ALGEBRA HW 2
CLAY SHONKWILER
1
Suppose that
A1
α1
B1
f1
/ A2
α2
g1
/ B2
f2
/ A3
α3
g2
/ B3
f3
/ A4
α4
g3
/ B4
f4
/ A5
α5
g4
/ B5
is a commutative diagram of R-moduels, with exact rows.
(a): Show that if α1 is surjective and α2 , α4 are injective, then α3 is
injective.
Proof. Suppose x, y ∈ A3 such that α3 (x) = b3 = α3 (y). Let b4 =
g3 (b3 ). Then
(α4 ◦ f3 )(x) = (g3 ◦ α3 )(x) = g3 (b3 ) = b4 = (g3 ◦ α3 )(y) = (α4 ◦ f3 )(y).
Since α4 injective and α4 (f3 (x)) = α4 (f3 (y)), f3 (x) = f3 (y); call this
element a4 . Then
α3 (x − y) = α3 (x) − α3 (y) = a4 − a4 = 0,
so, since the top row is exact, there exists a2 ∈ A2 such that f2 (a2 ) =
x − y. Let b2 = α2 (a2 ). Then
g2 (b2 ) = (g2 ◦ α2 )(a2 ) = (α3 ◦ f2 )(a2 ) = α3 (x − y) = b3 − b3 = 0,
so, since the bottom row is exact there exists b1 ∈ B1 such that
g1 (b1 ) = b2 . Since α1 is surjective, there exists a1 ∈ A1 such that
α1 (a1 ) = b1 . Let a02 = f1 (a1 ). Then,
b2 = g1 (b1 ) = (g1 ◦ α1 )(a1 ) = (α2 ◦ f1 )(a1 ) = α2 (a02 ).
Since α2 (a02 ) = α2 (a2 ) and α2 is injective, α20 = α2 . Since the top
row is exact,
x − y = f2 (a2 ) = f2 (a02 ) = (f2 ◦ f1 )(a1 ) = 0,
so x = y. Therefore, we conclude that α3 is injective.
(b): Show that if α5 is injective and α2 , α4 are surjective, then α3 is
surjective.
1
2
CLAY SHONKWILER
Proof. Let b3 ∈ B3 and let b4 = g3 (b3 ). Then, since α4 is surjective,
there exists a4 ∈ A4 such that α4 (a4 ) = b4 . Let a5 = f4 (a4 ). Then
α5 (a5 ) = (α5 ◦ f4 )(a4 ) = (g4 ◦ α4 )(a4 ) = g4 (b4 ) = (g4 ◦ g3 )(b3 ) = 0
since the bottom row is exact. Hence, since α5 is injective, a5 = 0.
Therefore, since the top row is exact, there exists a3 ∈ A3 such that
f3 (a3 ) = a4 . Let b03 = α3 (a3 ). Then
g3 (b03 ) = (g3 ◦ α3 )(a3 ) = (α4 ◦ f3 )(a3 ) = α4 (a4 ) = b4 = g3 (b3 ),
so g3 (b03 − b3 ) = b4 − b4 = 0, so there exists b2 ∈ B2 such that
g2 (b2 ) = b03 − b3 . Since α2 is surjective, there exists a2 ∈ A2 such
that α2 (a2 ) = b2 . Let a03 = f2 (a2 ). Then
α3 (a03 ) = (α3 ◦ f2 )(a2 ) = (g2 ◦ α2 )(a2 ) = g2 (b2 ) = b03 − b3 .
Hence,
α3 (a3 − a03 ) = α3 (a3 ) − α3 (a03 ) = b03 − (b03 − b3 ) = b3 ,
so we conclude that α3 is surjective.
(c): In particular, deduce that α3 is an isomorphism provided α1 , α2 , α4 , α5
are.
Proof. If α1 , α2 , α4 , α5 are all isomorphisms, then certainly α1 is surjective and α2 , α4 are injective, so α3 is injective by part (a). On the
other hand, this also means α2 , α4 are surjections and α5 is an injection, so α3 is a surjection by part (b). Therefore, α3 is a bijective
homomorphism and so is an isomorphism.
2
In the notation of problems 4 and 5 of Problem Set 1:
(a): Show that I ∩ J = (y − 4) and I + J = (1) in R.
Proof. In PS1#4(a) we showed that I = (x − 3, y − 4); a parallel
argument demonstrates that J = (x + 3, y − 4). Hence,
I ∩ J = (x − 3, y − 4) ∩ (x + 3, y − 4) ⊃ (y − 4).
