Compute the minimal polynomials over the field $\mathbb{Q}$ of the given numbers
- $\sqrt{2+i\sqrt{2}}$
- $\sqrt{1+ \sqrt{3}}$
- $5^\frac{1}{4}$
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Git Gud
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Mathematician
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Some of the observations made here http://math.stackexchange.com/questions/288036/intermediate-fields-of-splitting-field/288059#288059 might be useful. – Andreas Caranti Jan 27 '13 at 14:27
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3How to ask a homework question – Nate Eldredge Jan 27 '13 at 19:00
1 Answers
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Partial answer:
Let $\alpha =\sqrt{2+i\sqrt{2}}$. It holds that $$\begin{align} \alpha =\sqrt{2+i\sqrt{2}} &\Longrightarrow \alpha ^2 = 2 +i\sqrt{2}\\ &\Longrightarrow \alpha^2-2=i\sqrt{2}\\ &\Longrightarrow \alpha ^4 -4\alpha ^2 + 4=-2\\ &\Longrightarrow \alpha ^4 -4\alpha ^2 +6=0 \end{align}$$
It follows that $\alpha$ is a root of the polynomial $m_\alpha(t)$ whee $\displaystyle m_\alpha(t):=t^4-4t^2+6\in \mathbb{Q} \textbf{[}t\textbf{]}$.
Could it be that $m_\alpha$ is irreducible over $\mathbb{Q}$? Ask Eisenstein.
The others are similar.
Git Gud
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So for the third one, I get x^4-5=0, and using Eisenstein's criterion for p=5 we can show that it is irreducible over Q, also it is primitive so it is minimal? right? – Mathematician Jan 27 '13 at 14:37
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@WaqasAliAzhar $x^4 - 5$ is the minimal polynomial of $\displaystyle 5^{\frac{1}{4}}$ because it is irreducible over $\mathbb{Q}$ because of Eisentein, because it is a monic polynomial and because $\displaystyle (5^{\frac{1}{4}})^4-5=0$. I don't see how primitive polynomials related to this. – Git Gud Jan 27 '13 at 14:41
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Isn't it enough to show some polynomial monic and irreducible to show it is minimal? – Mathematician Jan 27 '13 at 14:44
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