Proposition 2

Let $A'$ be a commutative ring, $\varphi$ a homomorphism of $A$ into $A'$, $Q’$ the quadratic form on $E' = A' \otimes_A E$ deduced from $Q$ by extension of scalars ($§ 3$, no 4, prop. 3). There exists one and only one isomorphism $j$ of the algebra $A' \otimes_A C(Q)$ onto $C(Q')$ such that $j (1 \otimes \rho_Q(x)) = \rho_{Q'}(1 \otimes x)$ for every $x \in E$.

It suffices to prove that the algebra $C' = A' \otimes C(Q)$ and the mapping $1 \otimes \rho_Q$ of $E'$ into $C'$ form a solution of the same problem of universal mapping as $C(Q')$ and $\rho_{Q'}$. Now, let D’ be an algebra over $A'$ and let $f'$ be an $A'$-linear mapping of $E'$ into D’ such that $f'(x')^2 = Q'(x').1$ for every $x' \in E'$. The mapping $g : x \to f' (1 \otimes x)$ of E into D’ (considered as an A-module by means of the homomorphism $\varphi$) is A-linear and we have $g(x)^2 = Q' (1 \otimes x).1 = Q(x).1$ for every $x \in E$. There therefore exists one and only one A-homomorphism $\overline{g}$ of $C(Q)$ into D’ such that $\overline{g}(\rho_Q(x)) = f'(1 \otimes x)$. Consequently there exists one and only one $A'$-homomorphism $\overline{f}'$ of $C'$ into D’ such that $\overline{f}'(1 \otimes \rho_Q(x)) = f'(1 \otimes x)$ for every $x \in E$; by linearity it follows that $\overline{f}'((1 \otimes \rho_Q)(x')) = f'(x')$ for every $x' \in E'$. QED.