In the field theory book that I am reading I found the following theorem:
If a field extension $K/F$ is finite, then it is algebraic.
Proof: If $K/F$ were transcendental, there would exist $u\in K$ such that $[F(u):F]=\infty$, so by de tower theorem $[K:F]=[K:F(u)][F(u):F]=\infty$, which is a contradiction.
However, the following example was also given:
$\mathbb{C}/\mathbb{R}$ is algebraic
Proof: Since $\{1,i\}$ is a basis of $\mathbb{C}$ as an $\mathbb{R}$-vector space, every element of $\mathbb{C}$ can be written as $a+bi$, with $a,b \in \mathbb{R}$. We have that $(x-(a+bi))(x-(a-bi))=x^2 - 2ax + a^2 +b^2 \in \mathbb{R}[x]$ has $a+bi$ as a root and therefore every $a+bi \in \mathbb{C}$ is algebraic.
This made me think if this process could be generalized to give a contructive proof of the first theorem, i.e., if having a finite basis $\{k_1 , \dots\ , k_n\}$ of $K/F$ and expressing $k \in K$ as a linear combination of the basis we can construct some $p\in F[x]$ having $k$ as a root.
Consider, for any $k \in K$, the following $F$-endomorphism of $K$: $u_k:x \longmapsto kx$. Let $P(X) \in F[X]$ be its characteristic polynomial. Then it is easy to see that $0=P(u_k)=u_{P(k)}$ thus $P(k)=0$.