Description

2. Suppose that B = ⟨b1,…,bn⟩ as an A-module. We first use the determinant trick to prove (special case of) the Cayley-Hamilton theorem, following the lecture notes and Theorem 4.3 in Eisenbud’s book. The statement of the theorem with our notation is:
If φ ∈ EndA(B) then there is a monic polynomial
p(x) = xn + p1xn−1 + … + pn−1x + pn
such that p ◦ φ = 0.
For the proof, we let φ ∈ EndA(B) and note (as shown in the notes) that we can write φ(bi) = Paijbj, a sum of the generators with coefficients aij ∈ A. We consider B as an A[x]-module by µx = φ (multiplication by x is φ). Denoting the n × n identity matrix by I, the above means that
.
Multiplying on the left by the cofactor matrix of (xI − (aij)ij) we get
.
In other words, ∆bi = 0 for all i = 1,…,n, so with ∆ as our choice of p(x), we see that p ◦ φ = 0.
From this, if b ∈ B, then choosing φ = µb, multiplication by b, we have p ◦ µb = 0, ie. b is integral over A.
3. Suppose x,y ∈ B are integral over A, so A[x] is finitely generated over A, and since y is integral over A it is integral over A[x], so A[x,y] is finitely generated over A[x]. As A[x] is finitely generated over A, A[x,y] is finitely generated over A. Thus by problem 2 every element of A[x,y] is integral over A, so in particular xy and x−y are. As this is true for any two elements of B integral over A, the integral closure of A in B is a subring of B.
4. (a) Suppose x ∈ B is integral over A, so
xm + a1xm−1 + … + am−1x + am = 0
for some m ≥ 1 and ai ∈ a, where i = 1,…,m. So, moving the degree less than m terms over we see that
m−1 xm = X(−ak)xk ∈ aB.
k=1
√
Thus x ∈ aB.
√
(b) We view a√B as an A-module, and we denote B = ⟨b1,…,bn⟩ over A, so in particular aB is an A-submodule of ⟨b1,…,bn⟩. We then follow the proof of problem 2, except we take the coefficients aij to be in a instead of√ A.

This shows that each element of aB is integral over a, and so if x ∈ aB then xn is integral over a, thus so is x.
5. We drop the function composition notation to save space, so fg means f ◦ g. We will show that dn+1dn = 0. Let f = (fj)j∈Z ∈ HomnA(C•,D•), then
dn+1(dn(f)) = dn+1(dnD+•jfj + (−1)n+1fj+1djC•)j∈Z

| {z }
gj
= (dnD+1+• jgj + (−1)n+2gj+1djC•)j∈Z. It suffices to show that the term above is zero for each j. This term is

which is equal to

and since the composition of differentials is zero, rearranging the signs we see the above is
.
6. Suppose that f = (fj)j∈Z ∈ Z0(Hom•(C•,D•)), so
0 = d0(f) = (djD• ◦ fj − fj+1 ◦ djC•)j∈Z,
which is equivalent to saying the following diagram commutes for each j, ie. f is a cochain map.

So, Z0(Hom•(C•,D•)) ⊂ Hom(C•,D•), and since all the statements above work in the converse direction we have Z0(Hom•(C•,D•)) = Hom(C•,D•).
7. Let f = (fj)j∈Z ∈ B0(Hom•(C•,D•). So, f is a cochain map (problem 6) and f = d−1(g) for some g ∈ Hom−1(C•,D•). Note the −1 here denotes the index, not some inverse. As

this is the same as saying as f ∼ 0. So,
B0(Hom•(C•,D•) = {f ∈ Hom(C•,D•) | f ∼ 0}.
8. This proof is adapted from the proof of proposition A3.12 in Eisenbud’sbook. Let f,g ∈ Hom(C•,D•) with f ∼ g, so f − g ∼ 0. We first show that Hn(f − g) = Hn(f) − Hn(g), so it suffices to show Hn(f − g) = 0 (for all n).
This is immediate from the definition: let x ∈ Cn, then
Hn(f)(x + Bn(C•)) − Hn(g)(x + Bn) = f(x) − g(x) + Bn(D•)
= Hn(f − g)(x + Bn(C•)).
So, as f − g ∼ 0 there exists a k = (kn)n ∈ Hom−1(C•,D•) such that for all n. Thus if z ∈ Zn(C•) we have
(f − g)n(z) = (dnD−•1 ◦ kn)(z) + (kn+1 ◦ dnC•)(z)
= (dnD−•1 ◦ kn)(z) + kn+1(0)
= (dnD−•1 ◦ kn)(z) ∈ Bn(D•)
so Hn(f − g) is the zero map on homology for all n.