# 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. 1.

$m_{k}(x)$

2. 2.

$m_{k}(k)=0$;

3. 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 $m^{\prime}(x)$ 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 $m_{k}(x)$.

Title existence of the minimal polynomial ExistenceOfTheMinimalPolynomial 2013-03-22 13:57:24 2013-03-22 13:57:24 alozano (2414) alozano (2414) 7 alozano (2414) Theorem msc 12F05 FiniteExtension Algebraic  