then, for each integer $p > 0$, the $A$-module $\bigwedge^p E$ is isomorphic to the direct sum of the modules $A / a_H$, where for each $p$-element subset $H = \{ k_1, \ldots, k_p \}$ of $[1, n]$, the ideal $a_H$ is $\sum_{j=1}^p a_{k_j}$.

Let $x_k$ be the canonical generator of $A / a_k$, that is the image of the unit element of $A$, so that $E$ is the direct sum of the $Ax_i$ ($1 \leq i \leq n$). Then we know (III, p. 515, Prop. 10) that the exterior algebra $\bigwedge^n E$ is isomorphic as an $A$-module to the tensor product $\bigotimes_{i=1}^n (\bigwedge (Ax_i))$. Now $\bigwedge (Ax_i)$ is just the direct sum $A \oplus Ax_i$, since the exterior product of any two elements of $Ax_i$ is zero, and so $\bigwedge^p E$ is the direct sum of the modules $M_H = (Ax_{k_1}) \otimes \ldots \otimes (Ax_{k_p})$ as $H = \{ k_1, \ldots, k_p \}$ runs over the set of $p$-element subsets of $[1, n]$ (with $k_1 < \ldots < k_p$) ; now $M_H$ is known to be isomorphic to $A / a_{11}$ (II, p. 257, Cor. 4), which completes the proof.

We now see that, in the notation of Prop. 1, if the ideals $a_k$ form an increasing sequence, they are completely determined by the module $E$. More precisely :

Indeed, in the notation of Prop. 1, we have $a_p = a_{s(H)}$, where $s(H)$ denotes the greatest element of the subset $H$. Since $s(H) \geq p$ for every $p$-element subset $H$, and since $s(H) = p$ for $H = \{ 1, \ldots, p \}$, it follows that $a_p$ is the intersection of $a_r$, as $H$ varies over the set of $p$-element subsets of $(1, n)$; the ideal $a_p$ is thus indeed the annihilator of $\bigwedge^p E$, by Prop. 1.

(i) $x$ is indivisible ; (ii) *there exists a linear form $f$ on $L$ such that $f(x) = 1$ ; (iii) $x$ is nonzero and the submodule $Ax$ is a direct factor of $L$ ; (iv) $x$ is an element of some basis of $L$.

(i) $\Rightarrow$ (ii) : this follows from the definition. (ii) $\Rightarrow$ (iii) : let $f$ be a linear form on $L$ such that $f(x) = 1$ ; then $x \neq 0$ and the map $y \mapsto f(y)x$ is a projector of $L$, with image $Ax$. (iii) $\Rightarrow$ (iv) : let $L'$ be a complement of $Ax$ in $L$, and let $B'$ be a basis of $L'$ (VII, p. 15, Cor. 2) ; then $B' \cup \{x\}$ is a basis of $L$. (iv) $\Rightarrow$ (i) : trivial.

a) $L$ is the direct sum of $A(x/c)$ and the kernel $K$ of $f$. b) $M$ is the direct sum of $Ax$ and $K \cap M$. c) $g(M) \subset Ac$ for every linear form $g$ on $L$.

Put $y = x/c$ ; clearly $Ay \cap K = \{0\}$, since $f(y) = 1$. Furthermore, for each $u \in L$ we have

with $f(u)y \in Ay$ and $u - f(u)y \in K$ ; this proves a). Now note that for $u \in M$ we have $f(u) \in Ac$ : indeed, let $u \in M$, and let $d$ be a gcd of $f(u)$ and $c$ ; then there exist $\lambda, \mu \in A$ such that $d = Xf(u) + \mu c = f(Xu + \mu x)$ ; hence the content of the element $\lambda u + \mu x$ of $M$ divides $d$ ; by the maximality of $c$, this implies that $d$ is an associate of $c$, so $f(u) \in Ac$. Hence for all $u$ in $M$, we can write which shows b). Finally, let $g$ be a linear form on $L$ ; by a) there exists a scalar $\alpha \in A$ and a linear form $h$ on $K$ such that $g(u) = \alpha f(u) + h(u - f(u)\,y)$ ; thus by

b) we have $g(M) \subset Ac + h(K \cap M)$. To prove c), it is therefore sufficient to prove that $h(K \cap M) \subset Ac$ for every linear form $h$ on $K$, or equivalently for every linear form $h$ on $L$ such that $h(x) = 0$; now, if $u \in K \cap M$ and $d$ is a gcd of $h(u)$ and $c$, then there exist $A, \mu \in A$ with $d = \lambda h(u) + \mu c$; then $(f + h)(Au + \mu x) = d$, which implies as above that $h(u) \in Ac$, and c) follows.

