AtiyahMacDonald习题全解_1

Solutions to Atiyah and MacDonald’s Introduction to Commutative Algebra Athanasios Papaioannou CHAPTER 1 Rings and Ideals. 1.1 We see that x ∈ R implies x ∈ J (the Jacobson radical), hence 1 + xA ⊂ A×. In particular, 1 + x is a unit. We can now easily deduce that the sum of a nilpotent element and a unit is a unit itself. 1.2 We have the following: (i) If f(x) = a0 + a1x + . . . + anxn is a unit in A[x], let g(x) = b0 + b1x + . . . + bmxm be its inverse. We deduce that a0 is a unit. We use induction on n to prove that the coefficients are nilpotent. The case n = 0 is a tautology. If the proposition holds for n− 1, then we see that ar+1n bm−r = 0 (we just write down explicitly the relations that ensue from fg = 1 and then multiply each of them by increasing powers of an). In particular, this implies that am+1n b0 = 0 and, since b0 is a unit, we deduce that a m+1 n = 0. Hence an is nilpotent and we may apply the inductive hypothesis. The converse follows from exercise 1 and exercise 2, (ii). (ii) If f(x) is nilpotent, then we can apply induction to n to show that all its coefficients are nilpotent. The case n = 0 is a tautology. In the general case, it’s apparent that the leading coefficient will be amn for suitable m ∈ N hence an is nilpotent. Now the inductive hypothesis applies. Conversely, if all the coefficients of f(x) are nilpotent and d ∈ N is such that adi = 0, 0 ≤ i ≤ n (e.g. let d be the sum of the orders of all the orders of the coefficients), then we see that f(x)d = 0. (iii) If f is a zero divisor, then let g be a polynomial of minimal order, such that fg = 0. If g = b0+ b1x+ . . .+ bmxm is not of degree 0, then anbm = 0, hence ang is annihilates f but is of degree less than m, contradiction. Therefore, g is of degree 0; there is a ∈ A, such that af = 0. The converse is trivial. (iv) If fg is primitive, then so are f and g, too. The converse is just Gauss’s Lemma, in a more general context (the elementary argument still carries though). 1.3 The generalisation follows easily by induction. 1.4 If J denotes the Jacobson radical and R denotes the nilpotent radical, then J ⊃ R, since R is the intersection of all prime ideals, while J is the intersection of all prime and maximal ideals. Therefore, we only need to show J ⊂ R in A[x]. Indeed, if f(x) ∈ J, then 1−f(x)g(x) ∈ A×, for all g(x) ∈ A[x]. In particular 1−f(x)x is a unit, hence if f(x) = a0 + a1x+ . . .+ anxn, then a0, a1, . . . an are all nilpotent, hence by exercise 2, (ii) f(x) ∈ R. This completes the proof. 3 4 . 1.5 We have the following: (i) If f = ∑∞ n=0 anx n ∈ A[[x]] is invertible, then obviously a0 is a unit. Conversely, if a0 is a unit, then we may let b0 be such that a0b0 = 1 and then we may define bn, n ∈ N recursively by the explicit relations they have to satisfy. (ii) If f ∈ A[[x]] is nilpotent, then so is a0 ∈ A, which, raised to a power, is the constant term when f is raised to a power. Therefore, f − a0 = xg(x), g(x) ∈ A[[x]], will also be nilpotent, hence g(x) will also be nilpotent. But the constant term of g(x) is a1, which must be nilpotent, too. By this process, we show that all the coefficients will have to be nilpotent. The inverse is not true; a sufficient condition for it to be true would be the ring to be Noetherian. (iii) We easily see that 1− f(x)g(x) is a unit for all g(x) ∈ A[[x]] if and only if 1− a0b0 is a unit for all b0 ∈ A, hence if and only if a0 belongs to the Jacobson radical of A. (iv) The extension mapping sends any ideal a of A to the ideal ae which consists of f(x) = ∑ anx n, an ∈ a. Conversely, given any ideal b of A[[x]], bc consists of all coefficients in any element of b. It’s now clear that the contraction of a maximal ideal of A[[x]] is maximal too and that mc = (m, x). (v) This also follows immediately from the above. 