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This question already has an answer here:

Determine the Galois group of $X^4 - 2X^2 +2$ over $\Bbb Q$

EDIT : I want to address the particular polynomial and not the full general case. In order to solve this problem, I want to have a self-contained proof/answer and am not willing to reproduce the general argument when even it might not be required. If you can provide help overcoming this simple obstacle then please help, instead of just lazily labeling the question as duplicate !

The result that is being linked\suggested, it boils down to the same problem whether $\sqrt 2$ is in the field $\Bbb Q(\alpha)$ or not, so it's pointless to refer the link just by merely looking at the form of the polynomial!

Mt attempt :

Obtained roots of $X^4 - 2X^2 +2$ to be : $\pm \frac{1}{\sqrt 2}(\sqrt{1+\sqrt 2} \pm i \sqrt{\sqrt 2 -1})$ . Then considered $\Bbb Q(\alpha)$ where $\alpha=\frac{1}{\sqrt 2}(\sqrt{1+\sqrt 2} + i \sqrt{\sqrt 2 -1})$ .

So I need to show whether $\frac{1}{\sqrt 2}(\sqrt{1+\sqrt 2} - i \sqrt{\sqrt 2 -1}) \in \Bbb Q(\alpha)$ or not (then the Galois group will be of order 4 or 8 respectively) .

I have observed that $\sqrt 2 \in \Bbb Q(\alpha) \implies $ $\frac{1}{\sqrt 2}(\sqrt{1+\sqrt 2} - i \sqrt{\sqrt 2 -1}) \in \Bbb Q(\alpha)$

So, (1) How to show (if it is true) $\sqrt 2 \in \Bbb Q(\alpha)$ ?

(2) How to obtain the Galois group ? If (1) is true I was tempting to use computational tricks like the ones exhibited in this answer

Thanks in Advance for help!

A question has been asked on this polynomial before, but the OP was confused himself , I could not follow his arguments and most importantly the question did not had any answer.

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marked as duplicate by Dietrich Burde, dantopa, Parcly Taxel, YiFan, Eevee Trainer Mar 21 at 5:26

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Possible duplicate of Computing the Galois group of $x^4+ax^2+b \in \mathbb{Q}[x]$ $\endgroup$ – Dietrich Burde Mar 20 at 15:47
  • $\begingroup$ @DietrichBurde the result or whatever you are suggesting, it boils down to the same problem whether $\sqrt 2$ is in the field $\Bbb Q(\alpha)$ or not, so it's pointless to refer the link just by merely looking at the form of the polynomial! $\endgroup$ – reflexive Mar 20 at 15:56
  • $\begingroup$ The duplicates enable you to find the Galois group quickly. Isn't this what you want? And there are so many good and detailed answers, you cannot say "the OP was confused hiomself...and the question did not have an answer". $\endgroup$ – Dietrich Burde Mar 20 at 15:58
  • $\begingroup$ You are just using stronger results to prove some simple problems! More importantly, the result you suggested also boils down to the same question whether $\sqrt 2$ is in $\Bbb Q(\alpha)$ or not. So it's pointless to label my question as duplicate just because it has some specific form. I hope you get it now! $\endgroup$ – reflexive Mar 20 at 16:03
  • $\begingroup$ Then I suggest that you change the title and the question (or close this one), and do not ask about the Galois group, which has been answered several times here already, but specifically ask about containment of two fields, i.e., $\Bbb Q(\sqrt{2})\subseteq \Bbb Q(\alpha)$. $\endgroup$ – Dietrich Burde Mar 20 at 16:05
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Let $K=\Bbb Q(\alpha)$ be the field generated by the one root $\alpha\in\Bbb C$ specified in the OP.

Here is a pedestrian proof for $\sqrt 2\not \in K$. Assume for this that we have a tower of fields $\Bbb Q\subset \Bbb Q[\sqrt 2]=k\subset \Bbb Q[\alpha]=K$. (So $K:k$ is an extension of degree two, it is Galois.) Then the polynomial $X^4-2X^2+2$, seen as a polynomial in $k[X]$ splits as $$ X^4-2X^2+2 =\Big(X^2+(a+b\sqrt 2)X+(c+d\sqrt 2)\Big)\cdot\Big(\dots\Big)\ , $$ with $a,b,c,d\in\Bbb Q$.