Moreover the module $M'$ generated by the $(e_i)$ and the principal ideals $A\alpha_i$ are uniquely determined by the above conditions; the module $M'/M$ is the torsion submodule of $L/M$, and is isomorphic to the direct sum of the $A$-modules $A/A\alpha_i$; finally $L/M$ is the direct sum of $M'/M$ and a free module isomorphic to $L/M'$.

1) Existence of the $e_i$ and the $\alpha_i$.

If $M = \{0\}$ then the theorem is trivial. If $M \neq \{0\}$ then it follows from Prop. 3 that there exist an element $e_1$ of $L$, a nonzero element $a$, of $A$, and a submodule $L_1$ of $L$, such that $L$ is the direct sum of $Ae_1$ and $L_1$, such that $M$ is the direct sum of $A\alpha_1 e_1$ and the submodule $M_1 = M \cap L_1$ of $L$, and such that $g(M) \subset A\alpha_1$ for every linear form $g$ on $L$.

Now we can proceed by induction on the rank $n$ of $M$. Since $L_1$ is a free module (VII, p. 15, Cor. 2) and $M_1$ has rank $n-1$, there exist a basis $B_1$ of $L_1$, $(n-1)$ elements $e_2, \ldots, e_n$ of $B_1$, and nonzero elements $\alpha_2, \ldots, a$, of $A$ such that $(\alpha_2 e_2, \ldots, \alpha_n e_n)$ is a basis of $M_1$, and $\alpha_i$ divides $\alpha_{i+1}$ for $2 \leq i \leq n-1$. If $L'$ is the submodule of $L$, generated by the elements of $B_1$ distinct from $e_1, \ldots, e_n$, then $L$ is the direct sum of $L'$ and the module $M'$ generated by $e_1, \ldots, e_n$; now $(e_1, \ldots, e_n)$ is a basis for $M'$, and $(\alpha_1 e_1, \ldots, \alpha_n e_n)$ is a basis for $M$. It now remains only to show that $a_i$ divides $a_i$; but $A\alpha_2$ has the form $g(M_1)$, where $g$ is the linear form on $L$ defined by $g(e_2) = 1, g(e_i) = 0$ for $i \neq 2$, and $g(L') = (0)$; and we have seen above that $g(M_1) \subset A\alpha_1$.

2) Uniqueness properties.

Since the $a_i$ are nonzero, the module $M'$ is the set of $x \in L$ such that $\beta x \in M$ for some $\beta \neq 0$ in $A$; in other words $M'/M$ is the torsion submodule of $L/M$. This uniquely determines $M'$.

It is clear that $M'/M$ is isomorphic to the direct sum of the $n$ cyclic modules $A/A\alpha_i$ (II, p. 204, formula (26)). Let $r$ be the number of the ideals $A\alpha$, which are distinct from $A$: so that the first $n-r$ ideals $A\alpha_i$ are equal to $A$, and the last rare distinct from $A$. Then $M'/M$ is also isomorphic to the direct sum of the modules $A/A\alpha_n, \ldots, A/A\alpha_{n-r+1}$, where $A\alpha_n \subset A\alpha_{n-1} \subset \ldots \subset A\alpha_{n-r+1} \neq A$.

Thus the conditions of the Cor. to Prop. 2 (VII, p. 16) are satisfied: the ideals $Aa_i$ ($1 \leq i \leq n$) are thus uniquely determined by the module $M'/M$.

Since $L$ is the direct sum of $M'$ and $L'$, it follows that $L/M$ is the sum of $M'/M$ and $(L' + M)/M$, and this sum is direct since $(L' + M) \cap M' = M$; on the other hand $(L' + M)/M$ is isomorphic to $L'/(M \cap L')$ (I, p. 41, Th. 4, c)), that is to $L'$, which shows that $(L' + M)/M$ is a free module isomorphic to $L/M'$.

In the notation of Th. 1, if $L/M$ is torsion free, then $M = M'$, and $M'$ has a complement $L'$ in $L$. Conversely if $M$ has a complement $L'$ in $L$, then $L/M$ is isomorphic to $L'$, which is free (VII, p. 15, Cor. 2), and a fortiori torsion free.

