How does $p_0(t)\mid k(t)$? Let  $\alpha_i$ is a root of $k(t) \in \mathbb Z[t]$. Let $p_0(t)$ be the minimal polynomial of $\alpha_1$ over $\mathbb Z$. After this, it is written in a text that

Then, we have $p_0(t)\mid k(t)$. So, $\alpha_1$ is also a root of $k(t)$.

My question is how we know  $p_0(t)\mid k(t)$?
EDIT
Let me rephrase my problem.

the first statement:  $\alpha_i$ is a root of $k(t) \in \mathbb Z[t]$, this is according to a previous hypothesis.

the second statement:  Let $p_0(t)$ be the minimal polynomial of $\alpha_1$ over $\mathbb Z$, this is true because for $\alpha_1$, there exists a minimal polynomial.

the third statement:   We have $p_0(t)\mid k(t)$.


Now every algebraic number $\alpha_1$ has a minimum polynomial, say $p_0$, and  if polynomial $k(t)$ has root $\alpha_1$ then  $p_0$ divides $k(t)$ (in general every polynomial that$\alpha_1$ satisfies). But at this point we don't know $\alpha_i=\alpha_1$, then how we know $p_0(t)\mid k(t)$?

the fourth statement: &quot;So&quot;, $\alpha_1$ is also a root of $k(t)$

Now if we already used the fact that  $\alpha_1$ is also a root of $k(t)$ imply $p_0(t)\mid k(t)$ then how can we say $p_0(t)\mid k(t)$ implies  $\alpha_1$ is also a root of $k(t)$?
He has constructed $k(t)$ so that $\alpha_1$ is a root of $k(t)$ by showing that $h(g(\alpha_1))=0$ and naming $k(t)=h(g(t))$.  Then if $p_0(t)$ is the minimal polynomial of $\alpha_1$, the theorem you quote in the comments &quot;(for) every algebraic number α the minimum polynomial that α satisfies divides every polynomial that α satisfies.&quot; gives us $p_0(t)|k(t)$ because $k(\alpha_1)=0$
He has constructed $k(t)$ so that $\alpha_1$ is a root of $k(t)$ by showing that $h(g(\alpha_1))=0$ and naming $k(t)=h(g(t))$.  Then if $p_0(t)$ is the minimal polynomial of $\alpha_1$, the theorem you quote in the comments &quot;(for) every algebraic number α the minimum polynomial that α satisfies divides every polynomial that α satisfies.&quot; gives us $p_0(t)|k(t)$ because $k(\alpha_1)=0$