The factor with the root $\alpha$ (and its $(K:k)$-conjugate) was only shown. (The other factor does not have the root $\alpha$, so it is a different factor.) Consier the above relation as a product of polynomials in $k[X]$. Then we also have its conjugate polynomial as a factor. So from $$ \begin{aligned} X^4-2X^2+2 & =\Big(X^2+(a+b\sqrt 2)X+(c+d\sqrt 2)\Big)\cdot\Big(\dots\Big) \\ & =\Big(X^2+(a-b\sqrt 2)X+(c-d\sqrt 2)\Big)\cdot\overline{\Big(\dots\Big)} \end{aligned} $$ we get (second factor is different, unique factorization) $$ \begin{aligned} X^4-2X^2+2 =&\ \Big(X^2+(a+b\sqrt 2)X+(c+d\sqrt 2)\Big) \\ \cdot&\ \Big(X^2+(a-b\sqrt 2)X+(c-d\sqrt 2)\Big)\ , \end{aligned} $$ then we multiply, look at the term in $X^3$, get $a=0$, then at the term in $X^1$, get $d=0$, so our factorization is $$ X^4-2X^2+2 = \Big(X^2+b\sqrt 2 X + c\Big) \Big(X^2-b\sqrt 2 X + c\Big)\ , $$ so $c^2=2$. Contradiction.

This negates (1) in the OP, so (2) is out of charge.


To have a quick structure description of the Galois closure $L$ of $K=\Bbb Q(\alpha)$, let us start a parallel construction. Let $u=\zeta_8$ be the (cyclotomic) unit with minimal polynomial $$ X^4+1 $$ over $\Bbb Q$, and to fix ideas, we will fix it with the complex image $\frac 1{\sqrt 2}(1+i)$ in the first quadrant. Its conjugates are $u, u^3, u^5, u^7$, with complex images $\frac 1{\sqrt 2}(\pm 1\pm i)$. It is clear that $$ M=\Bbb Q(u)=\Bbb Q(\zeta_8)=\Bbb Q(\sqrt{-1},\sqrt 2) $$ is (the cyclotomic field of order $8$, degree $4$ over $\Bbb Q$,) Galois over $\Bbb Q$ with Galois morphisms $\phi_j$ determined by $u\to\phi_j u=u^j$. So $\phi_5$ corresponds to $u\to u^5=-u$, and $\phi_7$ to the complex conjugation $u\to u^7=\frac 1u=\bar u$, they commute, the structure of $\operatorname {Gal}(M:\Bbb Q)$ is $\Bbb Z/2\times \Bbb Z/2$.

($\sqrt2$ is in $M$: $\zeta_8^2=\zeta_4=i$, so $\Bbb Q[i]$ is a subfield, but then from $\zeta_8=(1+i)/\sqrt 2$ also $\Bbb Q[\sqrt 2]$ is a subfield, and by dimension reasons we have $M= \Bbb Q[\zeta_8]=\Bbb Q[\sqrt{-1},\sqrt 2]$. Then $\phi_5$ maps $i=u^2\to (-u)^2=u^2=i$, and thus $\sqrt 2=(1+i)/\zeta_8\to -\sqrt 2$. And $\phi_7$, the complex conjugation, does the other job, $i\to -i$, $\sqrt 2\to \sqrt 2$.)

