existence of the minimal polynomial
Proposition 1.
Let $K\mathrm{/}L$ be a finite extension^{} of fields and let $k\mathrm{\in}K$. There exists a unique polynomial^{} ${m}_{k}\mathit{}\mathrm{(}x\mathrm{)}\mathrm{\in}L\mathit{}\mathrm{[}x\mathrm{]}$ 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:
$$\psi :L[x]\to K$$ 
$$\psi (p(x))=p(k)$$ 
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 $L[x]$:
$$\mathrm{Ker}(\psi )=\{p(x)\in L[x]\mid p(k)=0\}$$ 
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, $p(x)\in \mathrm{Ker}(\psi )$.
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 $m(x)$:
$$\mathrm{Ker}(\psi )=(m(x))$$ 
Note that the only units in $L[x]$ are the constant polynomials, hence if ${m}^{\prime}(x)$ is another generator of $\mathrm{Ker}(\psi )$ then
$${m}^{\prime}(x)=l\cdot m(x),l\ne 0,l\in L$$ 
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 $\mathrm{Ker}(\psi )$ 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 $p(x)\in \mathrm{Ker}(\psi )$. 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 $\mathrm{Ker}(\psi )$, and, as we pointed out, there is a unique monic generator, namely ${m}_{k}(x)$.
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Title  existence of the minimal polynomial 

Canonical name  ExistenceOfTheMinimalPolynomial 
Date of creation  20130322 13:57:24 
Last modified on  20130322 13:57:24 
Owner  alozano (2414) 
Last modified by  alozano (2414) 
Numerical id  7 
Author  alozano (2414) 
Entry type  Theorem 
Classification  msc 12F05 
Related topic  FiniteExtension 
Related topic  Algebraic^{} 