qalgebra conmutativa, Apuntes de Matemáticas

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Adam Allan

Solutions to Atiyah Macdonald

Chapter 1 : Rings and Ideals

1.1. Show that the sum of a nilpotent element and a unit is a unit.

If x is nilpotent, then 1 − x is a unit with inverse

i=0 x i. So if u is a unit and x is nilpotent, then v = 1 − (−u−^1 x) is a unit since −u−^1 x is nilpotent. Hence, u + x = uv is a unit as well.

1.2. Let A be a ring with f = a 0 + a 1 x + · · · + anxn^ in A[x].

a. Show that f is a unit iff a 0 is a unit and a 1 ,... , an are nilpotent.

If a 1 ,... , an are nilpotent in A, then a 1 x,... , anxn^ are nilpotent in A[x]. Since the sum of nilpotent elements is nilpotent, a 1 x + · · · + anxn^ is nilpotent. So f = a 0 + (a 1 x + · · · + anxn) is a unit when a 0 is a unit by exercise 1.1.

Now suppose that f is a unit in A[x] and let g = b 0 + b 1 x + · · · + bmxm^ satisfy f g = 1. Then a 0 b 0 = 1, and so a 0 is a unit in A[x]. Notice that anbm = 0, and suppose that 0 ≤ r ≤ m − 1 satisfies

ar n+1 bm−r = arnbm−r− 1 = · · · = anbm = 0

Notice that

0 = f g =

m∑+n

i=

∑^ i

j=

aj bi−j

 (^) xi^ =

m∑+n

i=

cixi

where we define aj = 0 for j > n and bj = 0 for j > m. This means that each ci = 0, and so

0 = ar n+1 cm+n−r− 1 =

∑^ n

j=

aj ar n+1 bm+n−r− 1 −j = ar n+2 bm−r− 1

since m + n − r − 1 − j ≥ m − r for j ≤ n − 1. So by induction am n +1b 0 = 0. Since b 0 is a unit, we conclude that an is nilpotent. This means that f − anxn^ is a unit since anxn^ is nilpotent and f is a unit. By induction, a 1 ,... , an are all nilpotent.

b. Show that f is nilpotent iff a 0 ,... , an are nilpotent.

Clearly f = a 0 + a 1 x +... + anxn^ is nilpotent if a 0 ,... , an are nilpotent. Assume f is nilpotent and that f m^ = 0 for m ∈ N. Then in particular (anxn)m^ = 0, and so anxn^ is nilpotent. Thus, f − anxn^ is nilpotent. By induction, akxk^ is nilpotent for all k. This means that a 0 ,... , an are nilpotent.

c. Show that f is zero-divisor iff bf = 0 for some b 6 = 0.

If there is b 6 = 0 for which bf = 0, then f is clearly a zero-divisor. So suppose f is a zero-divisor and choose a nonzero g = b 0 + b 1 x + · · · + bmxm^ of minimal degree for which f g = 0. Then in particular, anbm = 0. Since ang · f = 0 and ang = anb 0 + · · · + anbm− 1 xm−^1 , we conclude that ang = 0 by minimality. Hence, anbk = 0 for all k. Suppose that

an−r bk = an−r+1bk = · · · = anbk = 0 for all k

Then as in part a we obtain the equation

m+∑n−r− 1

j=

am+n−r− 1 −j bj = an−r− 1 bm

k[x 1 ,... , xr ]. Then π(h) = 0 in k[x 1 ,... , xr ]. This condition is also sufficient for showing that h is not a primitive polynomial.

So if f g is not primitive, then π(f g) = 0 as above for some maximal m. But π(f g) = π(f )π(g) and k[x 1 ,... , xr ] is an integral domain so that π(f ) = 0 or π(g) = 0. In other words, either f is not primitive or g is not primitive. The converse is obvious.

1.4. Show that R(A[x]) = N(A[x]) for every ring A.

As with any ring N(A[x]) ⊆ R(A[x]). So suppose that f ∈ R(A[x]). Then 1−f x is a unit. If f = a 0 +.. .+anxn this means that 1 − a 0 x −... − anxn+1^ is a unit, so that a 0 ,... , an are nilpotent by exercise 1.2. By exercise 1.2 this means that f is nilpotent, and so f ∈ N(A[x]). Hence R(A[x]) ⊆ N(A[x]), giving the desired result.

1.5. Let A be a ring with f =

0 anx n (^) in A[[x]].

a. Show thatf is a unit iff a 0 is a unit.

Suppose f is a unit. Then there is g(x) =

0 bnx n (^) satisfying f g = 1. In particular, a 0 b 0 = 1, implying that a 0 is a unit. Conversely, suppose that a 0 is a unit. We wish to find bn for which f g = 1. This is equivalent to finding bn satisfying a 0 b 0 = 1 and

a 0 bn +

n∑− 1

i=

an−ibi = 0 for n > 0

So we define b 0 = a− 0 1 and

bn = −a− 01

n∑− 1

i=

an−ibi for n > 0

This constructively shows that f is a unit.

b. Show that each ai is nilpotent if f is nilpotent, and that the converse is false.

Suppose that f is nilpotent and choose n > 0 for which f n^ = 0. Then an 0 = 0. Hence a 0 is nilpotent, as is f − a 0. Now by induction we see that every an is nilpotent. The converse need not be true though. We can define

A = Z 4 × Z 8 × Z 16 × · · ·

and then let

a 0 = (2, 0 , 0 ,.. .) a 1 = (0, 2 , 0 ,.. .)...

Observe that aj ak = 0 for j 6 = k, and so

f n^ = an 0 + an 1 xn^ + an 2 x^2 n^ + · · · for all n > 0

Obviously each ak is nilpotent, and yet f is not nilpotent. The problem here is that there is no N for which aNk = 0 for all k. This issue does not occur when N(A) is a nilpotent ideal, as for instance when A is Noetherian.

c. Show that f ∈ R(A[[x]]) iff a 0 ∈ R(A).

