Algebra (Unknown 27). - Index of

20 Algebra 2
Thus, both in case l = 0 and l > 0 a 1 = a 11 divides all a ij and entries in the first
row and column except a 11 are zero.
Thus, M has a basis x 1 , . . . , x n such that N is generated by a 1 x 1 , . . . , a k x k .
Clearly k n. From ∑ c i (a i x i ) = 0 one deduces c i a i = 0 (since x i is a basis of
M ) and hence c i = 0. Thus, a 1 x 1 , . . . , a k x k form a basis of N .
Uniqueness. Let d i be a gcd of the i-rowed minors of (a ij ). Since the rows (resp.
columns) of T (a ij ) (resp. (a ij )T ) are linear combinations of rows (resp. columns) of
(a ij ), d i divides a gcd of the i-rowed minors of T (a ij ) and of (a ij )T . Since T is
invertible, we conclude that d i ((a ij ))A × = d i (C)A × . Since the i-rowed minors of C
is a 1 . . . a i , we deduce that
d i (C) = a 1 . . . a i u = a i d i−1 (C)v
for units u, v. Thus a i are equal up to a unit of A to d i ((a ij ))/d i−1 ((a ij )). So (a i )
are uniquely determined by the submodule N .
7.3. The Main Theorem on modules of finite type over a principal ideal domain. Let
A be a principal ideal domain. Let R ≠ 0 be an A-module of finite type. Then
R ≃ A/I 1 ⊕ · · · ⊕ A/I n
where I 1 ⊃ · · · ⊃ I n are proper ideals of A (some of which may be zero) uniquely
determined by R.
Proof. Let r 1 , . . . , r n be a minimal set of generators of R. By 2.6 there is a surjective
homomorphism f: (A) n → R so that R is isomorphic to (A) n /N where N is the
kernel of f . Let M = (A) n ; apply the previous theorem. Put a i = 0 for i > k and
I i = a i A. So the sequence of the ideals I 1 , . . . , I n is uniquely determined by R.
Define a map
g: ⊕A/a i A → (⊕x i A)/(⊕a i x i A), (b i + a i A) ↦→ (x i b i ) + ⊕a i x i A.
It is an isomorphism. Thus,
R ≃ (A) n /N ≃ (⊕x i A)/(⊕a i x i A) ≃ ⊕A/a i A = ⊕A/I i .
Example. Finitely generated abelian groups: every such group is isomorphic to
⊕ i Z/a i Z ⊕ (Z) n−k with a 1 dividing a 2 dividing a 3 , ..., dividing a k ≠ 0.
7.4. Corollary. Every module of finite type over a principal ideal domain is isomorphic
to the direct sum ⊕M i of modules M i where M i = A or M i = A/p m A where p is a
prime element of A.
Proof. If I 1 = (a 1 ) and a 1 = ∏ p r i
i with prime p i then by the Chinese Remainder
Theorem in 1.5 we have A/I 1 ≃ ∏ A i /p r i
i A.