then 1 a0b0 is a unitin A for all b0 2 A. Again from (i) it follows that 1 fg is a unit in A[[x]] forall g 2 A[[x]] with g = P1n=0 bnxn, so that f belongs to the Jacobson radical ofA[[x]].

iv) It is clear that mc 6= A for otherwise m (1 +P1n=1 anxn) = A[[x]] (see(i)). Let a be an element of A not in mc. Consider the ideal a = mc + (a). It isclear that m is a proper ideal of a + (x). Since m is maximal in A[[x]], a + (x)must be A[[x]], so that a = A and hence mc is maximal in A.

It is also easy to show that m mc + (x), and hence m = mc + (x) because ifx =2 m then m+ (x) would be a larger proper ideal of A[[x]].

v) Let p be a prime ideal of A. Consider the ideal q of A[[x]] generated by pand x. It is clear that qc = p. If fg 2 q, then a0b0 2 p, so that a0 2 p or b0 2 p,hence f 2 q or g 2 q, and therefore q is prime.6.

Proof. Let x be an element of the Jacobson radical of A. If (x) is a subset of thenilradical of A, then we are done. Otherwise, there would be an element e 2 (x)such that e2 = e 6= 0. Clearly, 1 e is a unit in A (Proposition 1.9). However,(1 e)e = 0 shows that it is absurd.7.

Proof. Let p be a prime ideal of A. Then A=p is an integral domain. By condition,every nonzero x + p 2 A=p satises xn + p = x + p for some n > 1, so thatxn1+p = 1+p, hence x+p is a unit, and therefore A=p is a eld or p is maximalin A.

8.

Proof. Let P be the set of prime ideals of A, with inclusion order, and it isnonempty because A 6= 0. For every chain (pi)i2I of ideals in P, let p =

Ti2I pi.

Then for xy 2 p, we have xy 2 pi for all i 2 I. If x =2 pi and y =2 pj for some iand j, then xy would not be an element of pi \ pj = pi or pj , so that x 2 p or


y 2 p, and hence p is a prime ideal. This implies p 2 P and p is a lower boundof the chain. Hence by Zorn's lemma P has a minimal element with respect toinclusion.

9.

Proof. If a = r(a), then a is the intersection of all prime ideals of A containing a(Proposition 1.14).

If a is an intersection of prime ideals, say (pi)i2I , then for any x such thatxn 2 a for some n > 0, we have x 2 pi for all i 2 I and hence x 2 a. This impliesthat r(a) = a.

10.

Proof. i) ) ii): If A has exactly one prime ideal, then it is the unique maximalideal which is equal to R (Proposition 1.8), so that every element of A is either aunit or nilpotent (Corollary 1.5).

ii)) iii): By condition, R is the unique maximal ideal, so that A=R is a eld.iii) ) i): If A=R is a eld, then R is maximal in A, so that R is the unique

prime ideal of A (Proposition 1.8).

11.

Proof. i) 2x = 4x 2x = 4x2 2x = (2x)2 2x = 0.ii) Since p is prime, A=p is an integral domain. Then for any nonzero x+ p 2

A=p, x2 + p = x + p implies x + p = 1 + p. This proves that A=p is a eld withtwo elements, and also that p is maximal in A.

iii) We will prove the result by induction on the number n of generators.The case n = 1 is trivially true. Suppose that the statement is true for n 1and consider the ideal a = (x1; x2; : : : ; xn). Then we have a = (y; xn) where(y) = (x1; : : : ; xn1). Let z = y + xn + yxn. It is clear that z 2 a. On the otherhand, y = yz 2 (z) and xn = xnz 2 (z). Therefore a = (z) is principal.

12.

Proof. Suppose e is an idempotent. If (e) = (1), then e = 1; otherwise, (e) iscontained in the unique maximal ideal m, and hence 1e is a unit (Proposition 1.9)and satises (1 e)2 = 1 e, so that 1 e = 1, and therefore e = 0.


Construction of an algebraic closure of a led (E. Artin)

13.

Proof. First, we show that a 6= (1). It suces to show that 1 =2 a. If 1 2 a, then

1 =nXi=1

aifi(xfi);

where ai 2 A0 = K[xf1 ; xf2 ; : : : ; xfn ] and fi 2 .To continue the proof, we consider the polynomial ring A0 with the lexico-

graphic monomial order. The degree, the leading term, the leading coecient,and the leading monomial of a polynomial f 2 A0 are denoted deg(f), LT(f),LC(f), and LM(f), respectively.

Since the lexicographic order is a well-ordering, we assume that the degree d =maxifdig is minimized where di = deg(aifi). On the other hand, however, it isclear that deg(1) = (0; : : : ; 0) < d. This implies that there are some cancellationsamong the polynomials fi, that is,

deg

Xi2I

LT(ai)fi(xfi)

!< d;

where I = fi : di = dg, or equivalently,Xi2I

LC(ai) = 0

since fi are all monic polynomials.

Let m = jIj, cj = LC(aij ), and pj = LM(aij )fij (xfij ), where fi1; i2; : : : ; imgis an arbitrary ordering of I. Then we have

mXj=1

cjpj = c1p1 +mXj=2

jX

k=1

ck j1Xk=1

ck

!pj

=m1Xj=1

(pj pj+1)jX

k=1

ck + pm

mXk=1

ck

=

m1Xj=1

(pj pj+1)jX

k=1

ck:


On the other hand, for any distinct j; j0,

pj pj0 = LM(ai)fi(xfi) LM(ai0)fi0(xfi0 )= xdd

0 LT(fi0(xfi0 ))fi(xfi) LT(fi(xfi))(fi0(xfi0 )

= xdd

0(LT(fi0(xfi0 )) fi0(xfi0 ))fi(xfi)

(LT(fi(xfi)) fi(xfi))fi0(xfi0 );

where d0 = deg(fi(xfi)) + deg(fi0(xfi0 )) and the second equality is because theleading terms of fi(xfi) and fi0(xfi0 ) are coprime. In other words, pj pj0 can berewritten as a linear combination of fi(xfi) and fi0(xfi0 ), whose each monomialhas a degree less than d.

Note that Xi2I

LT(ai)fi(xfi) =m1Xj=1

(pj pj+1)jX

k=1

ck;

so the degree d can be further reduced by an algebraic substitution, an obviouscontradiction to our minimal assumption of d, and therefore 1 =2 a.

Now that 1 =2 a, every element of K is not in a, and hence, not in m. Then K1is an extension eld of K under the map x 7! x+m from K to K1. If we identifyx with x+m, then we may say that K is a subeld of K1. Furthermore, for eachf 2 , f(xf ) 2 m, which implies f(xf +m) = f(xf ) +m = m.

Since Kn+1 is an extension eld of Kn for all n, it is easy to show that L is aeld. Each f 2 of degree n = deg(f) can be factored over K1 into irreduciblemonic polynomials of degrees at most n 1. If we repeat this process until overthe eld Kn, f splits completely into linear factors.

For any a; b 2 K, let f; g 2 be the polynomials such that f(a) = g(b) = 0.Then the eld K(a; b) generated by K and a; b is a nite dimensional extensionof K, so that all elements of K(a; b), including a b and a=b, are algebraic overK. This proves that K is a eld extension of K.

Finally, we will show that every polynomial in K[x] splits in K[x]. Let f =a0+a1+ +anxn 2 K[x], which splits over L and we denote its roots by r1, r2,. . . , rn. It is clear that the eld K(a0; : : : ; an; r1; : : : ; rn) is an algebraic extensionof K(a0; : : : ; an). Since a0; : : : ; an 2 K, K(a0; : : : ; an) is an algebraic extension ofK. Hence K(a0; : : : ; an; r1; : : : ; rn) is an algebraic extension of K, and thereforer1; : : : ; rn 2 K.14.

Proof. Let (ai)i2I be a chain of ideals in . Let a =Si2I ai. Then a is an ideal

and a 2 . Thus every chain in has an upper bound in , and hence hasmaximal elements by Zorn's Lemma.


