We first prove the case $r=2$ . The $\subseteq$ inclusion is clear, since the right side is a $T$ -invariant subspace that contains $v_1+v_2$ .

For the other inclusion, it is sufficient to show that $v_1,v_2 \in Z(v_1+v_2,T)$ . The idea is that the action of $T$ on $v_1 + v_2$ can "isolate" the two summands if their annihilator polynomials are coprime. Let's write $m_i$ for $m_{v_i}$ .

Since $(m_1,m_2)=1$ , there exist polynomials $p$ and $q$ such that \begin{equation}pm_1+qm_2=1\end{equation}this is Bézout's lemma (or the Euclidean algorithm, or the fact that $k[X]$ is a principal ideal domain).

Now $pm_1(T)$ is the projection from $Z(v_1,T) \oplus Z(v_2,T)$ to $Z(v_2,T)$ : \begin{equation}(pm_1)(T)v_1=p(T)m_1(T) v_1 = p(T)0 = 0\end{equation}(by assumption that $m_1$ is the annihilator polynomial of $v_1$ ) and \begin{equation}(pm_1)(T) = 1-(qm_2)(T)\end{equation}(by choice of $p$ and $q$ ), so \begin{equation}(pm_1)(T) v_2 = v_2 - q(T)m_2(T) v_2 = v_2 - q(T)0 = v_2\end{equation} Any subspace that is invariant under $T$ is also invariant under polynomials of $T$ . Therefore, the preceding equations show that $v_2 = (pm_1)(T)(v_1+v_2) \in Z(v_1+v_2,T)$ . By symmetry, we also get that $v_1 \in Z(v_1+v_2,T)$ .

For the last claim, we note that the annihilator polynomial $m$ of $Z(v_1,T) \oplus Z(v_2,T)$ is the least common multiple of $m_1$ and $m_2$ (that $m$ is a multiple of $m_1$ follows from the fact that $m$ must annihilate $v_1$ , and the set of polynomials that annihilate $v_1$ is the ideal generated by $m_1$ ). Since $m_1$ and $m_2$ are coprime, the lcm is just their product.

That concludes the proof for $r=2$ . If $r$ is arbitrary, we can simply apply the $r=2$ case inductively. We only have to check that the coprimality condition is preserved under applying the $r=2$ case to $i=1,2$ . But it is well-known that if $p,q,r$ (in $k[X]$ or in any principal ideal domain) are pairwise coprime, then $pq$ and $r$ are also coprime.