Corollary 5

— Let Y be an indeterminate and let $\tilde{u}$ denote the endomorphism of the $K(Y)$-vector space $K(Y) \otimes_K E$ induced from $u$ by extension of scalars from $K$ to the field $K(Y)$ of rational functions in $Y$ with coefficients from $K$. Then the endomorphism $Y \cdot 1 - \tilde{u}$ is invertible. Moreover, if $\chi_u'$ denotes the derivative of the polynomial $\chi_u$, then

The endomorphism $Y \cdot 1 - \tilde{u}$ is invertible because its determinant is the nonzero element $\chi_u(Y)$ of $K(Y)$. It follows that $\tilde{u}$ is substitutable in the rational function $r(X) = (Y - X)^{-1}$ in $K(Y)(X)$. The second assertion now follows from Cor. 2, by the relation