Conjugate of $\sqrt{a+\sqrt{b}}$

How can we compute the conjugate of $$\frac{1}{\sqrt{a+\sqrt{b}}}$$ ? I know that the conjugate of $$(a+\sqrt{b})$$ is $$(a-\sqrt{b})$$ but what about the global square ? For example how do we go from : $$\frac{\sqrt{2}}{2 \sqrt{2+\sqrt{2}}}$$ to $$\frac{\sqrt{2-\sqrt{2}}}{2}$$?

• See wikipedia. Commented Aug 21, 2021 at 18:40

Upon multiplying $$\sqrt{a+\sqrt{b}}$$ by itself you get the term that you know the conjugate of (i.e. the term $$a+\sqrt{b}$$. Now multiply again by $$(a-\sqrt{b})$$. Two multiplications were finally involved: $$(\sqrt{a+\sqrt{b}})(a-\sqrt{b})$$

• Thanks for your help. When doing this, I cant continue withe final multiplications for exemple : $(2-\sqrt{2} )(\sqrt{2+\sqrt{2}})$ Commented Aug 21, 2021 at 20:54
• You can keep it "as is" @ZchGarinch Commented Aug 22, 2021 at 13:30
• Ok but how can I reduce the nested radical here ? Commented Aug 22, 2021 at 21:49

The conjugates of $$\sqrt{a+\sqrt b}$$ are calculated as follows, for generic $$a,b\in\mathbb Q$$. The minimal polynomial of $$\sqrt{a+\sqrt b}$$ is $$(x^2-a)^2-b=0$$, by inspection. The other roots of this equation are $$\pm\sqrt{a\pm\sqrt b}$$.

Note that this is for generic $$a,b\in\mathbb Q$$. For instance, when $$a=3$$, $$b=8$$, we have $$\sqrt{a+\sqrt b}=1+\sqrt2$$, which only has a single conjugate $$1-\sqrt2$$.

P.S. The calculation above also shows that the Galois group is the order $$8$$ group $$\mathrm{Gal}\big(\mathbb Q(\sqrt{a\pm\sqrt b})/\mathbb Q\big)\cong D_8$$. This is because $$\mathbb Q(\sqrt{a\pm\sqrt b})/\mathbb Q(\sqrt b)$$ is a Kummer extension, with Galois group $$C_2\times C_2$$, and $$\mathbb Q(\sqrt b)/\mathbb Q$$ has Galois group $$C_2$$. Thus the Galois group $$G$$ splits as follows: $$1\to C_2\to G\to C_2\times C_2\to 1$$.

It is generated by $$\sigma\colon \sqrt{a+\sqrt b}\mapsto\sqrt{a+\sqrt b},\sqrt{a-\sqrt b}\mapsto-\sqrt{a-\sqrt b}$$ of order $$2$$ and $$\tau\colon\sqrt{a+\sqrt b}\mapsto-\sqrt{a-\sqrt b},\sqrt{a-\sqrt b}\mapsto\sqrt{a+\sqrt b}$$, which has order $$4$$.

• The parenthetical remark about the Galois group is not correct; the generic Galois group of this polynomial is in fact $D_8 \subset S_4$. Commented Aug 21, 2021 at 18:54
• @user952367 oh, my bad. Right, $\mathbb Q(\sqrt{a+\sqrt b},\sqrt{a-\sqrt b})/\mathbb Q$ has degree $8$ generically. Commented Aug 21, 2021 at 18:57