MATH 20A, FALL 207 HW 5 SOLUTIONS WRITTEN BY DAN DORE (If you find any errors, lease email ddore@stanford.edu) Question. Let R = Z[t]/(t 2 ). Regard Z as an R-module by letting t act by the identity. Comute Tor R k (Z, Z) and Extk R (Z, Z) for all k 0. Solution. We comuted a free resolution for Z on the revious assignment: fn+ R fn R f n f 2 R d R d 0 Z 0 Here, f n is multilication by t when n is odd and multilication by t + when n > 0 is even. Since R R M M for any R-module M, when we aly the functor ( ) R Z to this resolution (droing the last term), we get: δn+ Z δn Z δ Z Here, δ n = d n R id Z : R R Z R R Z. Under the isomorhism Z R R Z, which is given by n n, δ n : Z Z becomes the ma n (t ± ) n = n ± n. For n odd, this is n (t ) n = 0; for n > 0 even, this is n (t + ) n = 2n. So we can rewrite this comlex as: Z 0 Z 2 Z 0 Z So for k > 0 even, we have Tor R k (Z, Z) = ker(δ k)/ im(δ k+ ) = ker(δ k )/0 = ker(δ k ). Since δ k is multilication by 2, which is injective on Z, we have Tor R k (Z, Z) = 0 in this case. For k odd, we have im(δ k+ ) = 2Z and ker(δ k ) = Z, so Tor R k (Z, Z) = Z/2Z (viewed as an R-module by having t act as the identity). What about k = 0? Then Tor R 0 (Z, Z) = Z/ im(δ ) = Z. This makes sense: Z R/(t ), so Tor R 0 (Z, Z) = Z R Z = Z/(t ) Z = Z/0 = Z. To summarize: Z k = 0 Tor R k (Z, Z) = 0 k > 0, k is even Z/2Z k is odd All of these are given an R-module structure by having t act as the identity. Now, we can aly the contravariant functor Hom R (, Z) to our free resolution of Z, using the fact that Hom R (R, M) M for any R-module M: Z δ Z δ2 Z δ3 Via the natural isomorhism Hom R (R, M) M sending ϕ to ϕ(), the mas δ n : f f δ n become n (t ± ) m = m ± m. Thus, δ n = 0 when n is odd and δ n = 2 when n is even. So the comlex is: Z 0 Z 2 Z 0 When k > 0 is even, we have Ext R k (Z, Z) = ker δk+ / im δ k = Z/2Z. When k is odd, Ext R k (Z, Z) = ker δ k+ / im δ k = 0, and when k = 0, we have Ext R k (Z, Z) = ker δ0 = Z. These have R-module structures
via t acting by the identity, as before. To summarize: Z k = 0 Ext R k (Z, Z) = Z/2Z k > 0, k is even 0 k is odd Question 2. Let R = Z[ 30]. Regard F 2 as an R-module by letting 30 act by 0. Comute Tor R k (F 2, F 2 ) and Ext k R (F 2, F 2 ) for all k 0. Solution. Recall from last week that F 2 = R/I where I = (2, 30). Since I is rojective, as we know from class, we have a rojective resolution 0 0 0 I i R F 2 0. where i: I R is the inclusion. Alying ( ) R R/I and droing the last term, we get the comlex: which we can simlify to 0 0 0 I (R/I) i R (R/I) 0 0 0 0 I/I 2 0 R/I 0 Note that the ma is 0 since the image of i: I R is zero in R/I), so the kernels and images are even easier to comute, and we just get F 2 k = 0 Tor R k (F 2, F 2 ) = I/I 2 k = 0 k 2 The last thing to do is comute what I/I 2 is. Recall that I = {x + y 30 x 2Z, y Z}. Multilying out (2a + b 30)(2c + d 30) = (4ac 30bd) + (2ad + 2bc) 30 shows that I 2 {x + y 30 x 2Z, y 2Z} = (2), and we can guess that this might be equality. To show that this guess is correct, we just need to show that (2) I 2, i.e. that 2 I 2. This is easy: 2 2 = 4 belongs to I 2, and ( 30) ( 30) = 30 belongs to I 2, so 30 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 = 2 belongs to I 2. In articular, I 2 is an index-2 subgrou of I, and so I/I 2 F 2 (it is easy to check that this has the same R-module structure as the F 2 we started with). 2 Therefore: F 2 k = 0 Tor R k (F 2, F 2 ) = F 2 k = 0 k 2 If you wanted to rove it (you didn t have to) the easiest way is to show that I m is free for all maximal ideals m of R. Since R/I F 2 is a field, I is a maximal ideal, so I m = R m for m I. Then it s not so hard to see directly that for m = I, I m = mr m is rincial, i.e. that R m is a rincial ideal domain (this also follows from some general theory about rings like Z[ 30], called Dedekind domains). Then we wrote down a finite resentation for I, so by HW3 Q0, I is rojective. 2 For a maximal ideal m of a ring R, such as I above, the sace m/m 2 can be thought of as the cotangent sace of R at the oint m. It s a vector sace over the field R/m. This turns out to be a useful construction throughout commutative algebra and algebraic geometry. When a ring is sufficiently nice, such as Z[ 30], this dimension is the same for all m. 2
