Definition 2

We suppose that the ring $A$ is commutative. We say that a mapping $Q$ of $E$ into $A$ is a quadratic form on $E$ if 1) $Q(\alpha x) = \alpha^2 Q(x)$ for $\alpha \in A$ and $x \in E$; 2) the mapping $\Phi : (x, y) \to Q(x + y) - Q(x) - Q(y)$ of $E \times E$ into $A$ is a bilinear form. The bilinear form $\Phi$ (which is necessarily symmetric) is called the bilinear form associated with $Q$. If $\Phi$ is non-degenerate, we say that $Q$ is non-degenerate.

Since $Q(2x) = 4Q(x)$, it follows at once from 2) that

In particular, if $A$ is a ring of characteristic 2, the form $\Phi$ is alternating.

We shall say that two elements (resp. two subsets) of $E$ are orthogonal relative to $Q$ if they are orthogonal relative to the bilinear form associated with $\Phi$.

Let $(x_i)_{i \in I}$ be a family of elements of $E$ and $(a_i)_{i \in I}$ a family of elements of $A$ all but a finite number of which are zero. By induction on the number of indices $i$ for which $a_i \neq 0$, one easily shows that

the last summation being extended over the two-element subsets of I.

For every bilinear form $f$ on $E \times E$, we define a quadratic form $Q$ by putting $Q(x) = f(x, x)$; the bilinear form $\Phi$ associated with $Q$ is then defined by $\Phi(x, y) = f(x, y) + f(y, x)$ for $x, y$ in $E$. Moreover, if one supposes that the scalar 2 has an inverse $\frac{1}{2}$ in $A$, there exists one and only one symmetric bilinear form $f$ such that $Q(x) = f(x, x)$, namely $f = \frac{1}{2} \Phi$; the discriminant of $f$ with respect to a system $S = (x_1, \ldots, x_n)$ is also called the discriminant of $Q$ with respect to $S$. There is therefore in this case a one-to-one correspondence between the quadratic forms and the symmetric bilinear forms on $E$ (cf. exerc. 6).

In the case of a free module, we have in addition the following result: