# Field Theory - Algebraic Extensions - D&F - Section 13.2 - Exercise 7, page 530

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• Sep 28th 2013, 02:17 AM
Bernhard
Field Theory - Algebraic Extensions - D&F - Section 13.2 - Exercise 7, page 530
Exercise 7 in Section 13.2 Algebraic Extensions, page 530 of Dummit and Foote states the following:

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7. Prove that $\mathbb{Q} ( \sqrt{2} + \sqrt{3} ) = \mathbb{Q} ( \sqrt{2} , \sqrt{3} )$.

Conclude that $[\mathbb{Q} ( \sqrt{2} + \sqrt{3} ) \ : \ \mathbb{Q} ] = 4$.

Find an irreducible polynomial satisfied by $\sqrt{2} + \sqrt{3}$

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I am somewhat overwhelmed by this problem ... can someone advise me on an approach ... and, indeed, get me started?

Peter
• Sep 28th 2013, 04:06 AM
Idea
Re: Field Theory - Algebraic Extensions - D&F - Section 13.2 - Exercise 7, page 530
Let $u=\sqrt{2}+\sqrt{3}$

$\frac{1}{2}u^3-\frac{9}{2}u=\sqrt{2}$

$\frac{11}{2}u-\frac{1}{2}u^3=\sqrt{3}$

These equations show that $\sqrt{2},\sqrt{3}\in Q(u)$
• Sep 28th 2013, 12:02 PM
SlipEternal
Re: Field Theory - Algebraic Extensions - D&F - Section 13.2 - Exercise 7, page 530
For the irreducible polynomial, take powers.

\begin{align*}(\sqrt{2}+\sqrt{3})^2 & = 5+2\sqrt{6} \\ (\sqrt{2}+\sqrt{3})^3 & = ? \\ (\sqrt{2}+\sqrt{3})^4 & = ? \\ & \vdots \end{align*}

From those, you should be able to find a relationship such that if $x=\sqrt{2}+\sqrt{3}$, then $x^n+a_{n-1}x^{n-1}+\ldots + a_1x+a_0=0$. Hint: the problem tells you you don't need to go past $x^4$.