For Ext, we can aly the contravariant functor Hom R (, F 2 ) to the rojective resolution 0 0 I i R F 2 0 (droing the first term) to obtain 0 Hom R (R, F 2 ) i Hom R (I, F 2 ) 0 0 0 Note that Hom R (R, F 2 ) = F 2 has only one nonzero element, namely f : R F 2 sending x + y 30 to x mod 2. Since this restricts to 0 on I R, the ma i here is 0. So it remains only to comute Hom R (I, F 2 ). Note that as abelian grous I is isomorhic to Z 2, so there are only three nonzero grou-homomorhisms from I to F 2 : α(2a+b 30) = a mod 2; β(2a+b 30) = b mod 2; and γ(2a+b 30) = a+b mod 2. So we need only check which of these are R-linear. Since and 30 additively generate R, we need only check that they reserve multilication by these elements (and for this is automatic). Recalling that 30 acts by 0 on F 2, we just need to check whether nd so on. α ( 30 (2a + b 30) )? = 0 = 30 α(2a + b 30) α ( 30 (2a + b 30) ) = α( 30b + 2a 30) = 5b mod 2 = b mod 2 0 β ( 30 (2a + b 30) ) = β( 30b + 2a 30) = 2a mod 2 = 0 mod 2 = 0 γ ( 30 (2a + b 30) ) = γ( 30b + 2a 30) = 5b + 2a mod 2 = b mod 2 0 Therefore the only two R-linear homomorhisms from I to F 2 are the zero ma and β. So as an abelian grou Hom R (I, F 2 ) = F 2, and the fact that multilication by 30 annihilates β means this is the same R-module structure as always. We conclude: F 2 k = 0 Ext k R(F 2, F 2 ) = Hom R (I, F 2 ) = F 2 k = 0 k 2 Question 3. Let R = R[T ]. Let M = R 2, with R-module structure where T acts by ( ) 2 0. Let N = R with R-module structure where T acts by 0. Comute Tor R k (M, N) and Extk R (M, N) and Extk R (N, M) for all k 0. Solution. On the last assignment, we comuted a free resolution for M: 0 0 R d R d 0 M 0 Here, d is multilication by (T ) 2. If we aly the functor ( ) N to this, after droing the M term, we get the comlex: 0 0 N (T )2 N 0 We used the same reasoning as in Question to determine the comlex: R R N N, and the ma f 0 id N becomes the action by (T ) 2 under this isomorhism. Now, since T acts by 0 on N, the element T acts by (T ) n = n. In articular (T ) 2 n = ( ) 2 n = n, so the action of (T ) 2 on N is the identity ma. 3
Thus, we have Tor 0 R (M, N) = N/N = 0, Tor R (M, N) = 0/0 = 0, and of course the higher Tor s vanish since the comlex is 0 after the first two terms. So we have Tor k R (M, N) = 0 for all k. If we aly Hom R (, N) to the sequence, we get: 0 N (T )2 N 0 0 Again, by the same reasoning as in Question, Hom R (R, N) N and multilication by (T ) 2 becomes the action by (T ) 2 under this isomorhism. Since (T ) 2 acts by the identity ma on N, we get that Ext R k (M, N) = 0 for all k just as above. In order to comute Ext k R (N, M), we either need a rojective resolution of N or an injective resolution of M. In general, it s much easier to comute rojective resolutions, so that s what we ll do. We have a surjective ma π : R R sending (T ) to (T ). Since T acts by 0, the kernel is (T ). Since R is a domain, the ma R R given by multilication by T is injective, so we have a free resolution: 0 0 R T R π N 0 Alying the functor Hom R (, M) (after droing the N term) and using the same reasoning as before, we get the comlex: 0 M T M 0 0 So Ext 0 R (N, M) = ker T and Ext R (N, M) = coker T, where we think of T as the linear transformation ( 2 ) 0 acting on M. But T is an isomorhism (det T = ), so Ext k R (N, M) = 0 for all k. Question 4. Let R = C[T ]. Given λ C, let C λ denote C regarded as an R-module by letting T act by λ. Comute Ext k R (C λ, C µ ) for all k 0, for all λ, µ C. Solution. We need to comute a rojective resolution of C λ. Since C λ is generated by as a C-module and therefore as an R-module, we have a surjection π : R C λ sending (T ) to (T ) = (λ). So the kernel is {(T ) R (λ) = 0}. This is the rincial ideal (T λ): if (T ) R, we can write (T ) = (T λ)q(t ) + r with deg r < deg(t λ) =, so r is a constant. Then (λ) = r, so (λ) = 0 iff (T λ) (T ). So we have a free resolution: 0 0 R T λ R C λ 0 Now, droing the C λ term and alying the contravariant functor Hom(, C µ ), we get: 0 C µ T λ C µ 0 0 As before, we use Hom R (R, M) M, and that multilication by T λ in R corresonds under this isomorhism to the action of T λ. Since T acts by µ on C λ, we know that T λ acts by µ λ. So we have two cases: if µ = λ, then the ma is 0 and we have Ext 0 R (C λ, C λ ) = C λ and Ext R (C λ, C λ ) = C λ. We also clearly have Ext k R (C λ, C λ ) = 0 for k 2. C λ k = 0 Ext k R(C λ, C λ ) = C λ k = 0 k 2 4
