Download Selected Exercises 1. Let M and N be R

Selected Exercises
1. Let M and N be R-modules. Write HomR (M, N ) for the R-module homomorphisms
from M to N . Show that HomR (M, N ) has a natural abelian group structure. What
condition on R enables HomR (M, N ) to be viewed as an R-module ? Describe the scalar
multiplication in this case and verify the required axioms.
2. Let φ : B → C be an R-module homomorphism. Show that for any R-module M , there
are induced homomorphisms
φ̃ : HomR (M, B) → HomR (M, C)
and
φ∗ : HomR (C, M ) → HomR (B, M ).
Prove that M is projective if and only if for all surjective φ : B → C, φ̃ is surjective. Prove
that M is injective if and only if for all injective φ : B → C, φ∗ is surjective.
3. Let 0 → K → M → C → 0 be an exact sequence of R-modules. Prove that for
R-modules A and B, the induced sequences
0 → HomR (A, K) → HomR (A, M ) → HomR (A, C)
0 → HomR (C, B) → HomR (M, B) → HomR (K, B)
are exact. What more can be said if A is projective or B is injective ?
4. An R-module F is said to be free on the set X if there exists a set map i : X → F
with the following (universal) property : Given an R-module M and a set map j : X → M ,
there exists a unique R-module homomorphism f : F → M such that f i = j. Prove that
an R-module F is a free R-module if and only if F is free on X, for some set X.
5. Let V be an infinite dimensional vector space over a field k and set R := Homk (V, V ),
the ring of k-vector space homomorphisms from V to V . Of course, {1} is a basis for R
considered as a module over itself. Find a basis consisting of two elements. Conclude that
∼ R ⊕ · · · ⊕ R (n times), as R-modules.
for all n ≥ 1, R =
6. Set R := k[X, Y, Z, U, V, W ], ∆1 := Y W − ZV , ∆2 := ZU − XW and ∆3 := XV − Y U .
Let I denote the ideal generated by ∆1 , ∆2 , ∆3 . Show that the following sequence of R-
EXERCISES
module homomorphisms provides a free resolution of R/I :
X
U
1
B
B
BY
B
@
V
C
C
C
C
A
0
( ∆ 1 ∆2 ∆3 )
Z W
π
→ R/I −
→ 0.
0−
→ R2 −−−−−−−−→ R3 −−−−−−−−−−−−→ R −
7. Prove McCoy’s Theorem : Let A be an n × m matrix with coefficients in R. The
corresponding homogeneous system of linear equations has a non-trivial solution if and only
if there exists 0 6= r ∈ R such that r · δ = 0 for every m × m minor δ of A. Conclude that
if φ : Rm → Rn is a one-to-one map of free R-modules, then m ≤ n. (Here and in the next
five exercises assume that R is commutative.)
8. An R-module M is said to have a finite free resolution (henceforth, FFR) if it has a
resolution
φn
φ1
0 → Fn −→ Fn−1 → · · · → F1 −→ F0 → M → 0,
with each Fj a finitely generated free R-module. The Euler characteristic of M , denoted
Pn
χ(M ), is defined by the formula χ(M ) = i=0 (−1)i rank(Fi ). Prove that the Euler characteristic is well-defined.
9. Suppose that 0 → A → B → C → 0 is an exact sequence of R-modules, each admitting
an FFR. Show that χ(B) := χ(A) + χ(C).
10. Let M be an R-module admitting an FFR and S ⊆ R a multiplicatively closed set.
Show that χ(M ) = χ(MS ). Conclude :
(i) If the annihilator of M contains a non-nilpotent element, then χ(M ) = 0.
(ii) If R is a domain with quotient field K, then χ(M ) = dimK (V ), where V is the K
vector space K ⊗R M .
11. Let R be a Noetherian ring. Suppose that M is an R-module admitting an FFR.
Show that χ(M ) ≥ 0. Prove that χ(M ) = 0 if and only if the annihilator of M contains a
non-zero-divisor. (Hint : Contemplate the domain case first.)
12. Reprove the hard direction of McCoy’s theorem using Euler characteristics.
(Hint : Reduce to the case where R is Noetherian.)
13. Let {Mα }α∈A be a collection of R-modules. Show that
L
α
Mα is a projective R-module
if and only if each Mα is a projective R-module. (?) Can you find an example where each