1.6 It clearly suffices to show that every prime ideal in A is maximal. Let p be a prime ideal in A and let x be a non-zero element of A − p. Then the ideal (x) = {ax/a ∈ A} will contain an idempotent element e 6= 0, say a0x. This implies that a0x(a0x − 1) = 0 ∈ p, hence a0x(a0x − 1) = 0 in A/p, too. However, A/p is an integral domain, therefore e = a0x 6= 0 implies a0x = 1, or that x is a unit. Hence A/p is a field and this means that p is maximal. 1.7 Let p be a prime ideal of A. Form A/p, which will be an integral domain. Given any non-zero x ∈ A/p, there will be a suitable n ∈ N − {1}, such that xn = x or equivalently x(xn−1 − 1) = 0. This implies that xn−1 = 1, hence that x is invertible. Therefore, A/p is a field and thus p is a maximal ideal. 1.8 Every descending chain of prime ideals has a lower bound (the intersection of them all), hence by Zorn’s lemma the set Spec(A) has at least one minimal element (in order to apply the lemma, we order the set by ⊇ rather than ⊆). 1.9 Since r(a) is an intersection of prime ideals (those that contain a), we see that r(a) = a implies that a is an intersection of prime ideals. Conversely, if a is an intersection of prime ideals, then this intersection is contained in the intersection of prime ideals of r(a), hence r(a) ⊂ a, which shows that r(a) = a (the other direction is obvious by definition). 1.10 We have the following: (i) ⇒ (ii) Any maximal ideal in A (there is at least one), will be prime, hence it will coincide with the unique prime ideal a of A. Hence A is a local ring and R = a. If we consider A/R = A/a (which is a field), we deduce that every element of A is nilpotent or a unit. 1.11. 5 (ii) ⇒ (iii) This direction is obvious by the definition of a field. (iii) ⇒ (i) The nilpotent radical is maximal (and thus prime) if A/R is a field. However, R = ⋂ p∈Spec(A) p, hence R is included in the intersection on the right, hence every prime ideal contains R. But this implies that every prime ideal coincides with R (since R is maximal) and so there is only one prime ideal in A. 1.11 We have the following: (i) We just apply the given condition to x+ x, to obtain x+ x = (x+ x)2 = x2 + x2 + 2x = x+ x+ 2x, hence 2x = 0. (ii) Every prime ideal is maximal follows from exercise 7, while the second corollary follows from (i). (iii) It suffices to show that any ideal generated by two elements of A is in fact principal. Indeed, given a, y ∈ A, we claim that (x, y) = (x + y + xy). The direction (x, y) ⊃ (x + y + xy) is trivial. For the other inclusion, note that any element of (x, y) is of the form ∑ xmyn but given the conditions of idempotency, we see that the only elements that remain after the reductions in the sum will belong to (x+ y+ xy), hence the other direction. 1.12 If m is the unique maximal ideal of the ring A, then m contains all the non-units of A. If e ∈ A were idempotent, then e(e− 1) = 0, hence if e or e− 1 were units, then e would be 0 or 1. Otherwise, e ∈ m and e− 1 ∈ m, which imply 1 ∈ m, contradiction, since m is by definition a proper ideal of A. Construction of an algebraic closure of a field (E. Artin) 1.13 We will first show that a = ({f(xf )}f∈Σ) is not the unit ideal. Otherwise, given any polynomial p ∈ A it would be presentable as a finite sum in the form p = ∑ f∈Σ yff(xf ), where yf ∈ A. But 1 clearly cannot be represented in such a form, hence α 6= (1). If we now let m be the maximal ideal of A containing a, we observe that K1 = A/m is a field extension of K in which every f ∈ Σ has a root. Repeating the construction, we obtain K2 in which every f ∈ Σ has two roots (if possible), and similarly we obtain Kn for al n ∈ N. We deduce that L = ⋃∞ i=1Ki is a field extension which contains all the roots of every f ∈ Σ; its algebraic elements form an algebraic closure K for K. 1.14 The fact that Σ has a maximal element is a trivial application of Zorn’s lemma; we just need to show that every ascending chain of ideals has a maximal element. Now, assume that m is a maximal ideal of Σ and let xy ∈ m, pxy = 0, with p 6= 0. We claim that x ∈ m or y ∈ m. Assume the contrary. Then, m ⊂ (m, x) and (m, x) would still be an ideal of Σ, since its elements are clearly zero divisors. This furnishes a contradiction to the maximality of m. Therefore, every maximal ideal of Σ is prime. The prime spectrum of a ring 6 . 