If we have µ λ, the ma is multilication by µ λ, which is an isomorhism, so we have Ext 0 R (C λ, C µ ) = Ext R (C λ, C µ ) = 0, so Question 5. Let R = C[x, y]. Ext k R(C λ, C µ ) = 0 for all k 0. (a) Regard C as an R-module by letting x and y act by 0. Comute Tor R k (C, C) for all k 0. (b) Let I R be the ideal I = (x, y). We would like to understand I R I, so: Give a basis for I R I as a comlex vector sace. If you can also describe the R-module structure without too much ain, lease do. Solution. (a) We comuted in the last assignment (for R = R[x, y] and M = R with R-module action by sending x, y to 0; but nothing changes when you relace R with any other field) the following free resolution: 0 0 R d 2 ( ) y R 2 d R d 0 M 0 ( x y ) x Here, f 0 and f are given by f 0 (, q) = x + yq and f (r) = (yr, xr). Alying the functor R C, we get: 0 0 C 0 C 2 0 C 0 Here, we used the fact discussed earlier that R n R (R/I) (R/I) n, and a matrix for d id R/I is given by reducing a matrix for d mod I. But both the matrices ( y ) x and ( x y ) become zero when we tensor with C (since x and y act by 0 there), which is how we know these mas are 0. This tells us that C k = 0 Tor R k (F C 2 k = 2, F 2 ) = C k = 2 0 k 3 (b) [Note from TC: I think the simler way to do this is to first show that ker(i R I I 2 ) = Tor (I, R/I), and then to show Tor (I, R/I) = Tor 2 (R/I, R/I), which we just roved is C. Therefore once we find one element in this kernel (namely x y y x) we just need any set that descends to a basis for I 2. But the aroach below works as well.] Note that I is the kernel of the ma d 0 : R M in the revious art, so our free resolution for M gives us a free resolution for I: 0 R d ( ) y R 2 d0 ( x y ) x I 0 Using this free resolution, we can comute I R I = Tor R 0 (I, I), by alying the functor ( ) R I, which yields the comlex: δ I ( ) y I I x 5
Here, we are identifying R R I with I and R 2 R I with I I (this should not be referred to as I 2, since that tyically refers to the ideal I I). Thus, I R I coker δ. Note that the identification of this cokernel with I I is via the ma ( x y ): R 2 I, so we should think of I I as (x I) (y I). In articular, the ma π : I I I I = coker δ is given by sending (, q) to (x + y q). Similarly, we should think of δ as sending (x, y) I to (x y (x, y), y x (x, y)). Certainly x y y x = 0, and we ve seen that this actually generates the kernel of π. Let (, q) I I. Then we can write uniquely as (x, y) = x 0 (x) + y (x) + y 2 2 (x) + + y n n (x) for some olynomials i (x) R, and any choice of i R give an element of I. Letting (x) = a + x (x) with a C, we can then write uniquely as ( ) = x 0 (x) + ya + y x (x) + y 2 (x) + y 2 3 (x) + + y n n (x) =: x 0 (x) + ya + yq(x, y) Note that q(x, y) I, and that if x 0 (x) + ya + yq(x, y) = x 0 (x) + ya + yq (x, y) with q, q I, then we can rearrange this to give: y ( q(x, y) q (x, y) ) = x ( 0(x) 0 (x) ) + y ( a a ) Since the left hand side is divisible by y, the right hand side must be as well, so 0 = 0. This imlies that a a = q q I, which means that a = a, since I is a roer ideal. Thus, any element I may be written uniquely as = x 0 (x) + ya + yq(x, y) with q(x, y) I. Therefore, we can write any element of I I uniquely as (x 0 (x) + ya, r(x, y)) + (y q(x, y), x q(x, y)) with r(x, y), q(x, y) I, so the second term is in im δ. This means that sub-c-vector sace of I I given by elements of the form (x 0 (x) + ya, r(x, y)) mas isomorhically under π to I R I. We can take as a C-basis. {(x a, 0)} a {(0, x b y c )} b+c {(y, 0)} This says that a C-basis for I R I is given by the elements x x a with a, together with the elements y x b y c with b + c, as well as the element x y. Question 6. Prove that if M is torsion-free and finitely generated, then Tor k (M, X) = 0 for all k > 0 and any X. Solution. One way to rove this where the hyothesis that M is finitely generated is to show that torsion-free finitely generated modules over a PID are free. Since this is an imortant art of the structure theorem for finitely generated modules over a PID, I won t include the roof here. Alternatively, there is an elementary argument using the equational criterion for flatness. This was Q on HW3: a finitely resented module is rojective iff every linear deendence is trivial. 