Q
Mα is projective yet Mα is not projective ? (This ain’t easy.)
14. Set R := R[X, Y, Z]/(f ), where f := X 2 + Y 2 + Z 2 − 1. Set K equal to the kernel of the
map φ : R3 → R which takes the column vector (a, b, c)t (“t” for transpose) to ax + by + cz,
where x, y, z denote the images of X, Y, Z in R.
(i) Show that K can be generated by three elements by finding an explicit set of generators.
(ii) Show that K cannot be generated by two elements.
(iii) Repeat the construction of K starting with the ring R̂ := C[X, Y, Z]/(f ) and call
the resulting R̂-module K̂. Show that K̂ is a free R̂-module by explicitly finding a
basis.
15. Let {Nα }α∈A be a collection of R-modules. Show that
Q
Nα is an injective R-module
if and only if each Nα is an injective R-module. (?) Suppose that R is left Noetherian, i.e.,
L
each left ideal is finitely generated. Show that α Nα is injective if and only if each Nα is
injective.
16. Show that Zp∞ is an injective Z-module, for each prime p. For the rings R := Zn or
R := k[X]/(X n ), k a field, show that R is injective when considered as module over itself.
(?) More generally, if we set R := k[X1 , . . . , Xd ]/(X1n1 , . . . , Xdnd ), then R is injective as a
module over itself.
17. Let k be a field and set R := k[X1 , . . . , Xd ]. Let E denote k[X1−1 , . . . , Xd−1 ]. Make E
into an R-module as follows : If X1f1 · · · Xdfd ∈ R is a monomial in the Xi ’s, then
(X1f1 · · · Xdfd ) · (X1−e1 · · · Xd−ed ) = X1−e1 +f1 · · · Xd−ed +fd
if fi < ei for all i, 0 otherwise. Extend this to all of R and E in the obvious way. Show that
E is an injective R-module.
18. An R-module D is said to be divisible if for each d ∈ D and non-zero-divisor r ∈ R,
there exists d0 ∈ D such that d = rd0 . Show that an injective R-module is divisible.
19. Let R be a commutative ring without zero divisors (i.e., an integral domain). An Rmodule M is said to be torsion-free if whenever rx = 0, for r ∈ R and x ∈ M , then r = 0
EXERCISES
or x = 0. Show that a torsion-free divisible R-module is injective. Conclude that K is an
injective R-module, for any field K containing R.
20. Let R be a Noetherian commutative ring and Q an injective R-module. Fix an ideal
I ⊆ R and set
ΓI (Q) := {x ∈ Q | I n x = 0, for some n ≥ 0}.
Show that ΓI (Q) is an injective R-module.
21. Let A and B be R-modules. Set I := annR (A) and J := annR (B). Show that
I + J ⊆ annR (ExtnR (A, B)), for all n ≥ 0.
22. Let R := Z48 . Set A := Z12 and B := Z16 , considered as R-modules. Calculate
ExtnR (A, B), for all n ≥ 0.
23. Set R := k[X, Y ]. Calculate ExtnR (k, R), for all n ≥ 0.
24. Let R denote the ring in Example 7.12. Calculate a (minimal) free resolution of k as
an R-module. For each n ≥ 1, calculate the nth Betti number, i.e., the rank of the nth free
module in the minimal resolution of k. Calculate ExtnR (k, R), for all n ≥ 0.
25. For any Z-module M , show that :
(i) M is torsion-free if and only if Ext1Z (M, Z) is divisible.
(ii) M is divisible if and only if HomZ (M, Z) = 0 and Ext1Z (M, Z) is torsion-free.
26. If gl.dim.(R) = ∞, show that there exists an R-module M satisfying p.d.R (M ) = ∞.
27. Let M be an R-module and x ∈ R a central non-zero divisor satisfying x · M = 0.
Assume p.d.R/xR (M ) < ∞. Show that p.d.R (M ) = 1+p.d.R/xR (M ). For R := k[X1 , . . . Xd ],
conclude that p.d.R (k) = d.
28. Let 0 → A → B → C → 0 be an exact sequence of R-modules. Show
p.d.R (B) = max{p.d.R (A), p.d.R (C)}, with equality except when p.d.R (C) = 1 + p.d.R (A).
29. Formulate and prove results parallel to 27 and 28 for injective dimension.
30. Comparison theorem for injectives. Consider the diagram
f0
f1
f2
g0
g1
g2
0 −−−−→ A0 −−−−→ M0 −−−−→ M1 −−−−→ · · ·


