Theorem 1

Let $A$ be a (commutative) principal ring, $E$ a free $A$-module of finite dimension $n$ and $\Phi$ an alternating bilinear form on $E$. Then there exists a basis $(e_i)_{1 \leq i \leq n}$ of $E$ and an even integer $2r \leq n$, such that

$1^o$ $\Phi(e_1, e_2) = \alpha_1, \Phi(e_3, e_4) = \alpha_2, \ldots, \Phi(e_{2r-1}, e_{2r}) = \alpha_r$

where the $\alpha_i$ are elements $\neq 0$ of $A$, and where $\alpha_i$ divides $\alpha_{i+1}$ for $i = 1, \ldots, r-1$.

$2^o$ All the other elements $\Phi(e_i, e_j)$ where $i \leq j$ are zero.

The ideals $A \alpha_i$ ($i = 1, \ldots, r$) are uniquely determined by the preceding conditions. The submodule $E^0$ of $E$ orthogonal to $E$ is generated by $e_{2r+1}, \ldots, e_n$.

We shall proceed by induction on the dimension $n$ of $E$. The theorem is evident for $n = 0$. If $\Phi = 0$, the theorem is evident also; we may therefore suppose $\Phi \neq 0$. Let $f$ be the linear mapping $d_\Phi$ of $E$ into $E^*$ associated on the right with $\Phi$ ($§ 1$, no 1); then $f(E)$ is a non-zero submodule of the module $E^*$, which is a free module of dimension $n$. Let $A\alpha_1$ be the largest invariant factor of $f(E)$ with respect to $E^*$ (chap. VII, $§ 4$, no 2, th. 1); we know (loc. cit.) that there exists a basis $(e'_1, a'_2, \ldots, a'_n)$ of $E^*$ and an element $f(e_2) \in f(E)$ such that $f(e_2) = \alpha_1 e'_1$. Let $(e_1, a_2, \ldots, a_n)$ be the basis of $E$ (identified with the bidual $E^{**}$) dual to $(e'_1, a'_2, \ldots, a'_n)$; we have

Let $P$ be the submodule $Ae_1 + Ae_2$ of $E$. We shall see that $E$ is direct sum of $Ae_1, Ae_2$ and the orthogonal submodule $P^0$ of $P$. For this, it is enough to prove that, for every $x \in E$, there exist elements $\xi_1, \xi_2$ of $A$, uniquely determined, such that $x - \xi_1 e_1 - \xi_2 e_2 \in P^0$, that is to say such that

According to (1) these conditions are written

But we know (loc. cit.) that the image of $f(E)$ by every linear form on $E^*$ is contained in the ideal $A\alpha_1$, in other words all the values $\Phi(x, y) = \langle x, f(y) \rangle$ belong to $A\alpha_1$; hence the existence and uniqueness of $\xi_1$ and $\xi_2$. Thus $P^0$ is a free module of rank $n-2$; there exists therefore, in $P^0$, according to the induction hypothesis, a basis $(e_3, e_4, \ldots, e_n)$ satisfying the conditions of the statement. To prove that the basis $(e_1, \ldots, e_n)$ of $E$ thus obtained also satisfies these conditions, it is enough to prove that $\alpha_1$ divides $\alpha_2$; now this follows from the fact that all the values $\Phi(x, y)$ are multiples of $\alpha_1$. It is then clear that $e_{2r+1}, \ldots, e_n$ generate $E^0$. Finally, if $(e'_i)$ is the dual basis of $(e_i)$, we have $f(e_{2j-1}) = -\alpha_j e'_{2j}$ and $f(e_{2j}) = \alpha_j e'_{2j}$ for $j = 1, \ldots, r$ and $f(e_k) = 0$ for $k = 2r+1, \ldots, n$; the ideals $A\alpha_1, A\alpha_1, A\alpha_2, A\alpha_2, \ldots, A\alpha_r, A\alpha_r$ are therefore the invariant factors of $f(E)$ with respect to $E^*$, which proves their uniqueness (Chapter VII, $§ 4$, No. 2, Theorem 1).