Let m be a maximal ideal in and we assume that xy 2 m and x =2 m. Itis clear that (x) + m =2 , so that there exist some a 2 A and m 2 m suchthat ax + m is not a zero-divisor. For any b 2 A and n 2 m, since xy 2 m,abxy + axn+ bym+mn = (ax+m)(by + n) is a zero-divisor, so that by + n is azero-divisor for all b 2 A and n 2 m, and hence y 2 m. Therefore m is prime, andthe set of zero-divisors in A is a union of prime ideals.

Note 1.5 The trick in the proof of Ex. 14 can be encapsulated into a reusableresult:

Fact 1.2. In a ring A, let S be a subset of A that is closed under multiplication,and let m be an ideal of A that is maximal in Sc. Then m is prime.

b

Proof. For xy 2 m, if x =2 m, then (x) + m intersects S but (xy) + m = m. Thenin the quotient ring A=m, ax+m 2 S+m for some a 2 A but (ax+m)(by+m) =abxy + m = m =2 S + m for all b 2 A. Since S is closed under multiplication,so is S + m in A=m, and hence we have by + m =2 S + m for all b 2 A, so that(y) +m Sc and therefore y 2 m.

The prime spectrum of a ring15.

Proof. i) Since for any A B, V (A) V (B). It is clear that V (E) V (a) V (r(a)). On the other hand, if a prime ideal p 2 V (E), then E p, so that a pand r(a) r(p) = p, hence p 2 V (r(a)), and therefore V (E) V (r(a)).

ii) Obvious.iii) p 2 V (Si2I Ei) , p Ei for all i 2 I , p 2 V (Ei) for all i 2 I ,

p 2 Ti2I Ei.iv) V (a \ b) = V (r(a \ b)) = V (r(ab)) = V (ab).It is clear that V (ab) V (a) [ V (b). On the other hand, if a prime ideal

p 2 V (ab) and p =2 V (a), then there exists some a 2 a such that a =2 p but ab 2 pfor all b 2 b. This implies that b p and hence p 2 V (b) V (a) [ V (b).

16.

Solution. Let X(A) be the set of all prime ideals of the ring A.i) X(Z) consists of all principal ideals generated by prime numbers, as well as

the zero ideal. Given any E Z with E 6= f0g, let g = Qmi=1 pnii be the greatestcommon divisor of E, and then V (E) = f(p1); (p2); : : : ; (pm)g. In case g = 1,V (E) = ?; in case E = f0g, V (E) = X(Z).


Thus Spec(Z) consists of the empty set and all subsets of X(Z) such thattheir complements are nite, that is, Spec(Z) is the nite complement topologyof X(Z).

ii) Since X(R) = f(0)g, it is clear that Spec(R) = f?; X(R)g.iii) X(C[x]) consists of the zero ideal and all principal ideals generated by

polynomials xc where c 2 C. Thus Spec(C[x]) is the nite complement topologyof X(C[x]). In the plane of C, these open sets correspond to the empty set andall subsets of C whose complements are nite sets of points.

iv) X(R[x]) consists of the zero ideal and all principal ideals generated byirreducible polynomials. Thus Spec(R[x]) is the nite complement topology ofX(R[x]). In the plane of C, these open sets correspond to the empty set and allsubsets S of C whose complements are nite sets of points such that c =2 S if andonly if c =2 S.

v) X(Z[x]) consists of the zero ideal and all principal ideals generated byirreducible polynomials (including the prime constant polynomials). AlthoughZ[x] is not a principal ideal domain, it is still a unique factorization domain.Thus Spec(Z[x]) is the nite complement topology of X(Z[x]).

17.

Proof. First, for any prime ideal p 2 X, choose f =2 p, and then p =2 V (f), so thatp 2 Xf . Second, by property (i), Xf \Xg = Xfg for all f; g 2 A. Therefore, thesets Xf form a basis of open sets for the Zariski topology.

