I have a problem proving the following statement:
"Let $V$ be a $K$ Vector-space with finite dimension, let $T \in \text{End}_K(V)$, define $p,q,h \in K[X]$ ($K[X]$ is the polynomial ring with coefficient from $K$). Define
$$p := \mu_{T,x}, \hspace{3mm} q := \mu_{T,y}, \hspace{3mm} h := \mu_{T,x+y} \hspace{3mm} d := \text{gcd}(p,q)$$ ($\mu_{T,x}$ is the $T$-Annihilator of $x$, so it's defined as (this is wrong, see Edit) $\mu_{T,x} = \{ p \in K[X] \setminus 0: p(T)(x) = 0\}$) Now show: $$\frac{pq}{d^2} \mathbin{\Big|} h \text{ and } h \mathbin{\Big|} \frac{pq}{d} $$"
First off, I am not quite sure what "$p=\mu_{T,x}$" means. The $T$-Annilihator is a set of polynomials, but $p$ was defined was being in $K[X]$. Does this mean, that one could just take any polynomial in $\mu_{T,x}$ to prove this statement? In my opinion, this can not be right, since it would have some consequences. For example, I could define $x_0=1$, $y_0=-1$, $p(x) =(x-1)$, $q(x) = (x+1)$ and $h(x) = x$, then $p(x_0) = q(y_0) = h(x_0+y_0) = 0$, and $\text{gcd}(p,q) = 1$, but $pq$ does not divide $h$ (since $pq$ has more zeros than $h$). So what is actually meant by "$p=\mu_{T,x}$"? Does somebody have some advice for me how to tackle this problem?
Edit: Okay, I should have looked a little bit closer to our definitions, actually, $\mu_{T,x}$ was defined as being the minimal polynom $p$ with $p(T)(x) = 0$, this makes a lot more sense now, I think that I now have an idea how to prove this, I will give the solution when I'm ready.
Solution:
It's actually quite simple. First off, we notice that $\gcd(\frac{p}{d},\frac{q}{d}) = 1$. Also, by the definition of the gcd and divisibility, we find that there are $t_1, t_2 \in K[X]$, st. $dt_1 = p, dt_2 = q$. Also, $t_1$ and $t_2$ do not divide either of $p$ or $q$. It's sufficient enough to show $ \frac{p}{d} | h$, because of this Lemma:
Lemma 1: "$\gcd(a,b) = 1$, $a|c$ and $b|c$ $\implies$ $ab|c$"
We see: $$ 0 = hq(T)(x+y) = hq(T)(x) + hq(T)(y) = hq(T)(x)$$ This means, that $p | hq$ (since $p$ is the minimal polynomial that satisfies $p(T)(x) = 0$). We can rewrite $q$ with $q = t_2 d$, and because $p$ and $t_2$ have 1 as their biggest common divisor, we get: $$p | hq \implies p | ht_1d \implies p | hd \iff \frac{p}{d} | h$$ Doing the same thing for $q$ and using the Lemma 1, we get the first statement, the second statement is even easier, because: $$\frac{pq}{d}(T)(x+y) = \frac{q}{d}p(T)(x) + \frac{p}{d}q(T)(y) = 0$$
Because $h$ is minimal, $h$ must divide $\frac{pq}{d}$