MATH 807 Algebra II
PROBLEM SET 5
Due Wednesday, March 231
Problem 1. Let R be a PID. Prove that an R-module M is flat if and only if M is torsionfree.
Problem 2. Let R be an integral domain and x 6= 0 be an element of R. Compute
TorR
1 (R/(x), M ) for an arbitrary R-module M .
Recall that an R-module P is called projective if the functor2 HomR (P, −) is exact.
Because HomR (P, −) is easily seen to be left exact, P is projective if and only if for every
surjection of R-modules M N , the induced map HomR (P, M ) → HomR (P, N ) is also
surjective.
Problem 3. (Do this only if you have never done this before.)
Prove that HomR (P, −) is a left exact (covariant) functor for every R-module P . Prove
that HomR (−, I) is a left exact (contravariant) functor for every R-module I.
Problem 4. Prove that an R-module P is projective if and only if P is a direct summand
of a free R-module.
In other words, P is projective if and only if there exists an index set I such that
M
RI :=
R ' P ⊕ K,
i∈I
for some R-module K. Hint: Every R-module is a quotient of a free R-module. Consider such a surjection
φ : RI P . Let K = ker φ. If P is projective, you should be able to find a section of φ and use it to split3
the s.e.s.
0 → K → RI → P → 0.
For the other direction, you should use/prove the fact that Hom(−, −) commutes with finite direct sums, in
each variable.
Problem 5. Prove that a projective module is flat.
Hint: Use Problem 4 and the fact that ⊗ commutes with direct sums.
Problem 6. Let S be a multiplicative subset of a ring R. Suppose P is a projective Rmodule. Prove that S −1 P is a projective S −1 R-module.
Hint: Either use Problem 4, or the fact that for any R-module M and an S −1 R-module N , there is an
(obviously defined) isomorphism (of S −1 R-modules) HomR (M, N ) ' HomS −1 R (S −1 M, N ).
Problem 7. Prove the following:
Date: March 17, 2016.
1but I will accept submissions until the end of the Easter week.
2It is a functor from the category of R-modules to itself.
3
If you forgot the necessary and sufficient conditions for a s.e.s. to split, look up Proposition 25 on page
384 of Dummit and Foote, 3rd ed.
1
2
PROBLEM SET 5
Lemma 1 (Schanuel’s lemma). Given short exact sequences of R-modules
0 → K1 → P1 → M → 0,
0 → K2 → P2 → M → 0,
where P1 and P2 are projective, there exists an isomorphism
K1 ⊕ P 2 ' K2 ⊕ P 1 .
Hint: How do you produce maps K1 → K2 , P2 → K2 , and P2 → P1 ?
Proof. Let π1 : P1 → M and π2 : P2 → M be the surjective homomorphisms as above, and
let i1 : K1 → P1 and i2 : K2 → P2 be the corresponding injective homomorphism. Note that
ij (Kj ) = ker(πj ) for j = 1, 2.
Step 1: Since π2 : P2 → M is surjective, and P1 is projective, the homomorphism
π1 : P1 → M lifts to α1 : P1 → P2 in such a way that
π2 ◦ α 1 = π1 .
Similarly, there exists α2 : P2 → P1 such that
π1 ◦ α 2 = π2 .
Summarizing, we have a diagram with the two squares on the right commutative:
/
0
K1
i1
/
α1
P1U
0
/
K2
i2
/
P2
π1
/
M
/
0
/
M
/
0
α2
π2
Step 2: Consider 1 − α2 ◦ α1 : P1 → P1 . We have
π1 ◦ (1 − α2 ◦ α1 ) = π1 − π1 ◦ α2 ◦ α1 = π1 − π2 ◦ α1 = π1 − π1 = 0.
It follows that Im(1 − α2 ◦ α1 ) ⊂ ker(π1 ) = Im(i1 ). Since i1 is an injection, we see that
1 − α2 ◦ α1 is a homomorphism from P1 to K1 (if we slightly abuse notation to consider K1
a submodule of P1 ).
Similarly, 1 − α1 ◦ α2 is a homomorphism from P2 to K1 .
Step 3: Consider α1 ◦ i1 : K1 → P2 . We have that
π2 ◦ α1 ◦ i1 = π1 ◦ i1 = 0.
Hence we can regard α1 ◦ i1 as a homomorphism from K1 to K2 .
Similarly, α2 ◦ i2 is a homomorphism from K2 to K1 .
Putting things together: We can now define a homomorphism Ψ : K1 ⊕ P2 → K2 ⊕ P1
using the above defined homomorphism as follows:
Ψ(a, b) = α1 (i1 (a)) + b − α1 (α2 (b)), i1 (a) + α2 (b) .
We define Φ : K2 ⊕ P1 → K1 ⊕ P2 analogously by
Φ(c, d) = α2 (i2 (c)) + d − α2 (α1 (d)), i2 (c) + α1 (d) .
PROBLEM SET 5
3
The Moment of Truth: We compute
Φ(Ψ(a, b)) = α2 (i2 (α1 (i1 (a)) + b − α1 (α2 (b)))) + i1 (a) + α2 (b) − α2 (α1 (i1 (a) + α2 (b))),
i2 (α1 (i1 (a)) + b − α1 (α2 (b))) + α1 (i1 (a) + α2 (b)) .
To see that Φ(Ψ(a, b)) = (a, b), we need to carefully track any and all abuses of notation
that we have introduced. For example, α2 (i2 (α1 (i1 (a)))) = α2 (α1 (i1 (a))) if we recall that we
regarded α1 ◦ i1 to be a homomorphism from K1 to K2 and hence i2 ◦ α1 ◦ i1 = α1 ◦ i1 as a
homomorphism from K1 to P2 .
A R-module M is called finitely presented if there is an exact sequence
Rm → Rn → M → 0.
In other words, M is finitely presented if M is the cokernel of a map between two finite
free modules. When R is Noetherian, being finitely presented is equivalent to being finitely
generated, is equivalent to being Noetherian, because the kernel of any surjective map Rn →
M is going to be finitely generated by Noetherianness of Rn .
Problem 8. Suppose M is a finitely presented R-module and φ : Rn → M is a surjection
from a finite free module. Prove that ker(φ) is finitely generated.
Hint: Schanuel’s lemma.
Conclude using the argument given in class for finitely generated modules over Noetherian
rings that a finitely presented module M is flat if and only if M is free.
Problem 9 (Tensor product commutes with localization). Let S be a multiplicative subset
of a ring R. Prove that for any two R-modules M and N , there is an isomorphism of
S −1 R-modules:
S −1 M ⊗S −1 R S −1 N ' S −1 (M ⊗R N ).
Problem 10. Let S be a multiplicative subset of a ring R. Prove that for any two R-modules
M and N , and every i ≥ 0, there is an isomorphism
−1
S −1 TorR
(M,
N
)
' TorSi R (S −1 M, S −1 N ).
i
Hint: Use the definition of Tor and Problems 6 and 9.
Problem 11. Suppose I is an ideal in a ring R such that I is not finitely generated. (In
particular, R must not be Noetherian.) Prove that R/I is a finitely generated, but not
finitely presented R-module.
Hint: Schanuel’s lemma.