Proposition 2

Let $E$ be a quadratic $A$-algebra and $e$ its unit element. For all $u \in E$, let $T(u)$ be the trace of the endomorphism $m_u : x \mapsto ux$ of the free $A$-module $E$ (II, § 4, no. 3). Then the mapping $s$ defined by $s(u) = T(u).e - u$ is an automorphism of the algebra $E$ and $s^2(u) = u$ for all $u \in E$.

Let $(e, i)$ be a basis of $E$ of type $(\alpha, \beta)$; then $T(e) = 2$, when $s(e) = e$, and $T(i) = \beta$, whence $s(i) = \beta e - i$. Hence $(e, s(i))$ is a basis of $E$, whose type is given by (3) with $\gamma = \beta$ and $\delta = -1$, which again gives $(\alpha, \beta)$; it follows that $s$ is an automorphism of the algebra $E$. As $m_{s(u)} = s m_u s^{-1}$, the endomorphisms $m_u$ and $m_{s(u)}$ of the $A$-module $E$ have the same trace (II, § 4, no. 3, Proposition 3), whence

for all $u \in E$.

The automorphism $s$ is called conjugation of the $A$-algebra $E$ and $s(u)$ the conjugate of $u$.

If $u = \xi e + \eta i$, with $\xi, \eta$ in $A$, then $s(u) = (\xi + \beta \eta)e - \eta i$, whence

(4)

(5)

where we have written $N(u) = \xi^2 + \beta \xi \eta - \alpha \eta^2$. The elements $T(u)$ and $N(u)$ (or, when $A$ and $Ae$ are canonically identified, the elements $T(u)e$ and $N(u)e$) are called respectively the trace and norm of $u$.

When $\beta = 0$, the above formulae are simplified to

(6)

Clearly $T$ is a linear form on $E$*and* $N$ is a quadratic form on $E$ (IX, § 3, no. 4)*. As $E$ is commutative and associative, it follows from (5) that

(7)

For $u$ to be invertible in $E$, it is necessary and sufficient that $N(u)$ be invertible in $A$. For, as $N(e) = 1$, the necessity of the condition follows from (7), writing $v = u^{-1}$. Conversely, if $N(u)$ is invertible in $A$, it follows from (5) that $u$ is invertible and that

(8)

It can be proved that $N(u)$ is the determinant (§ 8, no. 1) of the endomorphism $m_u$ (cf. § 9 no. 3, Example 1).

The following proposition gives the structure of quadratic algebras over a commutative field: