So I have been doing some practice on finding minimal polynomials and I wanted to
1) Verify if my working is correct.
2) Ask for help and explanation for some part of the question.
The question is as follows:
Find the minimal polynomial $\alpha = \frac{1+i}{\sqrt{2}}$ over
i) $\mathbb{Q}$
ii) $\mathbb{Q}[i]$
iii) $\mathbb{Q}[\sqrt{2}\ ]$
iv) $\mathbb{Q}[\sqrt{-2}\ ]$
My attempts for (i), (ii) are as follows.
$\alpha = \frac{1+i}{\sqrt{2}}$
$2\alpha^2 = 2i$
$4\alpha^4 = -4$
$\alpha^4 + 1 = 0$
$f(x) = x^4 + 1$ for the field $\mathbb{Q}$
For the case of $\mathbb{Q}[i]$
$2\alpha^2 = 2i$
$\alpha^2 = i$
$f(x)=x^2 - i$
Did I make a mistake, if yes please point it out for me. Also, I'm a bit clueless on how to attempt part (iii) and part (iv).
So any help or insights regarding that will be deeply appreciated.
In $\mathbb Q(\sqrt2)$: we have $\sqrt2\alpha=1+i\Rightarrow \sqrt2\alpha-1=i\Rightarrow (\sqrt2\alpha-1)^2=-1\Rightarrow 2\alpha^2-2\sqrt2\alpha+2=0 \Rightarrow x^2-\sqrt2x+1=0 \in \mathbb Q(\sqrt2)[x]$ and $x^2-\sqrt2x+1=0 $ is an irreducible polynomial in $\mathbb Q(\sqrt2)[x]$ because if its reducible then $x^2-\sqrt2x+1=(x-\alpha)(x-\beta); \alpha,\beta\in \mathbb Q(\sqrt2) \Rightarrow x^2-\sqrt2x+1=x^2-(\alpha+\beta)x+\alpha \beta\Rightarrow \alpha+\beta =\sqrt2, \alpha \beta=1 $
Let be $\alpha=a+b\sqrt2, \beta=c+d\sqrt2 \Rightarrow (a+c)+(b+d)\sqrt2=\sqrt2$ and $(ac+2bd)+(ad+bc)\sqrt2=1 \Rightarrow$
$ a+c=0\Rightarrow a=-c $
$b+d=1\Rightarrow b=1-d $
$ad+bc=0\Rightarrow -cd+c-cd=0\Rightarrow d=\frac{1}{2}$ and $b=\frac{1}{2}$
then $ac=\frac{1}{2}\Rightarrow a^2=\frac{-1}{2}$ contradiction.
In $\mathbb Q(i\sqrt2)$: we have $\sqrt2\alpha=1+i\Rightarrow \sqrt2x-1-i=0\in \mathbb Q(i\sqrt2) $