Algebra (Unknown 27). - Index of

18 Algebra 2
are generators (resp. a basis) of M .
Proof. (1) If T ∈ M k (A) × then T T ′ = E for some T ′ ∈ M k (A). So det(T ) det(T ′ ) =
1, and det(T ) ∈ A × . Conversely, let det(T ) ∈ A × . Recall that the inverse matrix T ′
of a nonsingular matrix T can be found by a formula
T ′ = det(T ) −1 (a ij )
where the entry a ij of the adjugate to T matrix is (−1) i+j times the determinant of the
matrix obtained from T by cutting off the i th column and j th row. In particular, since
T ∈ M k (A), (a ij ) ∈ M k (A). Now, since det(T ) ∈ A × we conclude T ′ ∈ M k (A).
(2) Since T is invertible in M k (A), not only elements y 1 , . . . , y k are A-linearly
expressible via elements x 1 , . . . , x k , but x 1 , . . . , x k are A-linearly expressible via
elements y 1 , . . . , y k .
7.2. Theorem (On submodules of a free module). Let M be a free A-module of
finite rank n. Let N be a non-zero submodule of M . Then (i) N is free of rank k n;
(ii) M has a basis x 1 , . . . , x n such that a 1 x 1 , . . . , a k x k is a basis of N where a i
are non-zero elements of A such that (a 1 ) ⊃ (a 2 ) ⊃ · · · ⊃ (a k ) ≠ 0. This sequence of
ideals is uniquely determined by N .
Proof. Existence.
The module M is of finite type over the Noetherian ring A, so it is a Noetherian
A-module by Corollary 3 in 4.6. N is its submodule, so it is a Noetherian A-module
by Lemma 4.6.
Let x 1 , . . . , x n be a basis of M . Let y 1 , . . . , y m be generators of N . Each of
them can be written as a linear combination of x i with coefficients in A:
⎛
⎝ y ⎞
1
. . .
y m
⎠ = (a ij )
⎛
⎝ x ⎞
1
. . .
x n
⎠ , a ij ∈ A.
Due to the previous lemma one can (by passing to another generators of M and N )
multiply (a ij ) by invertible matrices in M m (A) on the left and invertible matrices in
M n (A) on the right. We aim to show that as a result of such permited transformations
the matrix (a ij ) can be transformed to the matrix of the form
⎛
⎞
a 1 0 . . . . . . . . .
0 a 2 . . . . . . . . .
. . . . . . . . . . . . . . .
C =
⎜ 0 0 0 a k . . . ⎟
⎝
⎠
0 0 0 0 0
0 0 0 0 0
with (a 1 ) ⊃ (a 2 ) ⊃ · · · ⊃ (a k ).