Commutative algebra - Department of Mathematical Sciences - old ...

6.3. CAYLEY-HAMILTON’S THEOREM 77
6.2.10. Proposition. Let F, F ′ be finite free modules. Then
(1) F ⊕ F ′ is free and rankR F ⊕ F ′ = rankR F + rankR F ′ .
(2) F ⊗R F ′ is free and rankR F ⊗R F ′ = rankR F · rankR F ′ .
(3) HomR(F, F ′ ) is free and rankR HomR(F, F ′ ) = rankR F · rankR F ′ .
6.2.11. Exercise. (1) Let R n → R n be a surjective homomorphism. Show that it is an
isomorphism.
6.3. Cayley-Hamilton’s theorem
6.3.1. Remark. Let R be a ring and f : M → M a homomorphism. By 2.1.13
view M as an R[X]-module, where Xx = f(x) for x ∈ M. The homomorphism
1.6.7, 2.6.9, R[X] → HomR(M, M), a ↦→ aM, X ↦→ f is a ring homomorphism.
The image is R[f] the smallest subring containing 1M, f. M is naturally a R[f]module
and the R[X]-module above is the restriction of scalars.
6.3.2. Proposition. Let A be an n × n-matrix and I the ideal generated by the
entries aij. The polynomial
det(X1n − A) = a0 + a1X + . . . an−1X n−1 + X n
has a0, . . . , an−1 ∈ I and gives the relation
as n × n-matrix.
a0(1n) + a1A + . . . an−1A n−1 + A n = 0
Proof. View R n as a module over the ring R[X], 6.3.1, with scalar multiplication
Xx = Ax, x ∈ R n
Let the n × n-matrix U = X(1n) − A over R[X] have cofactor matrix U ′ . The
relations above give U ′ Uej = 0, written out by 6.2.4
det U ej = 0
for all j. That is det U M = 0. By calculation
in R[X], with ai ∈ I.
det U = a0 + a1X + · · · + an−1X n−1 + X n
6.3.3. Proposition. Let I ⊂ R be an ideal and f : M → M a homomorphism
on a finite module generated by n elements. Suppose Im f ⊂ IM, then there exist
a0, . . . , an−1 ∈ I such that
in HomR(M, M).
a01M + a1f + . . . an−1f n−1 + f n = 0
Proof. Let x1, . . . , xn generate M and write
f(xj) =
i
aijxi
for an n × n-matrix A with entries aij ∈ I. View this over the ring R[X], 6.3.1.
Then
Xxj =
i
aijxi