3 3 In general, an R-module M is flat iff every linear deendence is trivial, without needing to worry about finite resentation hyotheses, and this is strictly weaker than rojectivity away from the finitely generated case (for examle, Q is flat over Z but not rojective). The equational criterion of flatness usually refers to this more general statement. For a fun exercise, see if you can rove the general criterion. The roof is not very hard, and uses similar ideas to ones aearing in this assignment, namely that flatness of a module M can be checked by showing that for every ideal I, I R remains injective after tensoring with M 6
We can show fairly easily that any finitely generated module over a PID is finitely resented. Indeed, let π : R k M be a surjection. We want to show that ker π R k is finitely generated. In fact, this is true for any submodule K R k by induction on k. When k =, this is the statement that any ideal of R is finitely generated. In fact, any ideal of R is generated by a single element, since R is a PID, so this settles k =. Now, we can induct on k. We have a short exact sequence: 0 (K R k ) K K/(K R k ) 0 Here, we embed R k into R k by sending a basis to the first n coordinates. By induction, K R k R k is finitely generated, and K/(K R k ) R k /R k R, so K/(K R k ) is also finitely generated. Therefore, K is finitely generated. Now, it suffices to show the equational criterion. Let a m + + a n m n be a linear deendence in M, and suose without loss of generality that a i 0 for each i. Because R is a PID, the ideal (a,..., a n ) is rincial, so it is of the form (r) for some r R. Since the a i are nonzero, r 0. Thus, for each i, a i = ra i for some a i R. So we can write: 0 = ra m + + ra nm n = r (a m + + a ) nm n Since M is torsion-free and r 0, we have a m + + a nm n = 0 Now, consider the ideal (a,..., a n) = (r ) for some r R. Then we know that r a i for all i, so rr ra i = a i for all i. Thus, (r) = (a,..., a n ) (rr ), so we have r = (rr )r for some r R; since R is a domain, this imlies that r r =, so r is a unit, i.e. (a,..., a n) = R. We can multily a trivializing relation for the linear deendence 0 = a m + + a nm n by r to get one for 0 = a m + + a n m n, so by renaming a i to a i, we may now assume that (a,..., a n ) = R. Thus, there are r i R such that r a + + r n a n =. Now, for each i, we can write: m i = (r a + + r n a n ) m i = r j a j m i + r i a i m i j i = r j a j m i r i (a m + + a i m i + a i+ m i+ + + a n m n ) j i = r j a j m i + i a j )m j j i j i( r Define v j = m j and b j i to be the coefficient of m j in the last equation above: if i j, b j i = ( r ia j ) and if i = j, then b i i = j i r ja j. Thus, m i = j bj i vj. 7
Now, it suffices to show that for each j, n i= a ib j i This concludes the roof. = 0. But this is: n a i b j i = a i ( r i a j ) + a j b j j i= i j = a j r i a i + a j r i a i i j i j = 0 Question 6. (relaces Q6) Prove that if M is torsion-free, then Tor k (M, X) = 0 for all k > 0 and any finitely generated X. Solution. We induct on the number of generators of X. If X is generated by a single element, then X R/I for some ideal I. Since R is a PID, I = (r) for some r R, so X R/(r). If r = 0, then X R is free, and therefore rojective, so Tor k (M, X) = 0 for all k and all m. Consider the exact sequence: 0 R r R X 0 Alying the functor M R ( ) and using the long exact Tor sequence, we get an exact sequence: Tor R (M, R) = 0 Tor R (M, X) M R R r M R R M R R/(r) 0 Thus, we may identify Tor R (M, X) with the kernel of the ma M R R M R R given by multilication by r on the second factor. We may identify M R R with M, so this is just the ma M M given by multilication by r. The kernel of this ma is then exactly the set of m M with r m = 0. Since M is torsion-free and r 0, this imlies m = 0. Thus, we