existence of the minimal polynomial

Proposition 1.

Let $K\mathrm{/}L$ be a finite extension of fields and let $k\mathrm{\in }K$. Thereexists 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 non-zero ideal. This fact relieson the fact that $K/L$ is a finite extension of fields, andtherefore it is an algebraic extension, so every element of $K$ isa root of a non-zero 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 idealdomain (see example of PID).Therefore, the kernel of $\psi$ is a principal ideal, generated bysome 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, $m_k$ isthe unique generator of $\mathrm{Ker}(\psi )$ which is monic.

By construction, $m_k(k)=0$, since $m_k$ belongs to the kernelof $\psi$, 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 thisideal, 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, aswe pointed out, there is a unique monic generator, namely$m_k(x)$.

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