Next, let us prove the properties:i) By denition, Xf \ Xg = (V (f) [ V (g))c. It follows that V (f) [ V (g) =

V ((f)(g)) = V ((fg)) (Ex. 15.(iv)), so that Xf \Xg = Xfg.ii) Xf = ? , V (f) = X , f is nilpotent (Proposition 1.8).iii) Xf = X , V (f) = ? , f is not contained in every maximal ideal, so

that f is a unit.iv) Xf = Xg , V (r((f))) = V (f) = V (g) = V (r((g))) (Ex. 15.(i)), r((f)) =

r((g)) ((Proposition 1.14).v) Since the open sets Xf form a basis, it is enough to assume that X is

covered by the open sets (Xfi : i 2 I). It follows that

V

[i2Iffig

!=\i2I

V (fi) = ?: (Ex. 15.(iii))

This implies the fi generate the unit ideal. Then there is a nite subset J of Isuch that

1 =Xi2J

gifi (gi 2 A):


Therefore the Xfi (i 2 J) cover X.vi) Suppose that Xf is covered by the open sets (Xfi : i 2 I). It follows that

V

[i2Iffig

!=\i2I

V (fi) V (f): (Ex. 15.(iii))

This implies f belongs to the radical of the ideal generated by fi. Then there issome m > 0 and a nite subset J of I such that

fm =Xi2J

gifi (gi 2 A):

Therefore the Xfi (i 2 J) cover Xf .vii) By (vi), it suces to show that a quasi-compact open subset of X is a

nite union of sets Xf , which is also obvious by the denition of quasi-compactsets and the fact that the sets Xf are a basis.

18.

Proof. i) Use (ii).ii) Since the closure of x is the intersection of all closed sets containing fxg,

we have

fxg =\

E:EpxV (E)

= V

0@ [E:Epx

E

1A (Ex. 15.(iii))= V (px):

iii) y 2 fxg , y 2 V (px) , px py.iv) Given any distinct px and py, the relations px py and py px cannot

be both true. Suppose, for example, that px 6 py. Then y =2 fxg by (iii), so thatfxgc is a neighorhood of y which does not contain x.Note 1.6 It is necessary to go further to consider the closure of a general subsetX 0 of X. We need some denitions rst:

Denition 1.3. Let A be a ring and X = Spec(A). The radical r(X 0) of X 0 Xis dened by

r(X 0) :=\x2X0

px:


Then, some useful properties follow:

Fact 1.4. For an ideal a A, r(V (a)) = r(a).Fact 1.5. For X 0 X, X 0 = V (r(X 0)).Fact 1.6. Let f : A ! B be a ring homomorphism, X = Spec(A), and Y =Spec(B). Then r(V (a))e r(V (a)e) and r(V (b))c = r(V (b)c), where a and b areideals of A and B, respectively.

b

19.

Proof. Let X = Spec(A) and R be the nilradical of A.): For fg 2 R, it is clear that Xfg = ? (Ex. 17.(ii)). Since X is irreducible

and Xfg = Xf \Xg (Ex. 17.(i)), we have Xf = ? or Xg = ?, so that f 2 R org 2 R, and therefore R is prime.

(: Let B be an arbitrary nonempty open set inX. Then V (r(Bc)) = Bc 6= X,so that R =2 Bc, or R 2 B. This implies that B = V (r(B)) = V (R) = X, that is,B is dense in X. Therefore X is irreducible.

20.

Proof. i) Since every nonempty open subset A of Y can be expressed as A = Y \Bwhere B is open in X, we have A Y \B, so that A Y \B Y because Y \Bis open in Y which is irreducible, hence A Y , and therefore Y is irreducible.

ii) Let X0 be an irreducible subspace of X. Let denote the set of all irre-ducible subspaces of X containing X0, ordered by inclusion. is not empty sinceX0 2 . Let (Yi)i2I be a chain of subspaces in . Let Y =

Si2I Yi, which clearly

contains X0. For every nonempty open subset Y \B of Y where B is open in X,Yi \ B is open in Yi and hence either Yi \B = ? or Yi \B Yi ultimately, sothat