If $E$ can be generated by $q$ elements then it is isomorphic to a quotient module $L/M$, where $L = A^q$ (II. p. 218). Since $M$ has finite rank $n \leq q$ (VII, p. 15, Prop. 1), the conditions of Th. 1 (VII, p. 18) are satisfied. Then in the notation of Th. 1 $L/M$ is isomorphic to a direct sum of a complement $L'$ of $M'$ in $L$ and the torsion module $M'/M$. The module $L'$ is free of finite rank $p = q - n$, so isomorphic to $A^p$. If $r$ is the smallest index such that $A\alpha_r \neq A$, then $M'/M$ is isomorphic to the direct sum of modules $A/A\alpha_i$ for $r \leq i \leq n$. The stated conditions will then be satisfied if we take $m = p + (n - r + 1)$, $a_k = (0)$ for $1 \leq k \leq p$, and $a_j = A\alpha_n , - ,$ for $1 \leq j \leq n - r + 1$. Uniqueness follows from the Cor. of VII, p. 16.

The torsion submodule of E has in general several distinct complements. For example, if $E = \mathbf{Z} \times (\mathbf{Z}/(2))$ then the torsion submodule of E is $\{0\} \times (\mathbf{Z}/(2))$; it has as a complement the submodule $\mathbf{Z} \times (0)$, and also the submodule consisting of all elements $(n, \overline{n})$, where $n$ runs through $\mathbf{Z}$ and $\overline{n}$ is the residue class of $n$ mod 2.

The condition that the module be finitely generated is essential. For example the additive group of the field of fractions K of A is torsion free as an A-module; however it is not a free module if $A \neq K$, for on the one hand any two elements of K have a common divisor, and on the other hand K is not a cyclic A-module, for otherwise $K = ab^{-1}A$ ($a, b \in A$), whence $b^{-2} = acb^{-1}$ ($c \in A$), so $b^{-1} = ac \in A$, and $K = A$.

By Th. 1 it is clear that $M \subset \alpha_1 L$; hence the coordinates of any element of M are multiples of $a_1$. On the other hand, there exists an element $x$ of M for which $a_1$ is a content in L. Expressing $x$ as a linear combination of the $x_i$, it follows that $a_1$ is an element of the ideal generated by the coordinates of the $x_i$. As these are all multiples of $a_1$, it follows that $a_1$ is indeed their gcd, and our assertion is proved in the case $m = 1$.

For general $m$, consider the m-th exterior power $\bigwedge^m M$ of M (III, p. 507). In the notation of Th. 1 there exists a basis $(a_i)$ for M where $a_i = \alpha_i e_i$ ($1 \leq i \leq n$); thus $\bigwedge^m M$ has a basis consisting of the elements $a_{i_1} \wedge \ldots \wedge a_{i_m}$, as $(i_1, \ldots, i_m)$ runs through the set of strictly increasing sequences of $m$ elements of $(1, n)$. Now the elements $e_{i_1} \wedge \ldots \wedge e_{i_m}$ belong to a basis $B_m$ of $\bigwedge^m L$. Hence the canonical map from $\bigwedge^m M$ into $\bigwedge^m L$ is an isomorphism from $\bigwedge^m M$ onto the submodule of $\bigwedge^m L$ having as a basis the elements $(\alpha_{i_1} \ldots \alpha_{i_m})\, e_{i_1} \wedge \ldots \wedge e_{i_m}$, and we identify this submodule with $\bigwedge^m M$. Since $\alpha_j$ is a multiple of $\alpha_k$ for $j \geq k$, the elements $\alpha_{i_1} \ldots \alpha_{i_m}$ are all multiples of $\delta_m = \alpha_1 \ldots \alpha_m$, and one of them is equal to $\delta_m$; thus $\delta_m$ is a gcd of the coordinates with respect to $B_m$ of the elements of a system of generators of $\bigwedge^m M$. The first part of the argument shows that then $\delta_m$ is a gcd of the set of coordinates of any system of generators of $\bigwedge^m M$, with respect to any basis of $\bigwedge^m L$. Taking the basis for $\bigwedge^m L$ induced from the basis $(u_j)$ of $L$, and the system of generators for $\bigwedge^m M$ consisting of exterior products of the $(x_i)$, the expression for the coordinates of these products in terms of determinants (III, p. 528, Prop. 9) gives the stated result.