On the other hand, suppose h ∈ I ∩ J. Then, by PS1#5(a), there
exist f, g ∈ R such that h = (x − 3)f + (y − 4)g and h = (x + 3)f +
(y − 4)g. Setting the right hand sides equal yields
(x − 3)f + (y − 4)g = (x + 3)f + (y − 4)g.
subtracting (y − 4)g and (x − 3)f from both sides yields
0 = (x + 3)f − (x − 3)f = 6f,
so f = 0 since there are no 6-torsion elements of R. Hence, h =
(y − 4)g ∈ (y − 4). Since our choice of h ∈ I ∩ J was arbitrary, we
ALGEBRA HW 2
3
see that I ∩ J ⊂ (y − 4). Having proved containment both ways, we
conclude that
I ∩ J = (y − 4).
Now, consider
−1
6 (x
− 3) ∈ I and 16 (x + 3) ∈ J. Then
−1
1
1 1
(x − 3) + (x + 3) = + = 1,
6
6
2 2
so I + J ⊃ (1). Since I + J ⊂ R = (1), we conclude that I + J =
(1).
(b): Let ∆ : I ∩ J → I ⊕ J be given by ∆(f ) = (f, f ), and let − :
I ⊕ J → I + J be given by −(f, g) = f − g. Show that the sequence
0
/ I ∩J
∆
/ I ⊕J
−
/ I +J
/0
is exact.
Proof. ∆ is injective: Suppose f ∈ I ∩ J such that ∆(f ) = 0. Then
0 = ∆(f ) = (f, f ),
so f = 0. Hence, ∆ is injective.
−1
− is surjective: Suppose f ∈ I+J = (1) = R. Then
6 (x−3)f ∈ I
1
−1
1
and 6 (x + 3)f ∈ J. Hence 6 (x − 3)f, 6 (x + 3)f ∈ I ⊕ J and
−1
1
−1
1
1
1
−
(x − 3)f, (x + 3)f =
(x − 3)f − (x + 3)f = f + f = f,
6
6
6
6
2
2
so we see that − is surjective.
im ∆ ⊂ ker −: Let f ∈ I ∩ J. Then
(− ◦ ∆)(f ) = −(f, f ) = f − f = 0,
so we see that im ∆ ⊂ ker −.
ker − ⊂ im ∆: Suppose (f, g) ∈ I ⊕ J such that
0 = −(f, g) = f − g.
Then adding g to both sides, we see that f = g, so
∆(f ) = (f, f ) = (f, g);
hence, we conclude that ker − ⊂ im ∆.
Therefore, we conclude that im ∆ = ker −, that ∆ is injective and
that − is surjective, and so (*) is exact.
(c): Show that the exact sequence (*) is split.
Proof. Since I + J = (1) = R (by part (a)), since R is certainly a
free R-module, and since all free R-modules are projective, we see
that I + J is a projective module. Therefore, since any short exact
sequence with a projective modules as the cokernel splits, this means
that (*) is split.
4
CLAY SHONKWILER
(d): Deduce that I ⊕ J ' (I ∩ J) ⊕ (I + J), and conclude that I ⊕ J
is therefore free of rank 2.
Proof. Since (*) splits and the middle term of any split short exact
sequence of R-modules is isomorphic to the direct product of the
kernel and cokernel, we see that
I ⊕ J ' (I ∩ J) ⊕ (I + J).
By part (a), I ∩ J = (y − 4) and I + J = (1), so it is certainly true
that (y − 4, 0) and (0, 1) generate (I ∩ J) ⊕ (I + J) ' I ⊕ J. On the
other hand, if f, g ∈ R such that
0 = f (y − 4, 0) + g(0, 1) = ((y − 4)f, g)
then g = 0 and (y − 4)f = 0, meaning f = 0. Hence, (y − 4, 0) and
(0, 1) are linearly independent and so form a basis for I ⊕ J, which
is, therefore, a free module of rank 2.
3
In the notation of the above problem:
(a): Explicitly find a section s of −, corresponding to a splitting of the
exact sequence (*).
Answer: As we saw in problem 2(a) above,
1
1 1
−1
(x − 3) + (x + 3) = + = 1,
6
6
2 2
so
−
−1
−1
−1
−1
(x − 3),
(x + 3) =
(x − 3) −
(x + 3) = 1.
6
6
6
6
Hence, we define s : I + J → I ⊕ J by
−1
−1
1 7→
(x − 3),
(x + 3)
6
6
and extend linearly. Since I +J = (1), this determines a well-defined
homomorphism. Furthermore, for any f ∈ I + J, f = f · 1, so
(− ◦ s)(f ) = −(s(f · 1)) = −(f s(1)) = f · −(s(1)) = f · 1 = f,
so − ◦ s = id and so s is a section of −.
♣
(b): Explicitly find an isomorphism α : (I ∩ J) ⊕ (I + J) → I ⊕ J
induced by the section in (a).
