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

$$p{m}_1+q{m}_2=1$$(1)

this is Bézout’s lemma (or the Euclidean algorithm, or the fact that $k[X]$ is a principal ideal domain).

Now $p{m}_1(T)$ is the projection from $Z(v_1,T)\oplus Z(v_2,T)$ to $Z(v_2,T)$:

$$(p{m}_1)(T)v_1=p(T)m_1(T)v_1=p(T)0=0$$(2)

(by assumption that $m_1$ is the annihilator polynomial of $v_1$) and

$$(p{m}_1)(T)=1-(q{m}_2)(T)$$(3)

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

$$(p{m}_1)(T)v_2=v_2-q(T)m_2(T)v_2=v_2-q(T)0=v_2$$(4)

Any subspace that is invariant under $T$ is also invariant under polynomials of $T$. Therefore, the preceding equations show that $v_2=(p{m}_1)(T)(v_1+v_2)\in Z(v_1+v_2,T)$. 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.