# Thread: Method for finding a Minimal Polynomial?

1. ## Method for finding a Minimal Polynomial?

Hello, I cannot seem to find in my notes nor in any text a method of finding minimal polynomials. Most minimal polynomials I've encountered are somewhat trivial, but I came across one that I may be tested on that I am unable to solve using a brute-force method:

$\displaystyle m_{\sqrt 2+i}$ over $\displaystyle \mathbb Q$

The reason I ask about a method of finding the min. poly. is because I am taking an oral test, and I don't want to be floundering about if I am asked something different than this one in particular.

Anyway, I would appreciate any help. Thanks!

P.s.: That's m_{\sqrt 2 +i}; sorry that the "i" is so tiny.

2. Originally Posted by Dark Sun
Hello, I cannot seem to find in my notes nor in any text a method of finding minimal polynomials. Most minimal polynomials I've encountered are somewhat trivial, but I came across one that I may be tested on that I am unable to solve using a brute-force method:

$\displaystyle m_{\sqrt 2+i}$ over $\displaystyle \mathbb Q$

The reason I ask about a method of finding the min. poly. is because I am taking an oral test, and I don't want to be floundering about if I am asked something different than this one in particular.

Anyway, I would appreciate any help. Thanks!

P.s.: That's m_{\sqrt 2 +i}; sorry that the "i" is so tiny.
$\displaystyle x=\sqrt{2}+i$

$\displaystyle x^2=(\sqrt{2}+i)^2 \iff x^2=2+2i\sqrt{2}-1 \iff x^2-1=2i\sqrt{2}$

$\displaystyle (x^2-1)^2=(2i\sqrt{2})^2$

$\displaystyle x^4-2x^2+1=-8$

$\displaystyle x^4-2x^2+9=0$

3. Originally Posted by TheEmptySet
$\displaystyle x=\sqrt{2}+i$

$\displaystyle x^2=(\sqrt{2}+i)^2 \iff x^2=2+2i\sqrt{2}-1 \iff x^2-1=2i\sqrt{2}$

$\displaystyle (x^2-1)^2=(2i\sqrt{2})^2$

$\displaystyle x^4-2x^2+1=-8$

$\displaystyle x^4-2x^2+9=0$
you also need to prove that $\displaystyle p(x)=x^4 - 2x^2 + 9$ is irreducible over $\displaystyle \mathbb{Q}$: we only need to show that it cannot be factored into two quadratic polynomials over $\displaystyle \mathbb{Z}.$ why?

suppose $\displaystyle p(x)=(x^2+ax+b)(x^2+cx+d),$ for some $\displaystyle a,b,c,d \in \mathbb{Z}.$ then we'll have $\displaystyle bd=9$ and $\displaystyle b+d+2=a^2,$ which is easily seen to have no solutions in $\displaystyle \mathbb{Z}.$

4. Because irreducible over $\displaystyle \mathbb Z$ implies irreducible over $\displaystyle \mathbb Q$.

$\displaystyle x^4-2x^2+9$ clearly does not have any degree 1 factors. Also, a degree 3 factor would imply a degree 1 factor, so if a factor exists, then it will be degree 2, which you have just shown is a contradiction.

Thank you EmptySet and NonCommAlg for presenting me with this truly powerful method!