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Let $\alpha = \sqrt{2+\sqrt{2}}$

a) find the minimal polynomial of $\sqrt{2}$ over the rationals

b) find the minimal polynomial of $\alpha$ over $\mathbb{Q}(\sqrt{2})$

c) determine $[\mathbb{Q}(\alpha):\mathbb{Q}]$

d) Find the minimal polynomial of $\alpha$ over rationals.

This is not a homework assignment. This is additional material not covered in my course since it is not for galois theory, only because we didn't get enough time to cover it, but I will be taking the next math class for this, and would appreciate help on this and explanations.

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Could you please make the post human readable by using latex? It's not that hard to do. Also, what is s? –  Alex B. May 1 '11 at 9:58
    
sorry s should be a 2 –  mary May 2 '11 at 7:00
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Does not change the answers, which assumed that $s$, whatever it was, lay in $\mathbb{Q}$; $2$ certainly does. The answers are still perfectly applicable, if you understand them. –  Arturo Magidin May 3 '11 at 3:19

2 Answers 2

up vote 3 down vote accepted

An alternative to (c) which uses the results in both (a) and (b) (given by Dennis's answer) is the following:

Since the minimal polynomial of $\sqrt{2}$ over $\mathbb{Q}$ is $x^2-2$, then $[\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=2$.

Since the minimal polynomial of $\sqrt{2+\sqrt{2}}$ over $\mathbb{Q}(\sqrt{2})$ is $x^2 - (2+\sqrt{2})$, then $[\mathbb{Q}(\alpha):\mathbb{Q}(\sqrt{2})]=2$.

Now use the fact that since $\mathbb{Q}\subseteq \mathbb{Q}(\sqrt{2})\subseteq \mathbb{Q}(\alpha)$, then $[\mathbb{Q}(\alpha):\mathbb{Q}]$ can be expressed in terms of $[\mathbb{Q}(\alpha):\mathbb{Q}(\sqrt{2})]$ and $[\mathbb{Q}(\sqrt{2}):\mathbb{Q}]$.

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a) You need to find a polynomial in $\mathbb{Q}[x]$ that vanishes at $\sqrt{2}$, and is irreducible. As you can see, $x^2-2$ vanishes at $\sqrt{2}$ and doesn't decompose over $\mathbb{Q}$

b) You need to specify what is $s$. Assuming that $s\in\mathbb{Q}(\sqrt{2})$, check that $x^2-\sqrt{2}-s$ is the required polynomial.

c) The degree of the extension is equal to the degree of the minimal polynomial of $\alpha$ over $\mathbb{Q}$

d) You can build it in two steps: $\alpha^2=s+\sqrt{2}$. Hence $(\alpha^2-s)^2=2$. So we get that $(x^2-s)^2-2=x^4-2sx+s^2-2$ vanishes at $\alpha$. Check that it's irreducible over $\mathbb{Q}$.

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I think you've misread c). –  Gerry Myerson May 1 '11 at 12:42
    
Yeah, it should be $\alpha$. Corrected –  Dennis Gulko May 1 '11 at 17:28
    
For (b), it's enough for $s$ to be in $\mathbb{Q}(\sqrt{2})$. –  Arturo Magidin May 1 '11 at 18:48
    
@Arturo: right, thanks. –  Dennis Gulko May 3 '11 at 8:47

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