Y \B =[i2I

Yi \B

=[i2I

(Yi \B)

[i2I

Yi \B

[

i:Yi\BYiYi

= Y;


and therefore Y is irreducible. Hence Y 2 , and Y is an upper bound of thechain. By Zorn's Lemma has a maximal element.

iii) From (i) it follows that the maximal irreducible subspaces of X are closed.Since fxg is irreducible for every x 2 X, X is covered by all of its maximalirreducible subspaces. The irreducible components of a Hausdor space are justthe singleton sets fxg since a subspace of a Hausdor space is Hausdor.

iv) Let Y be an irreducible component of X. Since Y is closed, we haveV (p) = Y where p = r(Y ). For fg 2 p, Xfg \ Y = ?. Since Y is irreducible andXfg \ Y = (Xf \ Y )\ (Xg \ Y ) (Ex 17.(i)), we have Xf \ Y = ? or Xg \ Y = ?,viz., f 2 p or g 2 p. This proves that p is prime, so that p 2 Y . For any x =2 Y ,consider the space Z = Y [ fxg. Then Z is not irreducible and hence there existtwo nonempty disjoint open subsets of Z, one of which contains x. This impliesthat p =2 fxg, so that px 6 p (Ex 18.(iii)), and therefore p is a minimal prime idealof A.

Conversely, let p be a minimal prime ideal of A. First, we show that Y = V (p)is an irreducible subspace. For any nonempty open subset B = Y \C of Y whereC is open in X, since Y \ Cc is closed in Y and B is not empty, it is clear thatp =2 Y \ Cc. Then B fpg = V (p) = Y . Second, we show that Y is maximal.For any Z Y , r(Z) p and hence is not prime. Then there are two elementsf; g 2 A n r(Z) such that fg 2 r(Z). We thus nd two open subsets Xf and Xgsuch that Xf \Z 6= ?, Xg\Z 6= ?, and (Xf \Z)\(Xg\Z) = Xfg\Z = ?. Thisimplies that Z is not irreducible, and hence Y is an irreducible component.

Note 1.7 The result of Ex. 20.(iv) can be divided into two parts:

Fact 1.7. The subspace V (a) of X is irreducible if and only if r(a) is a primeideal.

This fact can be easily proved by considering the natural ring homomorphismA! A=r(a), which induces a homeomorphism from Y = Spec(A=r(a)) onto V (a)(Ex. 21). Then, V (a) is irreducible , Y is irreducible , the nilradical r(Y ) isprime (Ex. 19) , r(Y )c = r(Y c) = r(V (a)) is prime , r(a) is prime.Fact 1.8. The irreducible subspace X 0 of X is maximal if and only if r(X 0) isminimal in X (with respect to inclusion).

b

21.

Proof. i) It is clear that (f)e = ((f)) and p(y) = pcy for y 2 Y . Then, y 2 Y(f)

, (f)e 6 py , (f) 6 pcy , (y) 2 Xf . This proves that 1(Xf ) = Y(f), andhence is continuous (Ex. 17).

ii) p 2 V (ae) , ae p , a pc , (p) = pc 2 V (a).


iii) It is clear that p 2 (V (b)) , r(V (b)c) p and p 2 V (bc) , bc p.Then it suces to show that bc p if and only if r(V (b)c) p, which is truebecause r(V (b)c) = r(V (b))c = r(b)c = r(bc).

iv) If is sujective, then induces a map 0 : V (ker()) ! Y given byp ! p= ker(). It is clear that 0 is the inverse of if is understood as amap from Y to V (ker()). Now it remains to show that 0 is also continuous.Equivalently, we need to prove that maps closed sets to closed sets. It sucesto show that (V (b)) = (V (b)) for all ideals b of B. It is obviously true bynoting that every p r(V (b)c) bc has the inverse image (p) b in Y .

v) (Y ) is dense in X , r(Y c) R , r(0)c = r(Y )c R , r(ker()) =r(0c) R.

vi) By the properties of mapping, the map ( )1 : 2C ! 2A satises( )1 = 1 1. The proof is complete by noting that the inverse of a ringhomomorphism respects prime ideals.

vii) All prime ideals of B are q1 = 0K and q2 = (A=p)0, where the two zeroideals are understood as ideals of A=p and K, respectively. Their images under

are p and 0, the only two prime ideals of A. Hence is bijective. However, 1

is not continuous because (V (q2)) = f0g, which is not closed and its closure isf0; pg.