know that Tor R (M, R/(r)) = 0 for any r R. Now, let x,..., x n generate X, and let X be the R-submodule of X generated by x,..., x n. We have a short exact sequence: 0 X X X/X 0 Here, X/X is generated by a single element, so it is isomorhic to R/(r) for some r R. Now, we can take the Tor long exact sequence to get: Tor k (M, X ) Tor k (M, X) Tor k (M, X/X ) But for any k, since both X/X and X are generated by fewer than n elements, we may assume by induction that Tor k (M, X ) = 0, Tor k (M, X/X ) = 0, so this reads: Thus, Tor k (M, X) = 0 for all k. 0 Tor k (M, X) 0 Question 7. Deduce from Q6, or from Q6, or rove directly: for any torsion-free M, Tor k (M, X) = 0 for all k > 0 and any X. 8
[If you give a self-contained direct roof for Q7, you will automatically get credit for Q6.] Solution. Both deductions from Q6, Q6 are similar. Let s show Q6 = Q7 first. Let M be an arbitrary torsion-free module. We want to show that Tor k (M, X) = 0 for all R-modules X, which is equivalent to showing that M is flat. We will show directly that if ϕ: X Y is an injective homomorhism, then id M R ϕ: M R X M R Y is injective. Now, let β = n i= m i x i be an arbitrary element of M R X and suose that ϕ(β) = 0. We want to show that (id M R ϕ) ( i m ) i x i = i m i ϕ(x i ) is non-zero. Since only finitely many elements of M aear in this sum, there is a finitely generated submodule M M which contains m,..., m n. Since submodules of torsion-free modules are torsion-free, we know that M is torsion-free and finitely generated. By Q6, this imlies that Tor k (M, X) = 0 for any R-module X, i.e. that M is flat. We have a commutative diagram: M R X id M ϕ M R Y ι id X M R X id M ϕ ι id Y M R Y Alying this to α = i m i x i M R X, we get that (ι id Y ) ( (id M ϕ)(α) ) = (id M ϕ) ( (ι id X )(α) ) = (id M ϕ)(β) = 0 Since we do not know if Y is flat, we cannot conclude immediately that γ := (id M ϕ)(α) = 0. However, we know that any finitely generated submodule of M is flat. Let M be a finitely generated submodule of M containing M. Then we can extend the above commutative diagram to: M R X id M ϕ M R Y ι id X ι id Y M R X id M ϕ M R Y ι 2 id X M R X Note that we have ι = ι 2 ι, so we know that id M ϕ ι 2 id Y M R Y 0 = (ι id Y )(γ) = (ι 2 id Y ) (ι id Y )(γ) If we can find some such M such that (ι id Y )(γ) = 0, then we have: 0 = (ι id Y )(γ) = (id M ϕ) (ι id X )(α) Since M is flat, id M ϕ is injective, so this means that (ι id X )(α) = 0. But β = (ι id X )(α) = (ι 2 id X ) ( (ι id X )(α) ), so this imlies β = 0, as desired. To see that there is a finitely generated submodule M M with M M such that (ι id Y )(γ) = 0, we recall the construction of the tensor roduct M R Y. This is defined as the free abelian grou A on the symbols m y with m M, y Y, modulo relations of the form ρ m,m 2,+ := (m + m 2 ) y m y m 2 y and ρ r,m, := rm y = m ry for all m, m 2, m M, y Y, and r R. 9
We can write γ = n i= m i y i, and its image in M Y is reresented by A = n i= m i y i A. Since it is 0 in M Y, this means that A is in the subgrou of A generated by ρ m,m 2,+, ρ r,m,. Thus, there are finitely many m i, mi 2, rj, m j such that A = N i= ρ m i,mi 2,+ + M j= ρ r j,m j,. We can then take M to be the R-submodule of M generated by the m i, mi 2, mj and the generators of M. This is a finitely generated submodule of M. Then, (ι id Y )(γ) is reresented by A A A, where A is the free abelian grou corresonding to M Y, which is clearly a subgrou of A (it is the free abelian grou on a subset of the generators). But by construction, the ρ m i,m i 2,+ and ρ r j,m j, are in A. Then, since A A is injective, the equation A = i ρ m i,mi 2,+ + j ρ r j,m j, holds in A, so A is in the kernel of A M R Y. This means that (ι id Y )(γ) = 0 as desired. The deduction that Q6 = Q7 is very similar. We need to show that for any ϕ: X Y, the ma (id M ϕ): M X M Y is injective. Let β = i m i x i be in the kernel. Taking X to be any finitely generated submodule of X containing the x i, we can define α := i m i x i, thought of as an element of X. Since the homomorhic image of a finitely generated module is finitely generated, we get a ma ϕ : X Y with Y Y finitely generated. This gives us a commutative diagram: M R X id