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

Question

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.

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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?

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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.

Answer 1

Hint: It suffices to show that $Z(x+y;T)$ contains both $x$ and $y$. To that end, note that the restrictions $\mu_{T,x}(T) \mid_{Z(y;T)},\mu_{T,y}(T) \mid_{Z(x;T)}$ are invertible.

Answer 2

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.