We have now the "hint", that may be useful below, $$ X^4-2X^2+2 = (X^2-1)^2-(\sqrt{-1})^2=(X^2-(1+u^2))(X^2-(1-u^2))\ . $$ So we consider $N=M(a)=\Bbb Q(u,a)$, the quadratic extension of $M$ that introduces $a=\alpha=\sqrt{1+u^2}=\sqrt{1+\sqrt{-1}}$ as a root of the first factor. All roots of $X^4-2X^2+2$ are then $$ a,\ -a,\ u^3a,\ -u^3a\ . $$ We have a Galois substitution $\Psi:N\to N$ (over $M$, thus also over $\Bbb Q$), determined algebraically by $\Psi:(u\to u,\ a\to-a)$, and we lift $\phi_j$ as Galois morphisms $\Phi_j$ of $N$ as follows: $$ \begin{aligned} \Phi_5:&\ u\to\phi_5u=u^5=-u\ ,& (X^2-(1+u^2))&\text{ stays, so we can take } &\ a&\to a\ ,\\ \Phi_7:&\ u\to\phi_7u=u^7=\frac 1u=\bar u\ ,& (X^2-(1+u^2))&\text{ changes, so chose } & a&\to -u^3a\ .\\ &\qquad\text{Recall also:} \\ \Psi:&\ u\to u\ , && & a&\to -a\ . \end{aligned} $$ ($\Phi_7$ is well defined, in the relation $a^2-(1+u^2)=0$ the LHS is moved to $(-u^3a)^2-(1+u^6)=u^6a^2-1-u^6=u^6(a^2-1) -1 = (-i)(1+i-1)-1=0$. (Use $X$ instead of $a$ for a pedant factorization.)

In particular $$ \boxed a \overset{\Phi_7}{\longrightarrow} \boxed {-u^3a} \overset{\Phi_7}{\longrightarrow} - u^{21}\; u^3a= \boxed{-a} \overset{\Phi_7}{\longrightarrow} \boxed {u^3a} \ , $$ so $\Phi_7$ permutes as a rotation the vertices of the square with the vertices marked as $a,-u^3a,-a,u^3 a$ in this cyclic order. Then $\Psi$ is the reflection w.r.t. the "virtual center" of the square, and $\Phi_5$ acts as the transposition of $\pm u^3 a$. (Symmetry w.r.t. the "virtula line" through $a,-a$, which are invariated.)

So $\Phi_7$ generates a Galois subgroup $\cong\Bbb Z/4$ and $\Phi_5$ implements / acts as a "twist" of it, the relation $$ \Phi_5\Phi_7\Phi_5=\Phi_7^{-1} $$ can be tested either geometrically (in the virtual picture) or algebraically on the generators, e.g. $$ \begin{aligned} \left( \frac1u\leftarrow \frac1{-u}\leftarrow -u\leftarrow u\right) &= \left( \frac1u\leftarrow u\right) \ , \\ (u^3 a\leftarrow -u^3a \leftarrow a \leftarrow a) &= (u^3 a\leftarrow a)\ . \end{aligned} $$


It is a good idea to investigate the situation using computer power. Here sage. The following code gives information on the given situation.

sage: R.<X> = PolynomialRing(QQ)
sage: K.<a> = NumberField( X^4 - 2*X^2 + 2 )
sage: L.<s> = K.galois_closure()
sage: K.is_galois()
False
sage: L.degree()
8
sage: RK.<Y> = PolynomialRing(K)
sage: RK
Univariate Polynomial Ring in Y over Number Field in a with defining polynomial X^4 - 2*X^2 + 2
sage: RL.<Z> = PolynomialRing(L)
sage: RL
Univariate Polynomial Ring in Z over Number Field in s with defining polynomial X^8 - 20*X^6 + 104*X^4 - 40*X^2 + 1156
sage: a.minpoly()
x^4 - 2*x^2 + 2
sage: s.minpoly()
x^8 - 20*x^6 + 104*x^4 - 40*x^2 + 1156
sage: factor( Y^2 - 2 )
Y^2 - 2
sage: # so sqrt(2) is not in K
sage: factor( Z^2 - 2 )
(Z - 1/328*s^6 + 15/328*s^4 + 3/82*s^2 - 155/164) * (Z + 1/328*s^6 - 15/328*s^4 - 3/82*s^2 + 155/164)
sage: # so sqrt(2) is in L
sage: G.<g> = L.galois_group()
sage: G.structure_description()
'D4'
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  • $\begingroup$ Without Computer programming, using galois theory can you please give the proof that D4 is the galois group $\endgroup$ – reflexive Mar 20 at 18:30
  • $\begingroup$ This is a good related question, hits the essence of the OP also in my view... i will insert the details for the $D_4$ structure, need time to (make the argument shorter and) type it... $\endgroup$ – dan_fulea Mar 20 at 18:34

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