Assume a 0 ∈ R(A) and suppose g ∈ A[[x]] with constant coefficient b 0. Then there is h ∈ A[[x]] satisfy- ing 1 − f g = 1 − a 0 b 0 + hx. Since 1 − a 0 b 0 is a unit in A, we see by part a that 1 − f g is a unit in A[[x]], so that f ∈ R(A[[x]]). On the other hand, if f ∈ R(A[[x]]) and b ∈ A, then 1 − f b is a unit in A[[x]]. Again by part a this means that 1 − a 0 b is a unit in A, so that a 0 ∈ R(A).

d. Show that the contraction of a maximal ideal m of A[[x]] is a maximal ideal of A, and that m is generated by mc^ and x.

By part c we have (x) ⊆ R(A[x]) ⊆ m since 0 ∈ R(A). Now if f = a + gx is in m then a = f − gx ∈ m since x ∈ m, so that a ∈ m ∩ A. In other words, m is generated by mc^ and x.

Notice that mc^ = m ∩ A, and that A/mc^ naturally embeds into A[[x]]/m via the map a + mc^7 → a + m. I claim that A/mc^ is a subfield of the field A[[x]]/m. So suppose that a + mc^6 = mc^ and choose f ∈ A[[x]] for which (a + m)(f + m) = 1 + m, so that af − 1 ∈ m. Write f = a 0 + gx for some g ∈ A[[x]] and observe that af − 1 = aa 0 − 1 + agx ∈ m, implying that aa 0 − 1 ∈ m since x ∈ m. So we see that aa 0 − 1 ∈ mc, and hence a + mc^ has the inverse a 0 + mc. This means that A/mc^ is a subfield of A[[x]]/m, and hence mc^ is a maximal ideal in A.

e. Show that every prime ideal p of A is the contraction of a prime ideal q of A[[x]].

Let q be the ideal in A[[x]] consisting of all

akxk^ for which a 0 ∈ p. If f g ∈ q with f =

akxk^ and g =

bkxk, then a 0 b 0 ∈ pzz. Hence, a 0 ∈ p or b 0 ∈ p, implying that f ∈ q or g ∈ q. So q is a prime ideal in A[[x]] and p = A ∩ q, so that p is the contraction of q.

1.6. Let A be a ring such that every ideal not contained in N(A) contains a nonzero nilpotent. Show that N(A) = R(A).

As always N(A) ⊆ R(A). Now suppose that N(A) ( R(A). By hypothesis, there is an idempotent e 6 = 0 in R(A). Now (1 − e)e = e − e^2 = 0. Since e ∈ R(A) we know that 1 − e is a unit in A, so that e = 0. But this contradicts our choice of e, showing that N(A) = R(A).

1.7. Let A be a ring such that every x ∈ A satisfies xn^ = x for some n > 1. Show that every prime ideal p in A is maximal.

For x ∈ A choose n > 1 satisfying xn^ = x. Then ¯x(¯xn−^1 − ¯1) = ¯0 in A/p. Since A/p is an integral domain we have ¯x = ¯0 or ¯xn−^1 = ¯1. In the second case ¯x is a unit in A/p since n > 1. This shows that A/p is a field, so that p is in fact a maximal ideal.

1.8. Let A 6 = 0 be a ring. Show that the set of prime ideals of A has minimal elements with respect to inclusion.

Suppose that pα are prime ideals for α ∈ I. Suppose further that I has a linear ordering ≺ for which pα ⊃ pβ whenever α ≺ β. Define p =

α∈I pα, and suppose that^ p^ is not prime. Then there are^ x, y^ for which^ xy^ ∈^ p, and yet x, y 6 ∈ p. Hence, there are α, β for which x 6 ∈ pα and y 6 ∈ pβ. But either α ≺ β or β ≺ α, implying that x 6 ∈ pβ or y 6 ∈ pα. Either case leads to a contradiction as pα and pβ are prime ideals containing xy. So p is a prime ideal, contained in every pα. This means, by Zorn’s Lemma, that the set of prime ideals in A has minimal elements.

1.9. Let a 6 = (1) be an ideal in A. Show that a = r(a) if and only if a is the intersection of a collection of prime ideals.

g 1 f 1 (xf 1 ) + · · · + gnfn(xfn ) = 1

Let K′^ be a field containing K and roots αi of fi, noting that each fi is a non-constant polynomial. Letting xfi = αi yields 0 = 1 in K′, an impossibility. Therefore, a is a proper ideal of A. Let m be a maximal ideal in A containing a. Define K 1 = A/m. Then K 1 is an extension field of K. For g ∈ K[x] let f ∈ Σ be an irreducible factor of g. Then f (xf + m) = f (xf ) + m = m, implying that f , and hence g, has a root in K 1. Hence, every polynomial over K has a root in K 1.

Now given the field Kn, choose an extension field Kn+1 of Kn so that every polynomial over Kn has a root in Kn+1. Proceed in this way to obtain Kn for all n ∈ N+, and let L =

n=1 Kn. Then^ L^ is an extension field of K and every polynomial over Σ of degree m splits completely over Km, and hence splits completely over L. Finally, let L¯ be the set of all elements in L that are algebraic over K. Then L¯ is algebraic over K and every monic polynomial over K can be written as g =

∏deg(g) k=1 (x^ −^ αi), where^ αi^ are the roots of^ g^ in L. But then each αi is algebraic over K and hence lies in L¯. So g has roots in L¯. This means that L¯ is an algebraic closure of K.

1.14. In a ring A, let Σ be the set of all ideals in which every element is a zero-divisor. Show that Σ has maximal elements and that every maximal element of Σ is a prime ideal. Hence, the set D of zero-divisors in A is a union of prime ideals.

