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existence of the minimal polynomial
Proposition 1.
Let $K/L$ be a finite extension of fields and let $k\in K$. There exists a unique polynomial $m_{{k}}(x)\in L[x]$ such that:
1. $m_{{k}}(x)$ is a monic polynomial;
2. $m_{{k}}(k)=0$;
3. If $p(x)\in L[x]$ is another polynomial such that $p(k)=0$, then $m_{{k}}(x)$ divides $p(x)$.
Proof.
We start by defining the following map:
$\psi\colon L[x]\to K$ 
$\psi(p(x))=p(k)$ 
Note that this map is clearly a ring homomorphism. For all $p(x),q(x)\in L[x]$:

$\psi(p(x)+q(x))=p(k)+q(k)=\psi(p(x))+\psi(q(x))$

$\psi(p(x)\cdot q(x))=p(k)\cdot q(k)=\psi(p(x))\cdot\psi(q(x))$
Thus, the kernel of $\psi$ is an ideal of $L[x]$:
$\operatorname{Ker}(\psi)=\{p(x)\in L[x]\mid p(k)=0\}$ 
Note that the kernel is a nonzero ideal. This fact relies on the fact that $K/L$ is a finite extension of fields, and therefore it is an algebraic extension, so every element of $K$ is a root of a nonzero polynomial $p(x)$ with coefficients in $L$, this is, $p(x)\in\operatorname{Ker}(\psi)$.
Moreover, the ring of polynomials $L[x]$ is a principal ideal domain (see example of PID). Therefore, the kernel of $\psi$ is a principal ideal, generated by some polynomial $m(x)$:
$\operatorname{Ker}(\psi)=(m(x))$ 
Note that the only units in $L[x]$ are the constant polynomials, hence if $m^{{\prime}}(x)$ is another generator of $\operatorname{Ker}(\psi)$ then
$m^{{\prime}}(x)=l\cdot m(x),\quad l\neq 0,\quad l\in L$ 
Let $\alpha$ be the leading coefficient of $m(x)$. We define $m_{{k}}(x)=\alpha^{{1}}m(x)$, so that the leading coefficient of $m_{{k}}$ is $1$. Also note that by the previous remark, $m_{{k}}$ is the unique generator of $\operatorname{Ker}(\psi)$ which is monic.
By construction, $m_{{k}}(k)=0$, since $m_{{k}}$ belongs to the kernel of $\psi$, so it satisfies $(2)$.
Finally, if $p(x)$ is any polynomial such that $p(k)=0$, then $p(x)\in\operatorname{Ker}(\psi)$. Since $m_{{k}}$ generates this ideal, we know that $m_{{k}}$ must divide $p(x)$ (this is property $(3)$).
For the uniqueness, note that any polynomial satisfying $(2)$ and $(3)$ must be a generator of $\operatorname{Ker}(\psi)$, and, as we pointed out, there is a unique monic generator, namely $m_{{k}}(x)$.
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