M ϕ M R Y M R X id M ι X id M ϕ M R Y id M ι Y Thus, γ := (id M ϕ )(α) is in the kernel of id M ι Y. Exactly as in the roof that Q6 = Q7 (note that we did not use torsion-freeness for this art of the roof), we can see that there is a finitely generated submodule Y Y with Y Y such that if ι : Y Y is the inclusion, (id M ι )(γ) = 0. We get a ma ϕ : X Y defined by ι ϕ, and this gives a commutative diagram: M R X id M ϕ id M ϕ M R Y id M ι M R Y M R X id M ι X id M ϕ M R Y id M ι Y Thus, we see that 0 = (id M ι )(γ) = (id M ϕ )(α) = 0. But since ϕ : X Y is the comosition of the injective mas ϕ : X Y and ι : Y Y, it is injective. Now, consider the exact sequence: 0 X Y coker ϕ 0 Since coker ϕ is a quotient of the finitely generated module Y, it is finitely generated, so Tor (M, coker ϕ ) = 0. Thus, taking the Tor long exact sequence, we get: 0 = Tor (M, coker ϕ ) M R X id M ϕ M Y M (coker ϕ ) 0 Thus, id M ϕ is injective, so the fact that (id M ϕ )(α) = 0 imlies α = 0, and thus β = (id M ι X )(α) = 0. 0
Question 8. Deduce from the revious question that for any M, Tor k (M, X) = 0 for all k > and any X. Solution. For any R-module M, there is a short exact sequence 0 K F M 0 with F free and K F. Since submodules of torsion-free modules are torsion-free, we know that K is torsion-free. Now, for any X, we can take the long exact Tor sequence. For any k 2, we have the following iece: 0 = Tor k (F, X) Tor k (M, X) Tor k (K, X) = 0 Here, we used Q7 and the fact that k to show that Tor k (K, X) = 0. Certainly we know that for a free module F, Tor k (F, X) = 0 as soon as k > 0. Thus, Tor k (M, X) = 0 for k >. Do at least one of the following questions. If you ve seen one of these questions before, lease at least try to do one of the others. Question 9A. Comute Ext Z (Q, Z). Solution. Consider the short exact sequence: We can take the long exact Ext(Q, ) sequence: 0 Z Q Q/Z 0 0 Hom(Q, Z) Hom(Q, Q) Hom(Q, Q/Z) Ext (Q, Z) Ext (Q, Q) = 0 We know that Ext (Q, Q) = 0, since Q is an injective Z-module. Now, Hom(Q, Z) = 0: if q Q, we can write q = nq for any n, so if f Hom(Q, Z) then f(q) = nf(q ), i.e. f(q) is divisible by n for all n, which is clearly imossible unless f(q) = 0. We also know that Hom(Q, Q) Q, since any Z-linear ma f from Q to Q is just multilication by an element of Q. So we have: Ext (Q, Z) Hom(Q, Q/Z)/Q Thus, it suffices to describe the grou Hom(Q, Q/Z). Let s start by describing the structure of Q/Z. For any rime, there is the subgrou Z[ ]/Z, consisting of elements of the form a with a and 0 a < k. Putting all of these subgrous together, we get a k ma from Z[ ]/Z = (Z[ ]/Z) to Q/Z. This ma is injective: if a m + b n = an+bm nm = 0 in Q/Z with n, m corime, then an+bm nm Z, i.e. nm an + bm, so n bm and m an. But since n, m are corime, this a means that m a and n b. Thus, Z[ ]/Z as: a m and b n k are in Z, so they are 0 in Q/Z. Now, we can write an element of + + a n kn n = N k k n n + a n kn n Thus, the above argument shows that an Z, so we may induct on n to show that the whole sum is in Z, kn n and therefore 0 in Z[ ]/Z. Now, we will show that Z[ ]/Z Q/Z is actually an isomorhism. To do this, let a q Q with q = q q 2 corime. Then we may write = aq + bq 2 for some a, b Z (e.g. by the Chinese Remainder
Theorem, or by the fact that Z is a PID, so the ideal (q, q 2 ) is (gcd(q, q 2 )) = ()). Then we can take the artial fraction decomosition: q q 2 = aq + bq 2 q q 2 = a q 2 + b q By breaking q into its rime factorization and reeatedly using this identity, we may write q as an element in the image of Z[ ]/Z. Remember that for any modules M i, i I for some set I, the direct sum i M i embeds into the direct roduct i M i as the set of elements such that all but finitely many factors are 0. So we will start by describing Hom (Q, ) Z[ ]/Z. By the universal roerty of roducts, a ma to the roduct is the same as a tule of mas to each factor, i.e. we have: Hom Q, Z[ ] /Z ( ] ) Hom Q, Z[ /Z Now, we ve broken the roblem u one roblem for each rime. Now, we want to characterize homomorhisms from Q to Z[ ]/Z. Such homomorhisms of course restrict to homomorhisms from Z[ ] to Z[ ]/Z, and in fact any such homomorhism f extends uniquely to Q. To see