It is clear by Σ is chain complete. Hence, Zorn’s Lemma tells us that Σ has maximal elements. Suppose that a ∈ Σ is not a prime ideal. Let x, y ∈ A − a satisfy xy ∈ a so that a ( (a : x). If (a : x) 6 ∈ Σ then there is z ∈ (a : x) so that z is not a zero-divisor. I now claim that (a : z) ∈ Σ. If w ∈ (a : z) then wz ∈ a, so that vwz = 0 for some v 6 = 0. Since z is not a zero-divisor vz 6 = 0, and hence w is a zero-divisor. Thus a ( (a : z) ∈ Σ since x ∈ (a : z) − a. This means that a is not a maximal element in Σ. So maximal elements in Σ are indeed prime ideals.

Now if D is the set of zero-divisors in A and x ∈ D then (x) ⊆ D, and hence (x) ∈ Σ. It is clear from Zorn’s Lemma that there is a maximal a ∈ Σ containing (x), so that x ∈ a ⊆ D. This means that D is the union of some of the prime ideals of A.

1.15. Suppose A is a ring and let Spec(A) be the set of all prime ideals of A. For each E ⊆ A, let V (E) ⊆ Spec(A) consist of all prime ideals containing E. Prove the following.

a. If a = 〈E〉 then V (E) = V (a) = V (r(a)).

Since E ⊆ a ⊆ r(a) we have

V (r(a)) ⊆ V (a) ⊆ V (E)

Suppose p ∈ V (E) so that E ⊆ p. Then a = AE ⊆ Ap = p and r(a) ⊆ r(p) = p. So we have V (r(a)) ⊆ V (E). We are finished.

b. V (0) = Spec(A) and V (1) = ∅.

Every prime ideal contains 0, and so V (0) = Spec(A). Also, no prime ideal equals all of A, by definition, and so V (1) = ∅.

c. If (Ei)i∈I is a family of subsets of A then V (

Ei) =

V (Ei).

Any ideal contains

Ei iff it contains each Ei.

d. For ideals a, b we have V (a ∩ b) = V (ab) = V (a) ∪ V (b).

By part a we have

V (a ∩ b) = V (r(a ∩ b)) = V (r(ab)) = V (ab)

Clearly a ∩ b ⊆ p whenever a ⊆ p or b ⊆ p. The converse holds since p is a prime ideal. So V (a ∩ b) = V (a) ∪ V (b).

1.16? Describe the following

a. Spec(Z)

It is not hard to see that Spec(Z) = {(0)} ∪ {(p) : p > 1 prime}.

b. Spec(R)

Since R is a field, it has precisely one prime ideal, namely (0).

c. Spec(C[x])

Since C is a field, C[x] is a PID, and so its nonzero prime ideals are of the form (p) for some monic irre- ducible polynomial p. The only monic polynomials that are irreducible over C are of the form p = x − c for some c ∈ C. Of course, the zero ideal is prime as well.

d. Spec(R[x])

Since R is a field, R[x] is a PID, and so its nonzero prime ideals are of the form (p) for some monic irreducible polynomial p. Since every odd polynomial has a root, no polynomial of odd degree at least three is irreducible. Suppose p is a monic irreducible polynomial of even degree 2d > 2. In C[x] write p(z) =

∏ 2 d i=1(z^ −^ αi). Letting^ α

∗ i be the complex conjugate of^ αi, we see that^ p(α

∗ i ) =^ p(αi)

∗ (^) = 0 since p ∈ R[x]. This means that p =

∏ 2 d i=1(z^ −^ α

∗ i ). So there is^ σ^ ∈^ Σ^2 d^ so that^ α ∗ i =^ ασ(i) for every^ i. Since p has no real roots, we cannot have σ(i) = i for any i. Also, α∗ σ(i) = αi so that σ^2 = id, and hence σ is a product of 2-cycles. Thus

p(z) =

∏^ d

i=

(z − αi)(z − ασ(i)) =

∏^ d

i=

(z − αi)(z − α∗ i ) =

∏^ d

i=

(z^2 − 2 Re(αi)z + |αi|^2 )

Since each of these quadratics is in R[x], we see that p is reducible in R[x], a contradiction. Consequently, the irreducible elements in R[x] are of the form x − a and x^2 + bx + c where b^2 − 4 c < 0. These elements correspond bijectively with the non-zero prime ideals in R[x].

e. Spec(Z[X])

Notice that Z[x] is factorial. If p is an irreducible polynomial over Z then (p) is a prime ideal in Z[x]. Since Z[x] is an integral domain we see that (0) is a prime ideal in Z[x] as well. Suppose p is a non-zero prime ideal in Z[x] that is not principal. Suppose p has the property that, given f, g ∈ p, either (f ) ⊆ (g) or (g) ⊆ (f ). From this I will derive a contradiction. Let f 1 ∈ p and choose f 2 ∈ p − (f 1 ), making use of the fact that p is not principal. Then (f 1 ) ( (f 2 ). We can choose f 3 ∈ p − (f 2 ). Then (f 2 ) ( (f 3 ). We proceed in this way to get a properly ascending sequence of ideals in p. This is impossible since Hilbert’s Theorem tells us that Z[x] is Noetherian. Therefore, there are nonzero f, g ∈ p with (f ) 6 ⊆ (g)

Suppose X =

Uα with each Uα open, and write Uα =

β∈Jα Xfα,β.^ Then^ X^ =^

Xfα,β so that ∅ =

V (fα,β ) = V (

fα,β ). This means that {fα,β } generates A. So we can write 1 =

aα,β fα,β with cofinitely many of the aα,β non-zero. Working backwards, we see that X is the union of the Xfα,β for which aα,β 6 = 0. So in turn, X is the union of finitely many Uα. Thus, X is compact.

f. Show that each Xf is compact.

Suppose that Xf ⊆

Uα and write Uα =

β∈Jα Xgα,β^. Then^ Xf^ ⊆^

Xgα,β. This gives us V (

gα,β ) ⊆ V (f ). Suppose a is the ideal generated by the gα,β. Then f ∈ r(a), so that there is an equation f n^ =

aα,β gα,β with cofinitely many of the aα,β non-zero. Let g 1 ,... , gn be the gα,β with aα,β 6 = 0. Then V (

⋃n 1 gi)^ ⊆^ V^ (f^

n) = V (f ) so that Xf ⊆ ⋃n 1 Xgi^. It follows that^ Xf^ is the union of finitely many Uα. Thus, Xf is compact.

g. Show that an open subspace of X is compact if and only if it is the union of finitely many of the basic open sets Xf.

Clearly, the union of finitely many Xf is open and compact. So suppose U is compact and open. Then since U is the union of some Xf , it is the union of finitely many Xf.

1.18. Show the following about X = Spec(A).

a. The set {p} is closed iff p is a maximal ideal.

If p is a maximal ideal, then V (p) = {p}, and so {p} is closed. If {p} is closed then {p} = V (E) for some E ( A. Let m be a maximal ideal containing p so that m ∈ V (E). Then m = p, so that p is a maximal ideal.

b. Cl({p}) = V (p)

Notice that Cl(p) ⊆ V (p) since V (p) is a closed set containing p and Cl(p) is the intersection of all closed sets containing p. Conversely, suppose that q is a prime ideal not in Cl(p), and choose a neighborhood U of q that does not intersect {p}. Then there is E ⊂ A for which X − U = V (E). Consequently, p ∈ V (E) and q 6 ∈ V (E). Since p contains E and q does not, we conclude in particular that q does not contain p. This means that q 6 ∈ V (p). So Cl(p) = V (p).

c. q ∈ Cl({p}) if and only if p ⊆ q.

Obvious from part b.

d. X is a T 0 space.

Suppose that p 6 = q. If p ( q then X − V (q) is an open set containing p but not containing q; otherwise p 6 ⊆ q and hence X − V (p) is an open set containing q but not containing p.

1.19. Show that Spec(A) is an irreducible topological space iff N(A) is a prime ideal in A.

Suppose that N(A) is not a prime ideal. Then there are f, g ∈ A for which f g ∈ N(A) and yet f, g 6 ∈ N(A). Since f and g are not nilpotent, we see that Xf and Xg are nonempty open sets. But Xf ∩ Xg = Xf g = ∅ since f g is nilpotent. Hence, Spec(A) is not irreducible.

Suppose that Spec(A) is not irreducible. Choose nonempty open U, V for which U ∩ V = ∅. Then there are f, g for which ∅ 6 = Xf ⊆ U and ∅ 6 = Xg ⊆ V. So f g is nilpotent since Xf g = Xf ∩ Xg = ∅. But neither f nor g is nilpotent. This means that N(A) is not a prime ideal.

1.20. Let X be a general topological space. Prove the following.

a. If Y is an irreducible subspace of X, then the closure Y¯ of Y in X is irreducible.

Suppose U and V are open in X, and that U ∩ Y¯ and V ∩ Y¯ are nonempty. Choose x ∈ U ∩ Y¯. Since U is a neighborhood of x, and since x ∈ Y¯ , we see that U intersects Y nontrivially. So U ∩ Y , and similarly V ∩ Y , are nonempty. Since Y is irreducible, U ∩ Y intersects V ∩ Y nontrivially, and therefore U ∩ Y¯ intersects V ∩ Y¯ nontrivially. Hence, Y¯ is irreducible as well.

b. Every irreducible subspace of X is contained in a maximal irreducible subspace.

Suppose that Σ consists of all irreducible subspaces of X and that Σ is partially ordered by inclusion. Let C = {Yα : α ∈ I} be an ascending chain in Σ. Define Y =

α∈I Yα, and suppose that^ U, V^ open in X are such that U ∩ Y and V ∩ Y are nonempty. There are α, β for which U ∩ Yα and V ∩ Yβ are nonempty. We may assume that α ≤ β. Notice then that U ∩ Yβ ⊇ U ∩ Yα is nonempty. Since Yβ is irreducible, we conclude that U ∩ Yβ and V ∩ Yβ intersect nontrivially. But then U ∩ Y and V ∩ Y intersect nontrivially. That is, Y is irreducible. So by Zorn’s Lemma, Σ has maximal elements. Thus, every irreducible subspace of X is contained in a maximal irreducible subspace of X.

c. The maximal irreducible subspaces of X are closed and cover X. What are the irreducible components of a Hausdorff space?

If Y is a maximal irreducible subspace of X, then Y = Y¯ since Y¯ is irreducible. In other words, Y is closed. If x ∈ X, then {x} is irreducible, and so x is contained in some maximal irreducible subspace of X. This means that X is covered by the irreducible components.

If X is a Hausdorff space and Y ⊆ X contains two distinct points x and y, then we can choose disjoint open U and V for which x ∈ U and y ∈ V. Then U ∩ Y and V ∩ Y are nonempty disjoint open sets in Y , implying that Y is not irreducible. So the irreducible components of a Hausdorff space are precisely the one point sets.

d. The irreducible components of Spec(A) are of the form V (p) for some minimal prime ideal p.

Let p be a prime ideal and suppose f ∈ A. Then Xf ∩ V (p) 6 = ∅ if and only if f 6 ∈ q for some prime ideal q ⊇ p, and this occurs if and only if f 6 ∈ p. Now assume that Xf ∩ V (p) and Xg ∩ V (p) are nonempty open subsets of V (p). Then f, g 6 ∈ p so that f g 6 ∈ p, and hence

p ∈ Xf g ∩ V (p) = (Xf ∩ V (p)) ∩ (Xg ∩ V (p))

This means that V (p) is an irreducible subspace of Spec(A). Now any irreducible subspace of Spec(A) is of the form V (r(a)) for some ideal a. Suppose r(a) is not prime. Then there are f, g 6 ∈ r(a) for which f g ∈ r(a). So there is p ∈ V (a) not containing f and there is q ∈ V (a) not containing g. This means that Xf ∩ V (r(a)) and Xg ∩ V (r(a)) are nonempty. But Xf g ∩ V (r(a)) = ∅ since every prime ideal containing r(a) contains f g. Hence, V (r(a)) is not irreducible. So the irreducible subspaces of X are precisely of the form V (p) for some prime ideal p. Further, V (p) is maximal among all sets of the form V (q), where q is prime, if and only if p is a minimal prime ideal. So we are done.

