proof of theorem about cyclic subspaces

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

 $pm_{1}+qm_{2}=1$ (1)

this is BÃÂ©zout’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 $Z(v_{2},T)$:

 $(pm_{1})(T)v_{1}=p(T)m_{1}(T)v_{1}=p(T)0=0$ (2)

(by assumption that $m_{1}$ is the annihilator polynomial of $v_{1}$) and

 $(pm_{1})(T)=1-(qm_{2})(T)$ (3)

(by choice of $p$ and $q$), so

 $(pm_{1})(T)v_{2}=v_{2}-q(T)m_{2}(T)v_{2}=v_{2}-q(T)0=v_{2}$ (4)

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.