22.

Proof. Let i be the canonical projection of A onto Ai. Let Xi = V (ker(i)). Itis clear that Xi is closed and is canonically homeomorphic with Spec(Ai) by theinduced map i (Ex. 21.(iv)).

For every p 2 X = Spec(A), it is clear that i(p) is either Ai or a prime idealof Ai. Since an ideal a of A can be written as a =

Pni=1 i(i(a)), where i is

the canonical injection of Ai into A, we conclude that there is only one i suchthat i(p) 6= Ai (if, for example, 1(p) and 2(p) were prime ideals of A1 and A2,respectively, then we may choose x = (a1; 1; 0; : : : ; 0) and y = (1; a2; 0; : : : ; 0) suchthat xy 2 p but x; y =2 p, where a1 2 1(p) and a2 2 2(p)). Therefore Spec(A) isthe disjoint union of Xi, and each Xi = (

Sj 6=iXj)

c is open.i) ) iii) If X is disconnected, then X = X1 [ X2 where X1 and X2 are

nonempty open (and closed) subsets. Let ai = r(Xi). Since a1[a2 is not containedin any maximal ideal of A, we have a1 + a2 = (1). Then there are a1 2 a1 anda2 2 a2 such that a1 + a2 = 1. Since a1a2 2 a1a2 a1 \ a2, the nilradical ofA, we also have an1a

n2 = 0 for some n > 0. Let e1 =

Pn1i=0

2ni

a2ni1 a

i2 and

e2 =Pn1

i=0

2ni

ai1a

2ni2 . It is clear that ei 2 ai, e1 + e2 = (a1 + a2)2n = 1, and

e1e2 = 0, so that ei are idempotents 6= 0; 1.iii) ) ii) Let e be an idempotent 6= 0; 1. Let a1 = (e) and a2 = (1 e).

Then a1 + a2 = (1) and a1 \ a2 = a1a2 = 0. Thus the natural homomorphism


A! A=a1A=a2 is a ring isomorphism (Proposition 1.10), that is, A = A1A2where Ai = A=ai.

ii) ) i) Already proved at the beginning.Finally, it follows from (iii) and Ex. 12 that the spectrum of a local ring is

connected.

23.

Proof. i) Since Xf \ X1f = Xf(1f) = X0 = ? (Ex. 17) and Xf [ X1f =V (f)c[V (1 f)c = (V (f)\V (1 f))c = V ((f; 1 f))c = V ((1))c = X (Ex. 15),Xf is closed.

ii)

n[i=1

Xfi =n[i=1

V (fi)c

=

n\i=1

V (fi)

!c= V (((fi)

ni=1))

c (Ex. 15)

= V ((f))c (Ex. 11.(iii))

= Xf :

iii) Let Y be a subset ofX, both open and closed. Since Y is open, we have Y =Si2I Xfi for some collection (fi)i2I of elements in A, so that X = Y

c [Si2I Xfi .Since X is quasi-compact (Ex. 17.(v)), X = Y c [Si2I0 Xfi where I 0 is nite, sothat Y =

Si2I0 Xfi . By (ii), Y = Xf for some f 2 A.

iv) Given any distinct x; y 2 X, we may assume that py 6 px and we choosean element f 2 py n px. Then x 2 Xf and y 2 V (f) = Xcf , and both Xf and Xcfare open. Therefore X is a compact Hausdor space (Ex. 17.(v)).

24.

