You are here
Homeexistence of the minimal polynomial
Primary tabs
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 $m_{{k}}(x)$.
∎
Mathematics Subject Classification
12F05 no label found Forums
 Planetary Bugs
 HS/Secondary
 University/Tertiary
 Graduate/Advanced
 Industry/Practice
 Research Topics
 LaTeX help
 Math Comptetitions
 Math History
 Math Humor
 PlanetMath Comments
 PlanetMath System Updates and News
 PlanetMath help
 PlanetMath.ORG
 Strategic Communications Development
 The Math Pub
 Testing messages (ignore)
 Other useful stuff
Recent Activity
new question: Prime numbers out of sequence by Rubens373
Oct 7
new question: Lorenz system by David Bankom
Oct 19
new correction: examples and OEIS sequences by fizzie
Oct 13
new correction: Define Galois correspondence by porton
Oct 7
new correction: Closure properties on languages: DCFL not closed under reversal by babou
new correction: DCFLs are not closed under reversal by petey
Oct 2
new correction: Many corrections by Smarandache
Sep 28
new question: how to contest an entry? by zorba
new question: simple question by parag