Minimal polynomial is invariant under field extensions I saw it's been answered before but I could not understand the answers as it delved into material that I did not study yet.
My question is as follows:
Suppose $p \in F[x]$ is the minimal polynomial of some matrix $A \in M_n(F)$ and $F \subseteq K$
I need to show that $p$ is also the minimal polynomial of $A$ over the field $K$. In other words: Show that the minimal polynomial remains the same when we move to a bigger field.
What I thought of:
Obviously $p$ is irreducible in $F$, suppose it is reducible in $K$, which means there are $q,t \in K, q,t \notin F$ such that $p=qt$, now we just need to show that $q(A) \neq 0$ and $t(A) \neq 0$...But why is that so?
Edit: Another way of presenting this question is "Show that if $p$ is the minimal polynomial of $A$ over the field $K$ while the entries of$~A$ are contained in a subfield $F \subseteq K$, then all the coefficients of $p$ are elements of $F$".
 A: Your question is answered here, but I've written out the solution below providing more detail. The idea is essentially to use the fact that for $T \in M_n(F)$, whenever $Tv = 0$ has a nontrivial solution $v$ in $K$, it has one over $F$ as well.
Lemma: If $v_1,\ldots,v_r \in F^n$ are linearly dependent over $K$, then they are also linearly dependent over $F$. (Here $F \subseteq K$.)
Proof: Contraposition. Assume $v_1,\ldots,v_r$ are linearly independent over $F$, and let $k_i \in K$ such that $\sum_i k_i v_i = 0$. Choose a basis $k_1',\ldots,k_s'\in K$ for $\text{Span}_F\{k_1, \ldots, k_r\}$. Then there exist $f_{ij} \in F$ such that $k_i = \sum_j f_{ij}k_j'$ for $i=1,\ldots,r$, and so
$$
0 = \sum_{i=1}^r k_i v_i = \sum_{i=1}^r \sum_{j=1}^s f_{ij}k_j'v_i.
$$
We claim that this implies $\sum_{i=1}^r f_{ij}v_i = 0$ for each $j$. To see this, note that by hypothesis $v_i = (v_{i,1},v_{i,2},\ldots,v_{i,n})^t$ with $v_{i,l} \in F$ for all $l$ (for all $i$). Hence for each $l$, we can identify $l$-th coordinates in the above equalities to see that
$$
\sum_{j=1}^s\left(\sum_{i=1}^r f_{ij}v_{i,l}\right)k_j' = \sum_{i=1}^r\sum_{j=1}^s f_{ij}k_j'v_{i,l} = 0.
$$
Since the $k_j'$ are linearly independent over $F$ by construction, we conclude that
$$
\sum_{i=1}^r f_{ij}v_{i,l} = 0
$$
for all $j$ and $l$. Hence for all $j$
$$
\sum_{i=1}^r f_{ij}v_i = (\sum_{i=1}^r f_{ij}v_{i,l})_{l} = 0.
$$
But $f_{ij} \in F$ for all $i,j$, whence the linear independence of the $v_i$ implies $f_{ij} = 0$ for all $i,j$, and therefore $k_i = \sum_{j=1}^s f_{ij}k_j' = 0$ for all $i$. It follows that $v_1,\ldots,v_r$ are linearly independent over $K$.

To apply this to your question, let $m_F$ and $m_K$ denote the minimal polynomials of $A$ over $F$ and $K$, respectively. Then $m_K \mid m_F$ is true by the definition of $m_K$ and the fact that $m_F(A) = 0$. Conversely, notice that for $r = \deg(m_K)$ the equation $m_K(A) = 0$ shows that $1, A, A^2, \ldots, A^r$ are linearly dependent over $K$, and therefore they must also be linearly dependent over $F$ by the above lemma. But this implies $p(A) = 0$ for some polynomial in $F[x]$ of degree at most $r$, whence $\deg(m_F) \leq r$. Combining this with $m_K \mid m_F$ and the fact that both $m_K$ and $m_F$ are monic, it follows that $m_K = m_F$.