Proof. By the properties of a Boolean lattice, it is easy to verify that

(a0)0 = a; (a _ b)0 = a0 ^ b0; (a ^ b)0 = a0 _ b0:

i) First, it is easy to observe that

a+ b = b+ a; a+ 0 = a; a+ 1 = a0; 2a = 0;ab = ba; a0 = 0; a1 = a; a2 = a:


Second, we show that the addition and multiplication are associative. It isclear that the multiplication is associative. As for the addition, we note that

(a+ b)0 = (ab) _ (a0b0)

and hence

(a+ b) + c = (((ab0) _ (a0b))c0) _ (((ab) _ (a0b0))c)= ((ab0c0) _ (a0bc0)) _ ((abc) _ (a0b0c))= (ab0c0) _ (a0bc0) _ (abc) _ (a0b0c);

which is invariant under any permutation of a, b, c. Hence the addition is asso-ciative.

Finally, we show that the multiplication is distributive over the addition.

(a+ b)c = ((ab0) _ (a0b))c= (ab0c) _ (a0bc)= (acb0 _ acc0) _ (a0bc _ c0bc)= (ac(bc)0) _ ((ac)0bc)= ac+ bc:

Therefore A(L) is a Boolean ring.ii) We dene the operations of the lattice, say L(A), by

a _ b := a+ b+ ab;a ^ b := ab;

a0 := 1 a:

It is clear that

a _ b = b _ a; a ^ b = b ^ a:Since a(a _ b) = a2 + ab + a2b = a and a(a ^ b) = a2b = ab = a ^ b, a _ b anda ^ b are an upper bound and a lower bound of a, b, respectively. For any c, dsuch that c a; b d, we have

c(a ^ b) = abc = ac = c; d(a _ b) = ad+ bd+ abd = a+ b+ ab = a _ b;

which implies that a_b and a^b are the least upper bound and the greatest lowerbound, respectively.

So far, we have shown that L(A) is a lattice. Now let us show that L(A) is aBoolean lattice.


First, since 0 = 0a and a = a1 for all a 2 A, 0 and 1 are just the least andgreatest element of L(A).

Second, we show that _ and ^ are distributive over each other. We have(a _ b) ^ c = (a+ b+ ab)c

= ac+ bc+ abc

= ac+ bc+ acbc

= (a ^ c) _ (b ^ c)and

(a _ c) ^ (b _ c) = (a+ c+ ac)(b+ c+ bc)= (ab+ ac+ abc) + (bc+ c+ bc) + (abc+ ac+ abc)

= ab+ c+ abc

= (a ^ b) _ c:Third, it is easy to see that

a _ a0 = a+ a0 + a ^ a0 = 1and

a ^ a0 = a(1 a) = a a2 = 0:The uniqueness of a0 is ensured by the dening properties of the complement. Ifb; c are two complements of a, then it is easy to show that b = b ^ c = c.

By the eorts above, we proved that L(A) is a Boolean lattice.

25.

Proof. Let L be a Boolean lattice. By Ex. 24, we obtain a Boolean ring, alsodenoted L. Let X = Spec(L). It follows from Ex. 23.(iv) that X is a compactHausdor space. Let L0 be the set of all open-and-closed subsets of X. ByEx. 23.(iii), the elements of L0 are just the sets Xf .

Under the ordinary set-inclusion order, X forms a Boolean lattice. Becausethe sup and inf operations induced by nite unions and intersections of Xf , aswell as the complement operation induced by Xcf , all yield elements in L

0, L0 is asub-lattice of X, and in fact, it is also a Boolean lattice.

Consider a map f : L ! L0 given by a 7! Xa. It is clear that f is surjective.Let us show that f is injective. If f(a) = f(b), then Xa = Xb, so that (a) =r((a)) = r((b)) = (b), hence a = cb and b = da, and therefore a = cb = cb2 = ab =ada = da = b. Finally, we show that f preserves the order. If a b, then a = ab,so that Xa = Xab = Xa \Xb, and hence Xa Xb. Therefore L is isomorphic toL0.


26.

Proof. Since the exercise has already provided the proofs of (i) and (ii), we onlyprove (iii).

iii) Since : X ! ~X is injective and surjective, there is a one-to-one corre-spondence between x 2 X and mx 2 ~X. Then, mx 2 (Uf ), x 2 Uf , f(x) 6= 0, f =2 mx , mx 2 ~Uf .

