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!


1 Answer 1


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, 2019 at 10:23
  • $\begingroup$ @TonyK No, it isn't. Thanks. $\endgroup$
    – Arthur
    Mar 17, 2019 at 10:24
  • $\begingroup$ Can you kindly give a detailed answer? I am totally new to Galois theory $\endgroup$
    – user422112
    Mar 17, 2019 at 10:31
  • $\begingroup$ Yeah got it, $\sigma(a)=b$, has order > 2, we are done! $\endgroup$
    – user422112
    Mar 17, 2019 at 10:44

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