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 non-zero 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 non-zero 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 $m'(x)$ is another generator of $\operatorname{Ker}(\psi)$ then $$m'(x)=l\cdot m(x), \quad l\neq 0,\quad l\in L$$ 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 $m_{k}(x)$ .