1.15 We have the following: (i) The relations V (E) = V (a) = V (r(a)) are obvious. (ii) Similarly, the relations V (0) = X = Spec(A) and V (1) = ∅ are obvious. (iii) Again, we have ⋂ i∈I V (Ei) = V ( ⋂ i∈I Ei) (iv) Similarly trivial are the relations V (a ∩ b) = V (ab) = V (a) ∪ V (b). These results show that the space Spec(A) of all prime ideals of A can be endowed with a topology - the Zariski topology - if we define the V (E) to be its closed sets. 1.16 We immediately see the following: Spec(Z) = {(p) : p ∈ Z is prime }. Spec(R) = ∅. Spec(C[x]) = {(p) : p ∈ C[x] is of degree 1}. Spec(R[x]) = {(p) : p ∈ R[x] is irreducible}. Spec(Z[x]) = {(p) : p ∈ Z[x] is irreducible}. 1.17 Given f ∈ A, we define Xf = {p ∈ Spec(A)/f /∈ p}. It’s obvious that X = X1, ∅ = X0 and O = X−V (E) =⋃ f∈E Xf , hence the set {Xf , f ∈ A} is a basis for the Zariski topology on Spec(A). We now have: (i) Xf ∩Xg = Xfg (obviously) (ii) Xf = ∅⇔ f ∈ R (obviously) (iii) Xf = X ⇔ f ∈ A× (obviously) (iv) Xf = Xg ⇔ r(f) = r(g) (obviously) (v) Note that for f, g ∈ A, Xf = Xg if and only if (f) = (g). In particular, Xf = X = X1 if and only if f ∈ A×. We also easily deduce (by de Morgan’s formula and exercise 15) that:⋃ i∈I Xfi = X({fi}i∈I) Therefore, if {Xfi}i∈I is an open cover of X (and it’s only those covers of X that we need to consider, by a standard proposition in point-set topology), then X({fi}i∈I) = X1, which implies that the {fi}i∈I generate the unit ideal. Therefore, there is a finite subset of indices J such that 1 = ∑ j∈J gjfj , where gj ∈ A. Now, obviously the {fj}j∈J generate the unit ideal, hence the {Xfj}j∈J are a finite subcover of X. (vi) This follows by exactly the same argument as above, but considering covers of the form {Xfi}i∈I , where Xfi ⊂ Xf . (vii) If an open subspace Y of X is quasi-compact, then considering a standard cover by sets of the form Xf , f ∈ A we see that Y must be a finite union of sets Xf . Conversely, if an open subspace Y of X is a union of a finite number of sets Xf , then any open cover {Xfi}i∈I of Y induces an open cover for each of the Xf (namely {Xf ∩Xfi}i∈I). By (vii), each of those will have a finite subcover and these subcovers yield a finite subcover of X. 1.18. 7 1.18 We have the following: (i) By the definition of the Zariski topology, {x} is closed in Spec(A) if and only if {x} = V (E) for some subset E of A, hence if and only if px is the only prime ideal that contains E, hence if and only if E = px and px is maximal (attaching any elements of A− E would generate the unit ideal). (ii) The relation {x} = V (px) is obvious by our remarks above. (iii) y ∈ {x} = V (px) if and only if py ⊃ px (iv) If x and y are distinct points of Spec(A), then either px * py or py * px; assume without loss of generality the latter. This is equivalent by our previous observations to x /∈ {y}, which implies that X −{y} is an open set that contains x but not y. 1.19 We claim that Spec(A) is irreducible if and only Xf ∩ Xg 6= ∅ for f and g non-nilpotent. Indeed, since {Xf}f∈A are a basis for the Zariski topology on Spec(A), we see that any two non-empty sets will intersect if and only if any two non-empty basis elements intersect. This is equivalent to Xfg = Xf ∩ Xg 6= ∅ if Xf , Xg 6= ∅. The latter condition is fulfilled if and only if f and g are not nilpotent (by exercise 17) hence the previous condition is equivalent to the following: there is a prime ideal p such that fg /∈ p if f, g are not nilpotent hence fg /∈ R if f /∈ R, g /∈ R . Therefore, X is irreducible if and only if the nilradical is prime. 