hy
0 −−−−→ A −−−−→ Q0 −−−−→ Q1 −−−−→ · · ·
where the top row is an exact sequence of R-modules and the bottom row is a complex
with each Qi injective. Show that there is a map of complexes, unique up to homotopy
equivalence, making the completed diagram commute. In other words, for each n ≥ 0, there
exists an R-module homomorphism ψn : Mn → Qn such that gn+1 ψn = ψn+1 fn+1 .
31. Prove the Snake Lemma, as stated in Lemma 10.2.
L
L
32. Let {Cα }α∈A be a collection of chain complexes. Show that Hn ( α Cα ) ∼
=
α Hn (Cα ),
for all n ∈ Z.
33. Let 0 → A → B → C → 0 be a short exact sequence of R-modules.
(i) Show that there exists a short exact sequence 0 → A → B → C → 0 of complexes,
such that A, B, C are projective resolutions of A, B and C, respectively. (Hint :
Start with resolutions for A and C first, and then try to build one for B.)
(ii) Let A0 , B 0 and C 0 be the complexes that result from deleting A, B and C from each
of the complexes A, B and C and applying HomR (−, D). Show that
0 → C 0 → B0 → A0 → 0 is a short exact sequence of complexes.
(iii) Write out the long exact sequence in homology associated to the short exact sequence
of complexes in (ii).
34. Let T be a tetrahedron with 4 vertices, 6 edges, and 4 two-dimensional faces. Thus,
the simplicial homology (with coefficients in R) of T is determined by the chain complex :
∂
∂
0 → R4 −
→ R6 −
→ R4 → 0.
Write down the matrices in this complex and verify computationally that H1 (T ) = 0 and
H2 (T ) ∼
= R.
= H0 (T ) ∼
35. A complex C is said to be split exact if it is exact and for each n ∈ Z, there exists a
map sn : Cn → Cn+1 such that d = dsn d. Show that C is a split exact chain complex if and
only if the identity map on C is null homotopic.
36. Let f : (C, d) −
→ (C 0 , d0 ) be a chain map between complexes. The mapping cone of f is
the complex M(f ) whose nth module is Cn0 ⊕ Cn−1 and whose nth boundary map is defined
by the equation
0 0 0
cn
d (cn ) + (−1)n−1 f (c0n−1 )
δn (
)=
.
cn−1
d(cn−1 )
EXERCISES
Show :
(i) (M(f ), δ) is a complex.
(ii) There exists a short exact sequence of complexes 0 −
→ C0 −
→ M(f ) −
→ C(−1) −
→ 0.
(iii) Write out the long exact sequence in homology associated to the short exact sequences of complexes in (ii). Show that the connecting homorphism is
f∗ : Hn (C) −
→ Hn (C 0 ).
(iv) M(f ) is exact if and only if f is a quasi-isomorphism.
d
n
Cn−1 −
→ · · · be a complex. For r ∈ Z, let C(r) denote the complex
37. Let C : · · · −
→ Cn −→
whose nth module is Cn+r and whose nth map is dn+r , for all n. In other words, C(r) is the
complex C “shifted” r units to the right or left. Show that Hn (C(r)) = Hn+r (C), for all n.
38. Let R be a ring, M the category of R-modules and T : M → M a left exact, additive
contravariant functor. Provide all the details required in showing that the right derived
functors {Rn T }n≥0 associated to T are well-defined, additive functors.
39. Formulate and prove a version of Corollary 11.10 for contravariant functors.
For the remainder of the exercises, assume that R is a commutative ring.
40. Let φ : Rn −
→ Rn and ψ : Rm −
→ Rm be R-linear maps between finitely-generated,
free modules. Let A and B respectively denote the matrices of φ and ψ with respect to the
standard bases.
(i) Describe C, the nm × nm matrix representing φ ⊗ ψ.
(ii) Calculate trace(C) and det(C).
41. Prove parts (i), (ii), (iv), (vii) in Proposition 12.3.
42. Let R = k[X, Y ], k a field. Calculate T oriR (k, k) for all i ≥ 0.
43. Let R and Q be the ring and module given in Example 7.12. Set J := Ry. Calculate
T ornR (Q, R/J), for all n ≥ 0.
44. Let A be an R-module and fix a presentation
φ
0−
→K−
→P −
→A−
→ 0,
with P a projective R-module and K := ker(φ). Prove that for any ideal I ⊆ R,
(i) T or1R (A, R/I) ∼
= (IP ∩ K)/IK.
(ii) Conclude that for all ideals J,
if and only if T or1R (R/I, R/J) = 0.
IJ = I ∩ J
(iii) Assume that R is a domain and A ⊆ F with F free and F/A torsion. Show that for
any torsion-free R-module B, T orR (F/A, B) ∼
= t(A ⊗R B).
1
45. Let M be a R-module and x ∈ R a non-zero-divisor on both M and R.
(i) If
ψ2
ψ1
ψ0
· · · → F2 −→ F1 −→ F0 −→ M → 0
is a free resolution of M , show directly that
ψ0
ψ0
ψ0
2
1
0
· · · → F2 /xF2 −→
F1 /xF1 −→
F0 /xF0 −→
M/xM → 0
is a free resolution of M/xM , considered as an R/xR-module. Here, the ψi0 ’s are
the obvious maps induced by the ψi ’s.
(ii) Conclude that T oriR (M, R/xR) = 0, for all i ≥ 1.
(iii) Now reprove (i) by calculating the T oriR (M, R/xR)’s an alternate way.
46. An R-module M is said to be flat if, whenever 0 −
→A−
→ B is exact,
0−
→ A ⊗R M −
→ B ⊗R M is exact. Show that :
(i) Projective modules are flat.
(ii) Hn (C) ⊗R M = Hn (C ⊗R M ), for any flat module M and complex of R-modules C.
47. Let A an R-module. A flat resolution of A is an exact sequence
F
:
··· −
→ F1 −
→ F0 −
→A−
→0
with each Fi a flat R-module. The flat dimension of A is the least integer n ≥ 0 (if it exists)
such that A has a flat resolution of length n. Thus, a flat module has flat dimension zero.
For an R-module A, show that the following statements are equivalent :
(i) A has flat dimension less than or equal to n.
(ii) T orjR (A, B) = 0, for all R-modules B and all j > n.
R
(iii) T orn+1
(A, B) = 0, for all R-modules B.
EXERCISES
48. Let C, D and E be chain complexes. Prove :
(i) C ⊗ D ∼
= D ⊗ C.
(ii) C ⊗ (D ⊗ E) ∼
= (C ⊗ D) ⊗ E.
(iii) C ⊗ (D ⊕ E) ∼
= (C ⊗ D) ⊕ (C ⊗ E).
49. Reprove Theorem 13.8 and Corollary 13.10 after replacing “projective” by “flat” in the
statement of the results.
50. Prove the Künneth Formula for Cohomology : Let P be a chain complex of projectives,
such that Bn (P) is projective for all n. Then for every R-module M and all n, there is a
(non-canonically) split exact sequence
0−
→ Ext1R (Hn−1 (P), M ) −
→ H n (HomR (P, M )) −
→ HomR (Hn (P), M ) −
→ 0.
51. Formulate and deduce the Universal Coefficient Theorem for Cohomology. Be sure
to state the special case for singular cohomology, recalling that if X is a topological space
with singular complex S(X), then the nth cohomology of X “with coefficients in M ” is, by
definition, H n (X; M ) := H n (HomZ (S(X), M )). For the intrepid : Prove the theorem for
cochain complexes, i.e., replace M by D, a cochain complex of R-modules.
52. Let A be a double complex with horizontal maps δ and vertical maps d. There is
Q
another total complex T π associated to A. The nth module in T π is Tnπ := i+j=n Aij and
π
→ Tn−1
is (still) defined by the equation
the nth boundary map ∂ π : Tnπ −
∂ π (aij ) = δ(aij ) + (−1)j d(aij ). Note that T π = T , if A is bounded above and to the right.
Now, assume that Aij = Z4 , for all i ≥ 0 and all j ∈ Z and that all maps in the complex are
multiplication by 2. Let B denote the “extension” of A to the entire plane, i.e., Bij = Z4 ,
for all i, j ∈ Z. Show :
(a) H0 (T π ) = Z2 (even though the rows of A are exact).
(b) Hn (T ) = 0, for all n ∈ Z.
(c) H0 (T π (B)) maps onto T π (even though all rows and columns of B are exact).
(d) T (B) is exact.
Compare the results of this exercise with Thereom 13.3.
53. Let x1 , . . . , xn ∈ R. Without using induction or Theorem 14.6, show directly that
K(x1 , . . . , xn ) ∼
= K(xσ(1) , . . . , xσ(n) ), for all σ ∈ Sn .
54. Let x1 , . . . , xn belong to the commutative ring R. Show that x1 , . . . , xn form a regular
sequence if and only if xe11 , . . . , xnen form a regular sequence for all ej ≥ 1, 1 ≤ j ≤ n. (If
x1 , . . . , xn form a regular sequence, you may use the fact that K(x) is exact, so that the 1−
boundaries in the Koszul complex generate the relations on x1 , . . . , xn .)
55. Let F be a free R-module with basis e1 , . . . en . Show that for each r ≥ 0,
Vr
F is a free
R-module with basis {ei1 ∧ · · · ∧ eir }, for 1 ≤ i1 < · · · < ir ≤ n.
56. Let φ : M −
→ N be an R-module homomorphism.
(i) Show that for each r ≥ 0, there exists an R-module homomorphism
Vr
Vr
∧r φ :
M−
→
N satisfying
∧r φ(m1 ∧ · · · ∧ mr ) = φ(m1 ) ∧ · · · ∧ φ(mr ),
for all m1 ∧ · · · ∧ mr ∈
Vr
M.
(ii) If M = Rm and N = Rn , calculate the matrix of ∧r φ with respect to the standard
bases. (Hint : Think “determinant”.)
57. Let x1 , . . . , xn ∈ R form a regular sequence. Use the Buchsbaum-Eisenbud Theorem
to show that the Koszul complex K(x) provides a free resolution of R/(x1 , . . . , xn )R.
58. Let x, y, z ∈ R be a regular sequence. Show that for all n ≥ 1, the complex
F(n)
:
φ3
φ2
φ1
0−
→ R3 −→ R6 −→ R4 −→ R
is exact, where φ1 = (xn+1 y n+1 z n+1 xn y n z n ),

