Proposition 2

Suppose that $A$ is commutative and that $E$ admits a basis $(e_i)_{i \in I}$. Then, for every quadratic form $Q$ on $E$, there exists a bilinear form $f$ on $E \times E$ such that $Q(x) = f(x, x)$ for every $x \in E$. For every family $(b_{ij})_{(i,j) \in I \times I}$ of scalars such that $b_{ij} = b_{ji}$ for $(i, j) \in I \times I$, there exists one and only one quadratic form $Q$ such that

where $\Phi$ denotes the bilinear form associated with $Q$; then $Q$ is given by the formula

the last summation extending over the subsets $\{i, j\}$ of $I$ having one or two elements.

En effet, comme the formula (16) is only a transcription of formula (14), the uniqueness of a quadratic form $Q$ satisfying (15) is proved. To prove its existence, let us first remark that there exists a family $(b'_{ij})$ of elements of $A$ such that $b'_{ii} = b_{ii}$ and that $b'_{ij} + b'_{ji} = b_{ij}$ for $i \neq j$; one obtains, for example, such a family by endowing $I$ with an ordered set structure (Theory of Sets, Chapter III, § 2, No. 3, Theorem 1) and by setting $b'_{ij} = b_{ij}$ for $i < j$ and $b'_{ij} = 0$ for $i > j$. Since the $e_i$ form a basis of $E$, there exists a bilinear form $f$ on $E \times E$ such that $f(e_i, e_j) = b'_{ij}$; by setting $Q'(x) = f(x, x)$ and denoting by $\Phi'$ the bilinear form associated with the quadratic form $Q'$, one obtains $Q'(e_i) = b_{ii}$ and $\Phi'(e_i, e_j) = f(e_i, e_j) + f(e_j, e_i) = b_{ij}$. This proves our second assertion. As for the first, it follows immediately, since, by virtue of the uniqueness, if a quadratic form $Q$ satisfies (15), one has $Q(x) = Q'(x) = f(x, x)$.

The module $E$ endowed with the structure defined by a quadratic form $Q$ is called a quadratic module. A homomorphism of the quadratic module $(E, Q)$ into a quadratic module $(E', Q')$ is a linear mapping $u$ of $E$ into $E'$ such that $Q = Q' \circ u$; if $\Phi$ and $\Phi'$ are the bilinear forms associated with $Q$ and $Q'$, one then has $\Phi(x, y) = \Phi'(u(x), u(y))$ for $x \in E,\ y \in E$; in other words $\Phi'$ is the inverse image of $\Phi$ by $u$ (§ 1, No. 1). One says that two quadratic forms $Q$ and $Q'$ on two $A$-modules $E$ and $E'$ are equivalent if the corresponding quadratic modules are isomorphic.

Let $(E_i, Q_i)_{i \in I}$ be a family of quadratic modules, and let $E$ be the direct sum of the modules $E_i$. The external direct sum of the quadratic modules $(E_i, Q_i)$ is the quadratic module obtained by endowing $E$ with the quadratic form $Q$ defined by $Q(\sum_i x_i) = \sum_i Q_i(x_i)$ for $x_i \in E_i$. One also says that the quadratic form $Q$ is the external direct sum of the quadratic forms $Q_i$.

If the forms $Q_i$ are non-degenerate, the same is true of $Q$.

Let $Q$ be a quadratic form on the $A$-module $E$; if $F$ is a submodule of $E$ and if $Q$ is constant on each class modulo $F$, the mapping $\overline{Q}$ of $E/F$ into $A$ deduced from $Q$ by passing to the quotient is obviously a quadratic form, and the canonical mapping of $E$ onto $E/F$ is a homomorphism for the structures of quadratic modules. For $Q$ to be constant on each class modulo $F$, it is necessary and sufficient that we have $Q(x + y) = Q(x)$ for $x \in E$ and $y \in F$, that is, denoting by $\Phi$ the bilinear form associated with $Q$, that we have $Q(y) + \Phi(x, y) = 0$ for $y \in F$ and $x \in E$. Taking $x = 0$, we see that we have $Q(y) = 0$ for $y \in F$, and therefore

$\Phi(x, y) = 0$ for $x \in E$ and $y \in F$. In other words, if we call the kernel of the quadratic module $(E, Q)$ the set $N$ of elements $x$ of $E$ such that $Q(x) = 0$ and $\Phi(x, z) = 0$ for all $z \in E$, for $Q$ to be constant on each class modulo $F$, it is necessary and sufficient that $F$ be contained in the kernel $N$ of $(E, Q)$. It is readily verified that $N$ is a submodule of $E$. For $Q$ to be constant on each class modulo $F$, it is therefore necessary and sufficient that $F$ be generated by elements of $N$.

It is immediately seen that the kernel of the quadratic module $|E/N$ is $\{0\}$.