# Determine the minimal polynomial of $\sqrt{2+\sqrt{2}}$ over $\Bbb Q$ and find its Galois group over $\Bbb Q$

Determine the minimal polynomial of $$\sqrt{2+\sqrt{2}}$$ over $$\Bbb Q$$ and find its Galois group over $$\Bbb Q$$.

Computed and obtained $$p(x)=x^4-4x^2+2$$ has $$\sqrt{2+\sqrt{2}}$$ as a root. $$p$$ is clearly monic and irreducible (Eisenstein), thus $$p$$ is the required minimal polynomial.

$$p$$ has roots : $$\pm \sqrt{2\pm \sqrt{2}}$$ . Since $$p$$ has no multiple root in its splitting field say, $$K$$ , thus $$K|\Bbb Q$$ is Galois.

$$|Gal(K| \Bbb Q)|= [K:\Bbb Q]=4$$

I don't seem to find any intermediate extension $$\Bbb Q \subset M \subset K$$ s.t. $$[K:M]=2$$ other than $$\Bbb Q(\sqrt{2})$$ . In that case, $$Gal(K| \Bbb Q)=\Bbb Z_4$$ . Here my argument is shaky .

Set $$a=\sqrt{2+\sqrt2}$$ and $$b=\sqrt{2-\sqrt2}$$. Clearly $$\Bbb Q(\sqrt2)$$ is fixed under $$a\mapsto -a, b\mapsto -b$$, as $$\sqrt2=a^2-2=2-b^2=ab$$ and this permutation has order $$2$$.

Hint: What about sending $$a$$ to $$b$$? Where does $$b$$ go? Which order does this element of the Galois group have? What is the fixed subfield?

More details: Note that $$a^2-2-ab=0$$. Any automorphism $$f$$ of $$K$$ must respect this, which is to say $$f(a)^2-2-f(a)f(b)=0$$ So, if $$f(a)=b$$, what do we get? $$b^2-2-bf(b)=0\\ -(2-b^2)-bf(b)=0\\ -\sqrt2=bf(b)$$ Among $$\pm a, \pm b$$, only $$-a$$ has this property. Which means that if $$f(a)=b$$, we must have $$f(b)=-a$$, meaning $$f$$ has order $$4$$. This means the Galois group must be $$\Bbb Z_4$$.

• You set $a$ and $b$ to the same thing. Is that right? – TonyK Mar 17 at 10:23
• @TonyK No, it isn't. Thanks. – Arthur Mar 17 at 10:24
• Can you kindly give a detailed answer? I am totally new to Galois theory – reflexive Mar 17 at 10:31
• Yeah got it, $\sigma(a)=b$, has order > 2, we are done! – reflexive Mar 17 at 10:44