Since f 2 C(X), Uf = f1(f0gc) is open. For any x 2 X, we have x 2 U1.For any x 2 Uf \ Ug, we have x 2 Ufg since Uf \ Ug = Ufg. Hence the sets Ufform a basis of the topology of X.

Since ~Uf = Xf \ ~X where Xf := Spec(C(X))nV (f) (see Ex. 17), ~Uf is open in~X with the induced topology. For any mx 2 ~X, mx 2 ~U1. For any mx 2 ~Uf \ ~Ug,we have mx 2 ~Ufg since ~Uf \ ~Ug = (Xf \ ~X) \ (Xg \ ~X) = Xfg \ ~X (Ex. 17.(i)).Hence the sets ~Uf form a basis of the topology of ~X.

Therefore is a homeomorphism.

27.

Proof. The exercise has already proved all the statements except the statementthat is surjective.

Just as the exercise mentioned, this is one form of Hilbert's Nullstellensatz, sowe will not provide the whole proof, and instead, we will give a short proof byutilizing the result of Ex. 7.14.

Let m be a maximal ideal of P (X). Let be the natural mapping fromk[t1; : : : ; tn] to P (X). Under the induced map

(see Ex. 21), Spec(P (X)) ishomeomorphic to V (I(X)). Then mc = (m) 2 V (I(X)).

Let V 0(S) denote the set of common zeros (in the corresponding domain) ofall polynomials in S, viz., an (ane algebraic) variety. Since X is a variety, weassume that X = V 0(S) for some S k[t1; : : : ; tn]. It is clear that V 0(mc) V 0(I(X)) V 0(S) = X and hence V 0(m) = V 0(mc). Since I(V 0(mc)) = r(mc) =mc 6= k[t1; : : : ; tn], V 0(mc) must be nonempty (Ex. 7.14). Let x be a point ofV 0(m). Then m mx, and hence m = mx.Note 1.8 Now that every point of X can be identied with a maximal ideal ofP (X), it may be helpful to imagine a variety X being the \crown of a subtree".More exactly, X can be identied with V (I(X))\Max(P (X)), a closed subset inthe subspace Max(P (X)) of Spec(P (X)). b

28.

Proof. Let (i)ni=1 and (j)

mj=1 be the coordinate functions of P (X) and P (Y ),

respectively. Then formally (and intuitively), we can denote P (X) and P (Y ) byk[1; : : : ; n] and k[1; : : : ; n], respectively.


Every regular : X ! Y induces a map : P (Y )! P (X) given by ! .Since is regular, that is, = (fj jX)mj=1 where fj 2 k[t1; : : : ; tn], we have

(j) = fj jX = fj(1; : : : ; n) 2 P (X):

It is then easy to verify that is well dened and is a k-algebra homomorphism.The map 7! thus dene a correspondence between regular mappings and

k-algebra homomorphisms. We will show that it is one-to-one and onto.If = (fj)

mj=1 6= = (gj)mj=1 over X, then there is at least one j such that

fj(1; : : : ; n) 6= gj(1; : : : ; n);

so that (j) 6= (j), hence 6= .On the other hand, given a k-algebra homomorphism , we dene

:= ((j))mj=1:

It is easy to see that () = ((1); : : : ; (m)) = for 2 P (Y ). Inparticular, 0 = 0, that is, g((x)) = 0 for all g 2 I(Y ) and x 2 X. Thisshows that is a map of X into Y and hence that is regular. The proof iscomplete.

Bibliography

M. F. Atiyah and I. G. MacDonald. Introduction to Commutative Algebra.Addison-Wesley Publishing Company, Reading, Massachusetts, 1969. v

D. Cox, J. Little, and D. O'Shea. Ideals, Varieties, and Algorithms. SpringerScience+Business Media, LLC, New York, third edition, 2007. 5

T. W. Hungerford. Algebra. Springer-Verlag, New York, 1974. 2

23

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