−y n+1

 −xn+1
φ2 = 
 0

0
−z n+1
0
−y n z n
0
0
−z n+1
0
−xn z n
0
0
x
y
n+1
−x
0
y
n+1
0
 0

 0
0



 yn
0 
 and φ3 = 

 z
−xn y n 



 0
z
0
xn
0
0
z
−x −y
zn 
0 


0 

.
y 


−x 
0
59. Let (R, P ) be a Noetherian local ring. Prove the Auslander-Buchsbaum formula : If M
is a finitely generated R-module and p.d.R (M ) < ∞, then
p.d.R (M ) + depth(P, M ) = depth(P, R).
EXERCISES
Hint : Use Nakayama’s lemma and induction on depth(P, R). Note that this Theorem confirms the lesson of the Buchsbaum-Eisenbud Theorem, that modules having finite projective
dimension (equivalently, non-trivial finite length projective resolutions) cannot exist in the
absence of regular sequences.
60. Let (R, P, k) be a Noetherian local ring admitting a regular sequence x1 , . . . , xd such
that for J := (x1 , . . . , xd )R, J is P -primary. Show that the following conditions are equivalent.
(i) dimk (ExtdR (k, R)) = 1.
(ii) λ(M ) = λ(ExtdR (M, R)), for finitely generated R-modules M with J ⊆ ann(M ).
(iii) (J : (J : I)) = I, for all ideals I ⊇ J.
(iv) inj.dim.R (R) = d.
Hint : Regard this as a d-dimensional version of Theorem 5.7. A local ring is Gorenstein if
it satisfies the conditions above.