1.20 We have the following: (i) If Y is an irreducible subspace if a topological space X, then Y is also irreducible, since by definition any neighbourhood of a boundary point will intersect Y (hence any two open sets in Y continue to intersect in Y ). (ii) We consider the set Σ of all irreducible subspaces of X; it’s not empty, since x ∈ Σ for all x ∈ X. Then, by an application of Zorn’s lemma in the usual fashion (any ascending chain of irreducible subspaces will be bounded by the union of all its elements which is irreducible itself) we guarantee a maximal element for Σ. (iii) The maximal irreducible components of X are obviously closed (otherwise their irreducible closures would strictly contain them, contradiction) and they cover X (we see that any point x of X is contained in a maximal irreducible subspace by applying Zorn’s lemma to the space Σx which comprises the irreducible subspaces of X that contain x). In a Hausdorff space each point is a maximal irreducible component. (iv) In the case of Spec(A) we note that the closed sets V (p), where p is a minimal prime ideal are irreducible (any two open sets will be of the form V (p)− V (E) and hence they will intersect) and that any two points x ∈ V (p1), y ∈ V (p2) can be separated by disjoint open sets. Therefore, the maximal irreducible components of Spec(A) are V (p), where p ∈ Spec(A) is minimal. 1.21 We have the following: (i) The following equivalences: q ∈ φ∗−1(Xf )⇐⇒ φ∗(q) ∈ Xf ⇐⇒ f /∈ φ∗(q) = φ−1(q)⇐⇒ q ∈ Yφ(f) yield that φ∗−1(Xf ) = Yφ(Xf ) Now φ ∗ is continuous, since the Xf form a basis for the Zariski topology. (ii) The following equivalences: q ∈ φ∗−1(V (a))⇐⇒ φ∗(q) ⊇ a⇐⇒ q ⊇ ae ⇐⇒ q ∈ V (φ(a)e) yield that φ∗−1(V (a)) = V (ae), as desired. 8 . (iii) The statement that φ∗(V (b)) = V (bc) follows in the same fashion as the previous one. (iv) By proposition 1.1, we know that φ∗(Y ) = V (ker(φ)) and φ∗ induces a bijective (and continuous, by the previous question) map from Y to V (ker(φ)). Thus we merely need to show that φ∗−1 is continuous. Let Y ′ = V (b) be any closed subset of Y and let a = φ−1(b). Then, the following equivalences: p ∈ φ∗(Y ′) = φ∗(V (b))⇐⇒ p = φ∗(q) ⊇ b⇐⇒ p = φ−1(q) ⊇ bc ⇐⇒ p ∈ V (bc) imply that φ∗(Y ′) = V (bc) and in particular that φ∗(Y ′) is closed when Y ′ is and therefore that φ∗ is a homeomorphism. In particular, the natural surjective projection map from A to A/R induces a homeomorphism between Spec(A) and Spec(A/R). (v) By the previous statement, we have φ∗(Y ) = φ∗(V (0)) = V (0c) = V (ker(φ)) thus φ∗(Y ) is dense in X ⇐⇒ φ∗(Y ) = V (ker(φ)) = X ←→ ker(φ) ⊆ p for all prime ideals p⇐⇒ ker(φ) ⊆ R. (vi) The desired result follows immediately by definition. (vii) By assumption, the two only prime ideals of A are p and 0, which implies that p is a maximal ideal of A and thus A/p is a field. This yields that the ring B = (A/p)×K will also have only two ideals, namely q1 = {(x, 0) : x ∈ A} and q2 = {(0, k) : k ∈ K}. The ring homomorphism φ : A −→ B defined by φ(x, x) is bijective (φ∗(q1) = 0 and φ∗(q2) = p) and continuous. However, φ∗ is not a homeomorphism. In the topological space Spec(B) = {q1, q2}, we have {q1} = V (q1) is closed as q1 * q2, but φ∗(q1) = 0 is not closed in Spec(A), since 0 is not a maximal ideal of A (by exercise 18). 1.22 A decomposition of A in the form A ' A1 ×A2 × . . .×An yields a decomposition Spec(A) ' Spec(A1)× Spec(A2)× . . .Spec(An) If we embed every space Spec(Ai) as Xi = (0, 0, . . . , Spec(Ai), 0, . . . , 0) in Spec(A), it’s a standard argument that the existence of the latter decomposition is equivalent to the decomposition of Spec(A) as a disjoint union of the Xi. Given a ring A, we have the following: (i) ⇒ (ii) This direction follows from our previous observation and the definition of connectedness. (ii) ⇒ (iii) If a decomposition of the form A ' A1 × A2 (where A1, A2 are non-trivial) existed, then a non-trivial idempotent element of A would be the pull-back of (1, 0). (iii) ⇒ (ii) If e ∈ A is a non-trivial idempotent, then X = Spec(A) decomposes as X1 unionsqX2, where X1 = {p ∈ X/e ∈ p} and X2 = {p ∈ X/e− 1 ∈ p}. It’s a trivial observation that X1 ∩ X2 = ∅ (as in our proof that a local ring possesses no non-trivial idempotents) and similarly trivial is the verification that X = X1 ∪X2. This decomposition implies that X is disconnected. 