this, we will use the following: Claim. The grou Z[ ]/Z is uniquely divisible by numbers corime to : for any α Z[ ]/Z and n N with n, there is a unique α Z[ ]/Z such that n α = α. Proof. We can write α = a k + Z with a, 0 a < k. Since, n, n and k are corime, so there are b, c Z with b k + cn =, so ab k + acn = a. Thus, we can write α as: α = a k + Z = abk k + acn k + Z = n ac k + Z Thus, we may take α = ac + Z. We want to show that α is unique, so let β be an element of Z[ k ]/Z with nβ = α. Then n(β α ) = 0, so it suffices to show multilication by n is injective. Now, let γ = m + Z Z[ nm l ]/Z. If nγ = 0, then Z, so l nm. Since n, this means that l m, so l γ = 0. Now, let f Hom(Z[ ], Z[ ]/Z). We want to show it extends uniquely to f Hom(Q, Z[ ]/Z). a Write any element of Q uniquely as k m with m, (k m, a) =, and m > 0. Let α = f( a ), which is k defined since a Z[ a ]. We can define f( k m ) as the unique element α such that m α = α. This gives a well-defined function from Q to Z[ f( ]/Z, and it is easy to see that it is additive and extends f. Moreover, it is ) unique since we need m a = f( a ). k m k Thus, we need to determine Hom(Z[ ], Z[ ]/Z). Let f 0 be such a homomorhism and consider α = n f 0 () Z[ ]/Z. We have α = a for some n 0 with a, so n α = a = 0 and m a 0 for m < n. Define f = n f 0, so f() = 0. Then, we define a sequence (m n ) := (f( )) n n for n. We have m n = m n, and n m n = f() = 0 for all n. On the other hand, given such a sequence (m n ) with m n Z[ ]/Z such that m n = m n and n m n = 0 for all n, we can define a homomorhism f : Z[ ] Z[ ]/Z with m n = f( n ) and f() = 0. To do this, we define f( a n ) = am n with a Z. If 2
we rewrite a a as, then since am n n+ n+ = am n, these definitions agree. This allows us to check that f is additive: if we have x = a and y = b, then n k ( ) a k + b n f(x + y) = f = a k m k+n + b n m k+n = am n + bm k = f(x) + f(y) n+k Now, we can describe the set of sequences (m n ) n with m n = m n and n m n = 0 for all n a bit differently. The second condition says exactly that m n = a for some a (erhas not corime to ). Since a n is only defined mod n, we can think of m n as living in Z/ n Z instead. Then m n = a = a, so the n n condition that m n = m n can be rehrased as saying that m n Z/ n is equal to m n mod n. Thus, the subgrou of Hom(Z[ ], Z[ ]/Z) with f() = 0 is isomorhic to the grou of sequences (m n) with m n Z/ n Z such that π n,n (m n ) = m n, where π n,n is the ma from Z/ n Z to Z/ n Z given by reducing mod n. Another name for this grou is Z, the -adic integers. Note that this is consist(z[ ]), since k m Z[ ] = k Z[ ] for m. Now, for any element f Hom(Z[ ], Z[ ]/Z) and any n, there is a unique f 0 Hom(Z[ ], Z[ ]/Z) with n f 0 = f: we can take f 0 (x) = f( x ), and this is unique since multilication by n on Z[ n ] is injective. Thus, Hom(Z[ ], Z[ ]/Z) has a unique structure of a Z[ ]-module. Since for any f Hom(Z[ ], Z[ ]/Z), there is some n such that n f() = 0, we can write f as f with f n () = 0. This shows that Hom(Z[ ], Z[ ]/Z) Z [ ] = Q, the -adic numbers as an abelian grou. Thus, we see that Hom(Q, Z[ ]/Z) Q. The submodule Hom(Q, Q/Z) Hom(Q, Z[ ]/Z) is given by elements (f ) such that for each x Q, f (x) = 0 for all but finitely many. This is the subgrou of (a ) Q such that for all but finitely many, a Z. To see this, let x = m n. For all nm, f (x) = a f () for some a Z with (a, ) =, by the definition of the isomorhism from Hom(Z[ ], Z[ ]/Z) Hom(Q, Z[ ]/Z) and the roof of Claim. Thus, f (x) = 0 iff f () = 0. Thus, we see that a sequence (f ) satisfies the condition that for all x, (f )(x) = 0 for all but finitely many x iff f () = 0 for all but finitely many x, iff the corresonding element (a ) Q is in Z for all but finitely many. We call the resulting grou A f Q := Z Q, where the Z stands for restricted roduct and it means the subset of the roduct where all but finitely many entries are in Z. This grou has a natural ring structure given by comonent-wise multilication, and is called the finite adele ring of Q, and is studied widely in number theory. 