Proposition 7

— Assume that f is monic of degree p; with the above notations we have¹

(30)

¹ The resultant $\operatorname{res}_{p,q}(f, g)$ is thus independent of $q$ when $f$ is monic of degree $p$. We then denote it simply by $\operatorname{res}(f, g)$.

Let us define an A-linear mapping $0$ of $S_q \times S_p$ into $S_{p+q}$ by $\theta(u, v) = uf + v$. Then $0$ maps the basis $B_1$ of $S_q \times S_s$ into the sequence

of elements of $S_{p+q}$; hence the matrix $M$, of $0$ with respect to the bases $B_1, B_2$ is lower triangular and its diagonal elements are equal to 1, whence $\det M = 1$.

It follows that $\theta$ is bijective and $\operatorname{res}_{p,q}(f, g)$ is equal to the determinant of the endomorphism $\varphi' = \varphi \circ \theta^{-1}$ of $S_{p-q}$. Explicitly, we have

for any pair $(u, v)$ in $S_q \times S_s$. Now we have $A[X] = S_{p+q} + (f)$ and $fS_q = S_s \cap (f)$, hence the canonical injection of $S_{p-s}$ into $A[X]$ defines by the passage to quotients an isomorphism $\gamma$ of $S_{p+q}/fS_q$ onto E. Let $\psi$ be the multiplication by $g(x)$ in E. Formula (31) shows that $\varphi'$ induces the identity on $fS_q$ and $\gamma^{-1}\psi\gamma$ on $S_{p+q}/fS_q$. We thus have $\det \varphi' = \det \psi$, whence (30) follows, because $\det \varphi' = \operatorname{res}_{p,q}(f, g)$ and $\det \psi = N_{E/A}(g(x))$, by definition.