1.21. Let φ : A → B be a ring homomorphism, with X = Spec(A) and Y = Spec(B). Define φ∗^ : Y → X by φ∗(q) = φ−^1 (q). Prove the following.

a. If f ∈ A then φ∗−^1 (Xf ) = Yφ(f ) and so φ∗^ is continuous.

Notice that Cl(φ∗(Y )) = Cl(φ∗(V (0))) = V (0c) = V (Ker(φ)). Consequently, φ∗(Y ) is dense in X if and only if V (Ker φ) = X, and this occurs precisely when Ker(φ) ⊆ N(A), and in particular when φ is 1-1.

f. Let ψ : B → C be another ring homomorphism. Show that (ψ ◦ φ)∗^ = φ∗^ ◦ ψ∗.

We have (ψ ◦ φ)∗(r) = (ψ ◦ φ)−^1 (r) = φ−^1 (ψ−^1 (r)) = φ∗(ψ∗(r)) for every r ∈ Spec(C).

g. Let A be an integral domain with only one nonzero prime ideal p, and suppose that K is the field of fractions of A. Define B = (A/p) × K and let φ : A → B by φ(x) = (¯x, x). Show that φ∗^ is bijective but not a homeomorphism.

First, A/p is a field since p is a maximal ideal in A. Now let q 1 consist of all (¯x, 0) ∈ B and let q 2 consist of all (0, x) ∈ B. Then q 1 and q 2 are maximal ideals in B since B/q 1 ∼= K and B/q 2 ∼= A/p. If q is another prime ideal of B, then q 1 q 2 = 0 is contained in q, and so q 1 ⊆ q or q 2 ⊆ q. So q 1 and q 2 are the only prime ideals of B. Hence, Spec(A) = { 0 , p} and Spec(B) = {q 1 , q 2 } are two-point spaces. It is easy to see that φ∗(q 1 ) = 0 and φ∗(q 2 ) = p, so that φ∗^ is a bijection. But φ∗^ is not a homeomorphism. After all, Spec(B) is Hausdorff since all prime ideals are maximal, but Spec(A) is not Hausdorff since 0 is a non-maximal prime ideal.

1.22. Suppose that A 1 ,... , An are rings and A =

∏n j=1 Aj^.^ Show that^ Spec(A)^ is the disjoint union of open (and closed) subspaces Xj , where Xj is canonically homeomorphic with Spec(Aj ).

Let πj : A → Aj and ij : Aj → A be the canonical maps. If q is a prime ideal in Aj , then π j− 1 (q) is a prime ideal in A. Conversely, suppose p is a prime ideal in A. Define ej = ij (1Aj ) so that

∑n 1 ej^ = 1A^ and^ ej^ ek^ = 0 if j 6 = k. Some ej∗^6 ∈ p since p 6 = A. For j 6 = j∗^ we have ej ej∗^ = 0 ∈ p so that ej ∈ p. From this we see that p = π− j∗^1 (q) for some ideal q in Aj∗^ , and it is easy to see that q is a prime ideal in Aj∗^.

Therefore, Spec(A) is the disjoint union of the subsets Xj , where Xj is the set of all π− j 1 (q), where q is a prime ideal in Aj. Notice that each Xj is closed since Xj = V (π− j 1 (0)). This also shows that each Xj is open since Xj =

k 6 =j X

c k.^ Since^ πj^ is surjective, exercise 1.22 tells us that^ π

∗ j : Spec(Aj^ )^ →^ Spec(A) is a homeomorphism of Spec(Ai) onto V (Ker(πj )) = V (π− j 1 (0)) = Xj. In particular, Xj and Spec(Aj ) are canonically homeomorphic.

Conversely, prove that the following are equivalent for any ring A. Deduce that the spectrum of a local ring is always connected.

a. X = Spec(A) is disconnected. b. A ∼= A 1 × A 2 where A 1 and A 2 are nonzero rings. c. A has an idempotent e 6 = 0, 1.

(a ⇒ c) We can write X = V (a)

V (b) where a and b are ideals in A. Then V (a ∩ b) = V (a) ∪ V (b) = X implying that a ∩ b ⊆ N(A). Also, ∅ = V (a) ∩ V (b) = V (a ∪ b), implying that A = 〈a ∪ b〉, and hence A = a + b. Now write 1 = a + b with a ∈ a and b ∈ b. Notice that ab ∈ a ∩ b ⊆ N(A) so that (ab)n^ = 0 for some n > 0. Now 1 = (a + b)n^ = an^ + bn^ + abx for some x ∈ A. Since abx ∈ N(A) we conclude that an^ + bn^ is a unit in A. Let u be the inverse of an^ + bn^ and notice that uanbn^ = 0 so that uan^ = uan(u(an^ + bn)) = (uan)^2 and similarly ubn^ = (ubn)^2. If uan^ = 0 then an^ = 0 and 1 = b(b−^1 + ax) ∈ b, which is not possible since V (b) 6 = ∅. So uan^ and ubn^ are nonzero. On the other hand, if 1 = uan^ = ubn^ then 1 = u(an^ + bn) = 2 so that 1 = 0. Hence, one of uan, ubn^ is a nontrivial idempotent.

(b ⇒ a) We already know that X = X 1

X 2 where Xi = Spec(Ai) is a non-empty open subset of X, since Ai 6 = 0. So X is disconnected.

(b ⇒ c) Take e = (0, 1) or e = (1, 0).

(c ⇒ b) Define non-zero subrings of A by A 1 = (e) and A 2 = (1−e). Then A = A 1 +A 2 since a = ae+a(1−e) for any a ∈ A. If x ∈ A 1 ∩ A 2 , then x = ae and x = b(1 − e) for some a and b. But ae = aee = b(1 − e)e = 0, and so x = 0. Therefore, A ∼= A 1 × A 2.

