conjugates

**Sampras** · January 8th 2010, 01:54 AM

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?

**tonio** · January 8th 2010, 01:56 PM

Originally Posted by Sampras

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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