1.23. 9 1.23 We have the following: (i) Each f ∈ A is idempotent hence Xf induces a disjoint decomposition as in the previous exercise. It’s now obvious that the sets X1 and X2 with the notation of exercise 22 are simultaneously closed and open. (ii) By the formula of exercise 17, and by the fact that a Boolean ring is always a Principal Ideal Domain, we deduce that there is f ∈ A such that X(f1,f2,...,fn) = Xf1 ∪Xf2 ∪ . . . ∪Xfn = Xf (iii) The hint in the book is a full proof; let Y ⊂ X be both open and closed. Since Y is open, it is a union of sets Xf . Since Y is a closed subspace of a quasi-compact space, it is quasi-compact too hence it is a finite union of sets Xf , say Xf1 , Xf2 , . . . , Xfn . Now, (ii) finishes the proof. (iv) X is (obviously) compact and Hausdorff. 1.24 There is nothing to be provided other than a tedious verification of the axioms. 1.25 Stone’s Theorem that every Boolean lattice is isomorphic to the lattice of open and closed sets of some compact Hausdorff space follows immediately from exercises 23 and 24. 1.26 We will just repeat the construction of the book, which shows that X ' Max(C(X)), by the map µ : X → Max(C(X)) given by x 7→ mx = {f ∈ C(X)/f(x) = 0}. Note that mx is always a maximal ideal, as the kernel of the surjective map that sends f to f(x) (whence C(X)/mx ' R is a field). (i) Let m be any maximal ideal in X. Then, let V (m) be the set of common zeroes of functions in m, namely V (m) = {x ∈ X/f(x) = 0 for all f ∈ m}. We claim that V (m) 6= ∅. Indeed, otherwise, for every x ∈ X there is fx ∈ m, such that fx(x) 6= 0. Since fx is continuous, there is a neighbourhood Ux of x on which fx does not vanish. By the compactness of X, a finite number of these neighbourhoods cover X, say {Uxi}i=1,2,...n. Then, f = ∑n i=1 f 2 xi ∈ m, but f does not vanish on any point of X, hence it’s a unit, hence m = (1), contradiction. Therefore, V (m) 6= ∅. Let x ∈ V (m). Then, ⊆ mx, which implies m = mx by the maximality of m. Hence m ∈ Imµ and µ is surjective. (ii) By Urysohn’s lemma, the continuous functions separate the points of C(X) and this implies that mx 6= my if x 6= y. Hence µ is injective. (iii) For f ∈ C(X), let Uf = {x ∈ X/f(x) 6= 0} and Uf = {m ∈ X/f ∈ m}. We obviously have µ(Uf ) = Uf . Since the open sets Uf (resp. Uf ) form bases of the topologies on X and X we deduce that µ is also continuous (as is µ−1). Therefore X is homeomorphic to Max(C(X)). Affine algebraic varieties 1.27 There is nothing to be proved in this exercise if we invoke the Nullstellensatz for the surjectivity of µ. 10 . 1.28 We have the following situation: Ψ : [φ : X → Y, regular]↔ Homk(P (Y ), P (X)), where Ψ is defined by φ 7−→ Ψ(φ) : (η 7→ η ◦ φ). We see that Ψ is injective because η(φ1) = η(φ2) for all η implies φ1 = φ2 (obviously; just let η be the natural projections). It’s also surjective; if Ψ is any k-algebra homomorphism P (Y )→ P (X), then Ψ = φ◦η, where η is the polynomial transformation that sends the values of φ on P (X) to the values of Ψ. CHAPTER 2 Modules. 2.1 Since m and n are coprime, there are integers a and b such that am + by = 1. Therefore, given x ⊗ y ∈ (Z/mZ) ⊗Z (Z/nZ), we see that x ⊗ y = 1(x ⊗ y) = am(x ⊗ y) + nb(x ⊗ y) = a(mx ⊗ y) + b(x ⊗ ny) = a(0⊗ y) + b(x⊗ 0) = 0, and since every generator is identically 0, so will the whole tensor product be. 2.2 If we tensor the exact sequence a incl−→ A pi−→ A/a −→ 0, we obtain a⊗AM incl⊗1−→ A⊗AM pi⊗1−→ (A/a)⊗AM −→ 0 and this induces an isomorphism between (A/a)⊗AM and the cokernel of incl⊗1, which is (A⊗AM)/(a⊗A M) 'M/aM , since given any ideal a of A a trivial argument shows that a⊗AM ' aM (we need M to be flat for this to be true). Hence, (A/a)⊗AM 'M/aM, as desired. The above proposition is true even if M is not a flat A