Exercise 1.12 shows that a local ring A has no idempotent e 6 = 0 or 1, so that Spec(A) is always connected by the above.

1.23. Let A be a Boolean ring. Prove the following.

a. For each f ∈ A, the set Xf is open and closed in Spec(A).

By definition, Xf = V (f )c^ is open. If p is a prime ideal, then f ∈ p or 1 − f ∈ p since f (1 − f ) = 0. It follows from this that Xf = V (1 − f ), so that Xf is closed in Spec(A).

b. If f 1 ,... , fn ∈ A then Xf 1 ∪ · · · ∪ Xfn = Xf for some f ∈ A.

Choose f , as in exercise 1.11, so that (f 1 ,... , fn) = (f ). Then V (f ) = V (

⋃n 1 (fj^ )) =^

⋂n 1 V^ (fj^ ), implying that Xf =

⋃n 1 Xfj.

c. If Y is both open and closed, then Y = Xf for some f ∈ A.

Since Y is closed in the compact space Spec(A), we see that Y itself is compact. Exercise 1.17 now says that Y is the union of finitely many sets of the form Xf. We now apply part b.

d. Spec(A) is a compact Hausdorff space.

Suppose that p, q are distinct prime ideals in X. We may suppose that there is f ∈ p − q. Then 1 − f ∈ q − p since f (1 − f ) = 0. So X 1 −f and Xf are open sets containing p and q, respectively. These sets are disjoint since X 1 −f ∩ Xf = X(1−f )f = X 0 = ∅. Therefore, X is compact Hausdorff.

1.24. Show that every Boolean lattice becomes a Boolean ring, and that every Boolean ring becomes a Boolean lattice. Deduce that Boolean lattices and Boolean rings are equivalent. A lattice L is a partially ordered set such that, if a and b are in L, then there is an element a ∧ b that is the largest element in the non-empty set {c ∈ L : c ≤ a and c ≤ b}, and there is an element a ∨ b that is the smallest element in the non-empty set {c ∈ L : c ≥ a and c ≥ b}. We say that L is Boolean provided that the following hold.

a. There is a smallest element 0 in L, and a largest element 1. b. For a, b, c ∈ L we have a ∨ (b ∧ c) = (a ∨ b) ∧ (a ∨ c) and also a ∧ (b ∨ c) = (a ∧ b) ∨ (a ∧ c). In other words, we have distribution. c. For each a there is a unique a′^ such that a ∧ a′^ = 1 and a ∨ a′^ = 0.

Lets make a few observations about ∧ and ∨. We first have

a ∧ 0 = 0 a ∨ 0 = a a ∧ 1 = a a ∨ 1 = 1

This implies that 0′^ = 1 and 1′^ = 0. Clearly a′′^ = a. We also have

a ∧ b = b ∧ a a ∨ b = b ∨ a a ∧ a = a a ∨ a = a

We have the associativity relations

a ∨ (b ∧ c) = a + (b ∧ c) + a(b ∧ c) = a + bc + abc = (a + 2ac) + (ab + bc + abc) + (ab + 2abc) = a(a + c + ac) + b(a + c + ac) + ab(a + c + ac) = (a + b + ab)(a + c + ac) = (a ∨ b)(a ∨ c) = (a ∨ b) ∧ (a ∨ c)

and similarly a ∧ (b ∨ c) = (a ∧ b) ∨ (a ∧ c). Now define a′^ = 1 − a so that a ∧ a′^ = a(1 − a) = 0 and a ∨ a′^ = a + (1 − a) + a(1 − a) = 1. If b ∈ A satisfies 0 = a ∧ b and 1 = a ∨ b = a + b + ab = a + b, then b = 1 − a = a′. So a′^ is unique. Thus, A is indeed a Boolean lattice.

Now suppose that we started with a Boolean lattice (L, ≤) and made it into a Boolean ring (L, +, ·), then made this ring into a new Boolean lattice (L, 4 ). If a ≤ b then ab = a ∧ b = a, so that a 4 b. If a 4 b then a = ab = a ∧ b, so that a ≤ b. Hence, (L, ≤) and (L, 4 ) are isomorphic Boolean lattices under the identity map id : L → L.

On the other hand, suppose we started with a ring (A, +, ·) and made it into a Boolean lattice (A, ≤), then made this Boolean lattice into a new Boolean ring (A, u, ×). Then a × b = a ∧ b = a · b and

a u b = (a ∧ b′) ∨ (a′^ ∧ b) = (a ∧ (1 − b)) ∨ ((1 − a) ∧ b) = a(1 − b) ∨ (1 − a)b = a(1 − b) + (1 − a)b + a(1 − b)(1 − a)b = a + b

Therefore, (A, +, ·) and (A, u, ×) are isomorphic rings Boolean rings under the identity map id : A → A. Suppose f : A → B is a ring isomorphism of Boolean rings. Let (A, ≤) and (B, 4 ) be the resulting Boolean lattices. The bijection f is order-preserving since a ≤ b implies that a = ab, and hence f (a) = f (a)f (b), implying that f (a) 4 f (b). This means that the two resulting lattices are isomorphic.

On the other hand, if (L, ≤) and ( L,¯ 4 ) are two Boolean lattices, isomorphic under f : L → L¯, then let (L, +, ·) and ( L,¯ +, ·) be the resulting Boolean rings. Notice that f −^1 : L¯ → L is order-preserving as well. It follows easily that f (a ∧ b) = f (a) Z f (b) and f (a ∨ b) = f (a) Y f (b). So f (a + b) = f (a) + f (b) and f (ab) = f (a)f (b). In other words, (L, +, ·) and ( L,¯ +, ·) are isomorphic Boolean rings. Summarizing, there is a bijective correspondence between (isomorphism classes of) Boolean rings and (isomorphism classes of) Boolean lattices.

