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

76 6. FINITE MODULES
(4) Let A be an n × n-matrix with cofactor matrix A ′ . Then the matrix Cramer’s
rule holds
AA ′ = A ′ A = det A(1n)
Where (1n) is the n × n identity matrix.
(5) A square matrix A is invertible if and only if det A is a unit in R.
Proof. More linear algebra homework.
6.2.5. Proposition. Let f : R n → R n be a homomorphism represented by an
n × n-matrix A.
(1) f is a surjective if and only if det A is a unit.
(2) f is injective if and only if det A is a nonzero divisor.
Proof. (1) If f is surjective, then a section is represented by a matrix B such that
AB = (1n). Then det A is a unit. Conversely by 6.2.4. (2) If det A is a nonzero
divisor, then by 6.2.4 A ′ A = det A(1n) gives an injective homomorphism, so f is
injective. If det A is a zero divisor, then there is number k < n such that Ann(1 −
minors) = · · · = Ann(k − minors) = 0 and 0 = b ∈ Ann((k + 1) − minors).
Assume that the k−minor from first k rows and columns ck+1 has ck+1b = 0. Let
cj(−1) k+1+j be the k-minor from first k rows and first k + 1 columns excluding
number j and put cj = 0, j > k + 1. Then A(cj) is a column with entries being
(k + 1) − minors so f((bcj)) = A(bcj) = 0 and f is not injective.
6.2.6. Proposition. Let R be a ring and f : R n → R m homomorphism.
(1) If f is surjective, then n ≥ m.
(2) If f is injective, then n ≤ m.
Proof. (1) If n < m let p : R m → R n be the projection onto first n coordinates.
Then f ◦p is surjective and represented by an m×m-matrix A with m-column zero.
A section to f ◦ p is represented by an m × m-matrix B such that BA = (1m). But
the product AB must have a zero m-column, so the contradiction gives n ≥ m.
(2) If n > m let i : R m → R n be injection onto first m coordinates. Then i ◦ f is
injective and represented by an n × n-matrix A with n-row zero. Then det A = 0
contradicting 6.2.5. So n ≤ m.
6.2.7. Proposition. (1) A finite free module has a finite basis.
(2) The number of elements in a basis for a finite free module is independent of
the basis.
Proof. Let F be finite free generated by n elements. If y1, . . . , ym is part of a
basis, then by projection F → R m there is a surjective homomorphism R n → R m .
Conclusion by 6.2.6.
6.2.8. Definition. The number of elements 6.2.7 in a basis for a finite free module
F is the rank, rankR F .
6.2.9. Proposition. If x1, . . . , xn generates a free module F of rank n, then they
constitutes a basis.
Proof. Choose a basis and an isomorphism f : R n → F . The homomorphism
g : R n → F, ei ↦→ xi is surjective. The composite f −1 ◦ g : R n → R n is
surjective and therefore by 6.2.5 an isomorphism. Then g is an isomorphism and
xi a basis.