Show that $V=Z(x;T)\oplus Z(y;T)$ and the $T$-annihilators $\mu_{T,x},\,\mu_{T,y}$ do not share any common divisors implies that $V$ is cyclic Provided that $V=Z(x;T)\oplus Z(y;T)$ where $Z(v;T)$ denotes the cyclic subspace and the corresponding $T$-annihilators $\mu_{T,x},\,\mu_{T,y}$ do not share any common divisors, show that $V$ is itself cyclic.

My approach was to first identify a possible cyclic vector, which was $x+y$ in this case. I then tried to show that every element of $V$ is an element of the cyclic vector space spanned by $T^jx+y,\ j\in\mathbb{N}\cup \{0\}$ but the problem seems to be the condition that the $T$-annihilators $\mu_{T,x},\,\mu_{T,y}$ do not share any common divisors.
How do I apply this or how do I continue?

Edit: Definition of the $T$-annihilator as in T-Annihilators and Minimal polynomial :

Definition: $T$-annihilator of a vector $\alpha$(denoted as  $p_\alpha$) is the unique monic polynomial which generates the ideal
such that $g(T)\alpha = 0$ for all $g$ in this ideal.


If the field is $k$, then $\pi_x:k[t]\to Z(x;T)$ by $p(t)\mapsto p(T)x$ is onto with kernel $\mu_{T,x}(t)$, so that $Z(x;T)\simeq \frac{k[t]}{\langle \mu_{T,x}(t)\rangle}$ and similarly for the other factor.
What you want to prove is that, if these minimal polynomials are coprime, then
$$
\frac{k[t]}{\langle \mu_{T,x}(t)\rangle}
\oplus
\frac{k[t]}{\langle \mu_{T,y}(t)\rangle}
\simeq
\frac{k[t]}{\langle \mu_{T,x}(t)\mu_{T,y}(t)\rangle};
$$
this is just the Chinese Remainder Theorem.