1.25. Deduce Stone’s Theorem, that every Boolean lattice is isomorphic to the lattice of open-and- closed subsets of some compact Hausdorff topological space. Suppose L is a Boolean lattice and make L into a Boolean ring A as in exercise 1.24. Then X = Spec(A) is a compact Hausdorff space. Let L consist of all subsets of X that are both open and closed. We order L by set-theoretic inclusion. L is clearly a partially ordered set. If Y, Y ′^ ∈ L then Y ∪ Y ′, Y ∩ Y ′^ ∈ L so that L is a lattice. The emptyset ∅ is the smallest element in L and full space X is the largest element of L. Also, if Y ∈ L then Y c^ is an open and closed subset of X, with Y ∩ Y c^ = ∅ and Y ∪ Y c^ = X, with Y c^ uniquely determined by these equations. This means that L is in fact a Boolean lattice. Exercise 1.23 tells us that Y ∈ L if and only if Y = Xf for some f ∈ L. So we have a surjective map ψ : L → L given by ψ(f ) = Xf. If f ≤ g then f = f g so that Xf = Xf ∩ Xg and hence Xf ⊆ Xg. This means that ψ is an order-preserving map. On the other hand, if Xf = Xg then

∅ = X 1 −f ∩ Xf = X 1 −f ∩ Xg = X(1−f )g

so that (1 − f )g ∈ N(A). But then 0 = [(1 − f )g]n^ for some n > 0 so that (1 − f )g = 0, and hence g = f g. Similarly, f = f g and hence f = g. This shows that ψ is an isomorphism of lattices.

1.26. Let X be a compact Hausdorff space, let C(X) consists of all continuous real-valued functions defined on X, and define X˜ as the set of all maximal ideals in C(X). We have a map μ : X → X˜ given by x 7 → mx, where mx consists of all f ∈ C(X) that vanish at the point x. Prove the following.

a. The map μ is surjective.

Suppose that m is a maximal ideal in C(X). Let V consist of all x ∈ X such that f (x) = 0 when- ever f ∈ m. If V is nonempty and x ∈ V , then m ⊆ mx, and so m = mx = μ(x) by maximality. So assume that V is empty. Then given x ∈ X there is f ∈ m for which f (x) 6 = 0. By continuity, there is a neighborhood Ux of x on which fx is nonzero. These neighborhoods cover X since V = ∅, and so by compactness there are {xi}n 1 so that X =

⋃n 1 Uxi.^ Let^ f^ =^

∑n 1 f^

2 xi and notice that^ f^ is a continuous function that is everywhere positive. But then f is a unit in C(X), having multiplicative inverse 1/f , and so m = C(X); a contradiction. Therefore, V is nonempty and m = μ(x) for some x ∈ V.

b. The map μ is injective.

Recall that every compact Hausdorff space is normal. Let x, y be distinct points of X. Since {x} and {y} are disjoint closed sets, we can apply Urysohn’s Lemma to deduce the existence of an f ∈ C(X) for which f (x) = 0 and f (y) = 1. Then f ∈ mx and f 6 ∈ my. So mx 6 = my. This shows that μ is injective.

c. The bijection μ is a homeomorphism when X˜ is given the subspace topology of Spec(C(X)).

Suppose f ∈ C(X) and define Uf = f −^1 (R∗) and U˜f = {m ∈ X˜ : f 6 ∈ m}. Every m ∈ X˜ is of the form mx for a unique x ∈ X. So f ∈ m if and only if f (x) = 0. It follows that μ(Uf ) = U˜f.

Now Uf is open in X since f is continuous. So suppose that U ⊆ X is open and that x ∈ U. By normality there is a neighborhood V of x such that Cl(V ) ⊆ U. By Urysohn’s Lemma there is f ∈ C(X) such that f (Cl(V )) = { 1 } and f (X \ U ) = { 0 }. But then Uf ⊆ Cl(V ) ⊆ U. This shows that {Uf }f ∈C(X) is a basis for the topology on X.

Notice that U˜f = X˜ ∩ Xf is open in subspace topology. This also shows that { U˜f }f ∈C(X) is a basis for the topology of X˜ since {Xf }f ∈C(X) is a basis for the topology of Spec(X) by exercise 1.17.

Now the fact that μ takes basis elements to basis elements shows that μ is a homeomorphism. Conse- quently, X and X˜ are homeomorphic topological spaces.

1.27. Let k be an algebraically closed field and X an affine variety in kn. Show that there is a natural bijection between the elements of X and the maximal ideals of P (X), where P (X) = k[t 1 ,... , tn]/I(X) is the coordinate ring of X.

Let x ∈ X and consider the map k[t 1 ,... , tn] → k given by f 7 → f (x). That is, consider the map given by evaluation at x. This map is surjective since k[t 1 ,... , tn] contains all of the constant functions. If f −g ∈ I(X) then f (x) = g(x) since x ∈ X, and so the map k[t 1 ,... , tn] → k induces a surjective map P (X) → k. The kernel of this map is a maximal ideal, denoted by mx. We now have a map μ : X → Max(P (X)) given by μ(x) = mx. If mx = my and x = (x 1 ,... , xn) while y = (y 1 ,... , yn), then ti − xi ∈ my for every i as

Chapter 2 : Modules

2.1. Show that Zm ⊗Z Zn is the zero ring if gcd(m, n) = 1.

Choose integers s and t for which sm + tn = 1. Then the identity element of Zm ⊗Z Zn satisfies

[1]m ⊗ [1]n = [sm + tn]m ⊗ [1]n = [tn]m ⊗ [1]n = tn · [1]m ⊗ [1]n = [1]m ⊗ tn · [1]n = 0

Therefore our whole ring Zm ⊗Z Zn = 0.

