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Math Help - conjugates

  1. #1
    Senior Member Sampras's Avatar
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    conjugates

    Find the conjugates in  \mathbb{C} of  \sqrt{1+\sqrt{2}} over  \mathbb{Q}(\sqrt{2}) .

    So  \text{irr}\left(\sqrt{1+\sqrt{2}}, \mathbb{Q}(\sqrt{2})\right) = x^2-\left(\sqrt{1+\sqrt{2}}\right) .

    Thus the conjugate is  -\sqrt{1+\sqrt{2}} .

    This is correct?
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  2. #2
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    Quote Originally Posted by Sampras View Post
    Find the conjugates in  \mathbb{C} of  \sqrt{1+\sqrt{2}} over  \mathbb{Q}(\sqrt{2}) .

    So  \text{irr}\left(\sqrt{1+\sqrt{2}}, \mathbb{Q}(\sqrt{2})\right) = x^2-\left(\sqrt{1+\sqrt{2}}\right) .

    Thus the conjugate is  -\sqrt{1+\sqrt{2}} .

    This is correct?

    I suppose you meant \text{irr}\left(\sqrt{1+\sqrt{2}}, \mathbb{Q}(\sqrt{2})\right) = x^2-\left(1+\sqrt{2}\right) , and then yes: the one you write is the conjugate.

    Tonio
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