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 .

Thanks in advance for help!


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 $\Bbb Z_4$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.