2.2. Let A be a ring with ideal a and A-module M. Show that A/a ⊗A M ∼= M/aM.

Tensoring the short exact sequence of A-modules

0 //a

j (^) // A

π (^) // A/a // 0

with M yields the exact sequence of A-modules

a ⊗A M

j⊗ (^1) // A ⊗A M

π⊗ (^1) // A/a ⊗A M // 0

Since the map f : A ⊗ M → M given by f (a ⊗ m) = am is an isomorphism of A-modules, we can define g = (π ⊗ 1) ◦ f −^1 : M → A/a ⊗ M. Then Im(g) = Im(π ⊗ 1) = A/a ⊗ M and Ker(g) = f (Ker(π ⊗ 1)) = f (Im(j ⊗ 1)) = aM. So we have an isomorphism ¯g : M/aM → A/a ⊗ M of A-modules.

2.3. Let (A, m, k) be a local ring. Show that, if M and N are finitely generated A-modules satisfying M ⊗A N = 0, then M = 0 or N = 0.

For every A-module P define a k-vector space Pk = k ⊗A P. Then Pk and P/mP are isomorphic by exercise 2.2. Now suppose that M and N are finitely generated A-modules for which M ⊗N = 0, so that (M ⊗N )k = 0. Then

Mk ⊗k Nk = (M ⊗A k) ⊗k (N ⊗A k) ∼= M ⊗A (k ⊗k (N ⊗A k)) ∼= M ⊗A (k ⊗k (k ⊗A N )) ∼= M ⊗A ((k ⊗k k) ⊗A N ) ∼= (M ⊗A N )k

Therefore Mk ⊗k Nk = 0. Since Mk and Nk are k-vector spaces, we see that Mk = 0 or Nk = 0. So either mM = M or mN = N. By Nakayama’s lemma, either M = 0 or N = 0.

2.4. Suppose Mi are A-modules and let M =

i Mi. Prove that^ M^ is flat iff each^ Mi^ is flat.

I claim that, for every A-module N , the A-modules N ⊗

Mi and

(N ⊗ Mi) are isomorphic. Define φ : N × M →

(N ⊗ Mi) by φ(n, (xi)) = (n ⊗ xi). Then φ is A-bilinear and so induces a homomorphism Φ : N ⊗ M →

(N ⊗ Mi) for which Φ(n ⊗ (xi)) = (n ⊗ xi). Suppose now that ji : Mi → M corresponds to canonical injection. The map n ⊗ xi 7 → n ⊗ ji(xi) is a homomorphism of N ⊗ Mi into N ⊗ M. Consequently, Ψ :

(N ⊗ Mi) → N ⊗ M by Ψ((ni ⊗ xi)) =

ni ⊗ ji(xi) is a homomorphism. It is easy to show that Φ and Ψ are inverse to one another, and so are isomorphisms.

Suppose now that f : N ′^ → N is injective and consider the mapping f ⊗ 1 : N ′^ ⊗ M → N ⊗ M. As above, N ′^ ⊗ M is isomorphic with

(N ′^ ⊗ Mi) under Ψ′, and

(N ⊗ Mi) is isomorphic with N ⊗ M under Φ.

Therefore, f ⊗ 1 is injective if and only if the induced map g = Φ ◦ (f ⊗ (^1) M ) ◦ Ψ′^ from

⊕ (N^ ′^ ⊗^ Mi) to (N ⊗ Mi) is injective.

N ′^ ⊗

α Mα

f ⊗ (^1) M −−−−→ N ⊗

x^ α^ Mα  Ψ′

yΦ ⊕ α(N^

′ (^) ⊗ Mα) −−−g−→ ⊕ α(N^ ⊗^ Mα) Notice that g((nα ⊗ xα)) = (f (nα) ⊗ xα). Put differently g = (f ⊗ (^1) α) where 1α is identity on Mα. Therefore, g is injective if and only if each of its coordinate functions f ⊗ (^1) α is injective. Hence, M is flat if and only if each Mα is flat.

2.5. Prove that A[x] is a flat A-module for every ring A.

Let Mi be the A-submodule of A[x] generated by xk. Then Mi = Axi^ ∼= A so that Mi is flat. Consequently, A[x] is a flat A-module since A[x] =

0 Mi.

2.6. For any A-module M , let M [x] denote the set of all polynomials in x with coefficients in M. Then M [x] is an A[x]-module. Show that M [x] ∼= A[x] ⊗A M as A[x]-modules.

It is clear that as A-modules A[x] ∼=

i=0 Ax

i. Therefore, we have the isomorphism of A-modules

A[x] ⊗A M ∼=

⊕^ ∞

i=

(Axi^ ⊗A M ) ∼=

⊕^ ∞

i=

M xi^ = M [x]

Here the isomorphism θ is given by θ(

aixi^ ⊗ m) =

(aim)xi. All we have to do now is verify that θ is A[x]-linear. Omitting indices we compute

θ

a′ ixi ·

aixi^ ⊗ m

= θ

a′ ixi ·

aixi

⊗ m

= θ

xn^

aia′ n−i ⊗ m

aia′ n−im

xn

=

a′ ixi^ ·

(aim)xi

=

a′ ixi^ · θ

aixi^ ⊗ m

Hence, θ is an isomorphism of A[x]-modules.

2.7. Let p be a prime ideal in A and show that p[x] is a prime ideal in A[x]. If m is a maximal ideal in A, must m[x] be a maximal ideal in A[x]?

Is π : A → A/p denotes the natural map, then π induces a map A[x] → (A/p)[x] given by

∑ akxk^7 → π(ak)xk. This map is surjective and has kernel p[x]. So A[x]/p[x] ∼= (A/p)[x]. But (A/p)[x] is an integral domain since A/p is an integral domain. So p[x] is a prime ideal in A[x]. If m is a maximal ideal in A, then A[x]/m[x] ∼= (A/m)[x] with A/m a non-zero field. So (A/m)[x] is not a field, implying that m[x] is not a maximal ideal in A[x].

2.8. Suppose that M and N are flat A-modules. Show that M ⊗A N is a flat A-module.

Let S 0 be an exact sequence. We may tensor S 0 with M to get an exact sequence S 1 , and we may tensor S 1 with N to get an exact sequence S 2. But the tensor product is associative, and so the sequence S 2 is