Let $L_0 = f^{-1}(0)$ be the kernel of $f$; the quotient $L/L_0$ is isomorphic to the module $f(L)$, which is a submodule of $L'$ and so free (VII, p. 15, Cor. 2); thus $L_0$ has a complement $L_1$ in $L$ (II, p. 218, Prop. 21), and the restriction of $f$ to $L_1$ is an isomorphism from $L_1$ onto $f(L) = M'$. If the ideals $A \alpha_i$ ($1 \leq i \leq r$) are the invariant factors of $M'$ in $L'$, then Th. 1 of VII, p. 18 shows that there exists a basis $(e'_j)$ ($1 \leq j \leq n$) of $L'$ such that $(\alpha_i e'_i)$ ($1 \leq i \leq r$) is a basis of $M'$. Since the restriction of $f$ to $L_1$ is an isomorphism from $L_1$ onto $M'$, there exists a basis $(e_i)$ ($1 \leq i \leq r$) of $L_1$ such that $f(e_i) = \alpha_i e'_i$. This basis extends to a basis $(e_s)$ ($1 \leq k \leq m$) of $L$ by taking $(e_s)$ ($r+1 \leq s \leq m$) to be a basis of the kernel $L_0$.

Given that two matrices X and $X'$ are equivalent if there exist invertible square matrices P and Q, of orders n and m, over A, such that $X' = PXQ$, Cor. 1 is just Prop. 5 expressed in terms of matrices.

In the notation of Prop. 5 and Cor. 1, the (nonzero) ideals $Au$, are called the invariant factors of the linear map f, or of the matrix X. It then follows immediately from Cor. I that:

Note that when A is a field we can take the $\alpha_i$ to be equal to 1, and then we recover Prop. 13 of II, p. 360.

If X is the matrix of the linear mapping f with respect to an arbitrary basis of L and an arbitrary basis of L', then the columns of X are the coordinate vectors, with respect to the basis of L', of elements of L' which form a system of generators for $f(L)$. The following result is thus an immediate consequence of Prop. 4.

Let $b$ and $c$ be two ideals of $A$, such that $A = b + c$ and $b \cap c = q$. Then $bc \subset b \cap c = q \subset \mathfrak{p}$; if $x \notin \mathfrak{p}$ and $x \in c$, then $xb \subset p$, so $b \subset \mathfrak{p}$ (I, p. 316, Prop. 4); hence either $b \subset \mathfrak{p}$ or $c \subset p$. Suppose for example that $c \subset \mathfrak{p}$, so $b + p = A$; then there exist $x \in b$ and $y \in \mathfrak{p}$ such that $1 = x + y$; let $n \in \mathbf{N}$ be such that $y^n \in q$; then $1 = (x + y)^n$. so $1 \in xA + y^nA \subset b + q \subset b$, so $b = A$. Lemma 2 now shows that $A/q$ is indecomposable.

Now suppose that $A$ is a principal ideal domain; by VII, p. 2, Prop. 2, the prime ideals of $A$ are the ideals $(p)$, where $p$ is an irreducible element of $A$, and the ideal $0$; by the previous proposition, the modules $A$ and $A/(p^n)$, for $p$ irreducible and $n > 0$, are indecomposable. Since every cyclic module is a direct sum of modules of this type (VII, p. 3, Prop. 4) and since every finitely generated $A$-module is a direct sum of cyclic modules (VII, p. 19, Th. 2), we deduce:

a) $M$ is indecomposable if and only if it is isomorphic to $A$ or to a module of the form $A/(p^n)$, where $p$ is an irreducible element of $A$ and $n > 0$ is an integer.

b) $M$ is a direct sum of a finite family of indecomposable submodules.

Part b) of the above proposition can be made more precise as follows:

The existence of the integers $m(0)$ and $m(p^n)$ ($p \in P, n > 0$) follows from Prop. 8. The integer $m(0)$ is uniquely determined: it is the rank of the free module which is the quotient of $M$ by its torsion submodule. Finally, the $p$-primary component of $M$ is isomorphic to the direct sum of the $(A/(p^n))^{m(p^n)}$; since the family of ideals $(p^n)$ ($n \geq 1$) is totally ordered by inclusion, the uniqueness of the $m(p^n)$ follows from the Cor. to Prop. 2 of VII, p. 16.