## A generating set for the kernel of a given module homomorphism

Let $V$, $T$, $A$, et cetera be as in this previous exercise.

Show that $\mathsf{ker}\ \varphi$ is generated by $D = \{\omega_1,\ldots,\omega_n\}$.

Let $\zeta \in \mathsf{ker}\ \varphi$. Now $\varphi(\zeta) = 0$, and by this previous exercise, we have $\zeta \in \mathsf{span}_{F[x]}(\omega_1,\ldots,\omega_n) + \mathsf{span}_F(\xi_1,\ldots,\xi_n)$. Recall that the $\omega_i$ are in $\mathsf{ker}\ \varphi$, so that $0 = \varphi(\zeta) = \sum b_i v_i$ for some $b_i \in F$. Since the $v_i$ are a basis for $V$ over $F$, we have $b_i = 0$ for all $i$. Thus $\mathsf{ker}\ \varphi \subseteq \mathsf{span}_{F[x]}(\omega_1,\ldots,\omega_n)$. The reverse inclusion is immediate, and so the $\omega_i$ are a generating set for $\mathsf{ker}\ \varphi$ over $F[x]$.