∆
−
Answer: Since 0 → I ∩ J → I ⊕ J → I + J → 0 is exact and
split by s, we know that the following diagram has exact rows and
ALGEBRA HW 2
5
the squares commute:
0
/ I ∩J
/ (I ∩ J) ⊕ (I + J)p2
i1
/ I +J
/0
/ I +J
/0
(∆,s)
0
/ I ∩J
∆
/ I ⊕J
−
where i1 is the first inclusion, p2 is the second projection and (∆, s) :
(I ∩ J) ⊕ (I + J) → I ⊕ J given by
(f, g) 7→ ∆(f ) + s(g)
is an isomorphism. Hence, α = (∆, s) is the desired isomorphism
and is given explicitly by
α(f (y − 4), g) = ∆(f (y − 4)) + s(g)
−1
−1
(x − 3),
(x + 3)
= (f (y − 4), f (y − 4)) + g ·
6
6
(x − 3)g
(x + 3)g
=
(y − 4)f −
, (y − 4)f −
.
6
6
♣
(c): Compare this isomorphism α : (y − 4) ⊕ (1) → I ⊕ J to the one in
problem 5 of Problem Set 1.
Answer: In PS1#5, we demonstrated that, for (i, j) ∈ I ⊕ J,
(i, j) = ((x − 3)g + (y − 4)f, (x + 3)g + (y − 4)f )
uniquely for some f, g ∈ R. Now,
(i, j) = ((x − 3)g + (y − 4)f, (x + 3)g + (y − 4)f )
= ((y − 4)f, (y − 4)f ) + ((x − 3)g, (x + 3)g)
−1
−1
= ((y − 4)f, (y − 4)f ) − 6g
(x − 3),
(x + 3)g
6
6
= ∆((y − 4)f ) + s(−6g)
so, up to scaling the g term by −6, this is the same isomorphism as
the one calculated above. Since, for any h ∈ R, h = −6g for some
g ∈ R (since we’re working over the field R), we could just as well
have defined α by
α((y − 4)f, −6g) = ((x − 3)g + (y − 4)f, (x + 3)g + (y − 4)f ),
which is the same isomorphism given in PS1#5.
♣
6
CLAY SHONKWILER
4
Let P be a finitely generated projective R-module.
(a): Show that there is a finitely generated free R-module F , and an
i
π
R-module K, such that 0 → K → F → P → 0 is exact.
Proof. Let {p1 , . . . , pn } be a generating set for P . Let F = Rn and
define π : F → P by
ei 7→ pi ,
where the ei are the elements of the standard basis for F . If a ∈ P ,
then
a = a1 p1 + . . . + an pn = a1 π(e1 ) + . . . + an π(en ) = π(a1 e1 + . . . + an en ),
so π is surjective. Let K = ker π and let i : K → F be the standard
inclusion. Then i is injective and, by definition, i(K) = K = ker π.
Hence,
0
/K
i
/F
/P
π
/0
is exact.
(b): Show that there exists a homomorphism j : F → K such that i
i◦j
π
is a section of j, and that the sequence F → F → P → 0 is exact.
i
π
Proof. Since 0 → K → F → P → 0 is exact and P is projective,
we know there exists a splitting s : P → F . Since this is a split
exact sequence of R-modules, it must be the case that F ' K ⊕ P
and, furthermore, that the following diagram has exact rows and the
squares commute:
0
/K
i1
0
/K
i
/ K ⊕P
p2
/P
/0
π
/P
/0
(i,s)
/F
where i1 is the first inclusion, p2 is the second projection and (i, s) :
K ⊕ P given by
(k, p) 7→ i(k) + s(p)
is an isomorphism. Specifically, since the squares commute,
i = (i, s) ◦ i1 .
Now, define j : F → K by
x 7→ (p1 ◦ (i, s)−1 )(x)
where p1 : K⊕P → K is the first projection and (i, s)−1 : F → K⊕P
is the inverse of (i, s) (and is, therefore, also an isomorphism). Now,
if k ∈ K
(j ◦ i)(k) = ((p1 ◦ (i, s)−1 ) ◦ ((i, s) ◦ i1 ))(k) = (p1 ◦ i1 )(k) = p1 (k, 0) = k,
ALGEBRA HW 2
7
so j ◦ i = idK and so i is a section of j.
Now, consider the sequence
F
i◦j
/F
π
/P
/0
We already know π is surjective, so to show that this sequence is
exact we need only show that im (i ◦ j) = ker π. Since i is a section
of j, j : F → K must be surjective and so
im (i ◦ j) = (i ◦ j)(F ) = i(j(F )) = i(K) = im i = ker π.
Hence, the above sequence is, indeed, exact.
(c): Conclude that P is a finitely presented R-module.
Proof. Suppose P is not finitely presented.
(d): Give an example of a finitely generated R-module M (for some
R) that is not projective and is not even finitely presented.
DRL 3E3A, University of Pennsylvania
E-mail address: [email protected]