4 Finally, we see that Ext (Q, Z) A f Q /Q, where the ma Q F is given by sending q Q to the ma (f ): Q Z[ ]/Z with f multilication by q for each. This corresonds to the element (ι (q)) A f Q, with ι Q Q defined by sending a ( with a, m to k m m a (mod ) n) k n (where m is an inverse to a mod n, which exists for each n but deends on n). Since the denominator of q is only divisible by finitely many rimes, we see that ι (q) Z for all but finitely many, so this in fact lands in A f Q. π d n If M is a Z-module, note that d n imlies nm dm, so there is a quotient ma : M/nM M/dM (it descends from the identity M M, so in symbols it s just m m). Define consist(m) to be the submodule of n N M/nM defined by consist(m) : { (m n M/nM) n N d n = πn(m d } n ) = m d 4 The full adele ring A Q is A f Q R: sometimes it is useful to think of R as being the rime at infinity. This ring has a locally comact toology coming from the locally comact toologies on R and Q, and many imortant results in number theory can be reformulated in terms of this toological ring. 3
This makes consist an additive functor from Z-modules to Z-modules (you do not have to rove this). Question 9B. Is consist an exact functor? Prove your answer is correct. Solution. Since Q/nQ = 0 for all n N, consist(q) n N Q/nQ = 0, so consistent(q) = 0. Since Z Q is injective, in order to show that consist is not an exact functor, it suffices to show that consist(z) 0. This will be clear from the descrition in Question 9C, but for now note that there is an injective ma Z consist(z) defined by sending m Z to (m (mod n)) n N. Certainly, if d n, then πn(m d (mod n)) = m (mod d), so the image of this ma is contained in consist(z). The ma is injective since if m (mod n) = 0 for all n N, then n = 0. Question 9C. consist(z) has a natural ring structure (for examle, it is a subring of n N Z/nZ); you do not have to rove this. Describe the commutative ring Q Z consist(z). (You have some flexibility here in what your descrition should be, but don t just rehrase the definition.) Solution. First, we will use the Chinese remainder theorem: Z/nZ Z/ k Z Z/km m Z for n = k km m its rime factorization. Let d i = k i i. Then the mas πd i n : Z/nZ Z/k i i Z corresond to the i-th rojection mas in the roduct decomosition Z/ k Z Z/km m Z. So if (m n ) consist(z),,..., m km m ) in this roduct descrition, so the collection of m l for a rime and l > 0 then m n = (m k comletely determine (m n ), and conversely, any collection of the m l which are consistent with resect to the π d n where n, d are both owers of the same rime define an element of consist(z). In other words, consist(z) consist (Z), where we define consist (Z) to be the set of sequences (m n) with m n Z/ n Z such that π k n(m n) = k for all k n. This is even an isomorhism of rings, since the ring structure on consist(z) is defined by comonent-wise multilication (i.e. (m n ) (m n) = (m n m n), and it s easy to check this reserves consistency, since the πn d are ring homomorhisms), and the Chinese remainder theorem gives an isomorhism of rings. Note that it is equivalent in the definition of consist (Z) to require that π n n (m n) = m n for all n, since the n are linearly ordered by divisibility. Now, consist (Z) is usually referred to as Z, the -adic integers. Thus, we see that consist(z) Z as rings. Let s see that consist(z) Z Q A f Q, the finite adele ring defined in the solution to Question 9A. Essentially, this is true because tensoring with Q is the same thing as adjoining n for all n N, and n has only finitely many rime divisors. More recisely, we define a homomorhism consist(z) Z Q A f Q by sending (a ) q to (ι (q)a ), with ι : Q Q the embedding defined in Question 9A. Since (a ) Z and for all but finitely many, ι (q) Z, we see that the image of this ma lands in A f Q. To see that it is an isomorhism, note that if we have a tensor of the form (a ) m n + (b ) m n, we can rewrite this as (mn a ) nn + (nm b ) nn = (mn a + nm b ) nn Thus, any element of Z Q may be written as (a ) n. Then the ma is certainly injective, since ι ( n )a is only 0 when a is 0. It is also surjective: given a finite adele (a ) A f Q, let,..., m be the finitely many rimes with a Z, and assume k i i a i (b ) := n (a ) Z. Thus, we ma (b ) n to (a ). Z for each i. Then let n := i k i i, and let 4