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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.

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    Your ideas are correct, but to prove that these are actually the minimal polynomials you need to show that they are irreducible. For iii) and iv): What degree will the minimal polynomials have? What are their possible roots? – 2017-02-22

1 Answers 1

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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) $

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    In fifth line, $\alpha, \beta \in \Bbb Q(\sqrt 2)$. – 2017-02-22
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    Also $d = \frac 12$. – 2017-02-22
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    @Vikrant Desai, thank you very much. I edit my answer. – 2017-02-22