Let $L=\mathbb{Q}(\sqrt{2},\sqrt{3},\alpha)$ with $\alpha=\sqrt{(2+\sqrt{2})(3+\sqrt{3})}$

Show that:

i) $\alpha$ is a primitive element

ii) $L/\mathbb{Q}$ is a Galois extension.

iii) $L/\mathbb{Q}\simeq Q_8$

iv) There is no irreducible polynomial of the form $p(x)=x^4-ax^2+b\in\mathbb{Q}[x]$, which has a root in $L$

Next I put my solution omitting some calculation details and topics such as proving that the extensions obtained by each root are in effect automorphisms.

Let $K:=\mathbb{Q}(\sqrt{2},\sqrt{3})$


Affirmation 0 . $\alpha$ is a primitive element of $L/\mathbb{Q}$

Proof: I did something similar to what is in this topic $\mathbb{Q}(\sqrt2,\sqrt3,\sqrt{(2+\sqrt{2})(3+\sqrt{3})})$ is Galois over $\mathbb{Q}$


Affirmation 1. $\alpha^2$ is not a square in $K$.

Proof. If $\alpha^2=c^2$ some $c\in K$. Let $\varphi\in \text{Aut}(K/\mathbb{Q})$ given by $\varphi(\sqrt{2})=\sqrt{2}$ and $\varphi(\sqrt{3})=-\sqrt{3}$ then $\alpha^2\varphi(\alpha^2)=6(2+\sqrt{2})^2$ therefore $6=\left(\frac{c\varphi{c}}{2+\sqrt{2}}\right)^2$ Because $c\in K$ then $c=a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}$ then $c\varphi(c)=(a+b\sqrt{2})^2-3(c+d\sqrt{2})^2\in\mathbb{Q}(\sqrt{2})$ Therefore $c\varphi(c)\in\mathbb{Q}(\sqrt{2})$ Therefore $6=\left(\frac{c\varphi{c}}{2+\sqrt{2}}\right)^2$ with $\frac{c\varphi{c}}{2+\sqrt{2}}\in \mathbb{Q}(\sqrt{2})$ Therefore $\sqrt{6}\in\mathbb{Q}(\sqrt{2})$ a contradiction.

Therefore $\alpha\not\in K$

Afirmation 2. $[L:\mathbb{Q}]=8$

Indeed, $[L:K][K:\mathbb{Q}]=[L:K][\mathbb{Q}(\sqrt{3})(\sqrt{2}):\mathbb{Q}(\sqrt{3})][\mathbb{Q}(\sqrt{3}):\mathbb{Q}]$ and $[L:K]=2$ with $\text{Irr}_{K,\alpha}(x)=x^2-\alpha^2$, $[\mathbb{Q}(\sqrt{3})(\sqrt{2}):\mathbb{Q}(\sqrt{3})]=2$ with $\text{Irr}_{\mathbb{Q}(\sqrt{3}),\sqrt{2}}(x)=x^2-2$ and $[\mathbb{Q}(\sqrt{3}):\mathbb{Q}]=2$ with $\text{Irr}_{\mathbb{Q},\sqrt{3}}(x)=x^2-3$

Affirmation 3. $|\text{Aut}(L/\mathbb{Q})|=8$.

Indeed, by afirmation 2, $\text{Irr}_{K,\alpha}(x)=0$ implies $x\in \left\{\pm \alpha\right\}$ (in $L$) therefore exists two extension $\alpha\mapsto \pm \alpha$ Analogously, $\text{Irr}_{\mathbb{Q}(\sqrt{3}),\sqrt{2}}(x)=0$ implies $x\in\left\{\pm \sqrt{2}\right\}$ therefore exists two extension $\sqrt{2}\mapsto \pm \sqrt{2}$

$\text{Irr}_{\mathbb{Q},\sqrt{3}}(x)=0$ implies $x\in\left\{\pm \sqrt{3}\right\}$ therefore exists two extension $\sqrt{3}\mapsto \pm\sqrt{3}$

Therefore exists 8 $\mathbb{Q}$-automorphisms, this are: $\alpha\mapsto \pm \sqrt{ (2+\pm\sqrt{2})(3\pm\sqrt{3})}$

Therefore $|\text{Aut}(L/\mathbb{Q})|=8=[L:\mathbb{Q}]$ therefore $L/\mathbb{Q}$ is a Galois extension.

Why $\pm\alpha=\pm\sqrt{ (2+\sqrt{2})(3+\sqrt{3})},\, \pm \beta=\pm\sqrt{ (2-\sqrt{2})(3+\sqrt{3})}$, $ \pm\gamma=\pm\sqrt{ (2+\sqrt{2})(3-\sqrt{3})},\, \pm\delta=\pm\sqrt{ (2-\sqrt{2})(3-\sqrt{3})}$ (conjugates of $\alpha$) are in $L$?

This is because

$\beta=\frac{\alpha \beta}{\alpha}$ and $\alpha\beta=\sqrt{2}(3+\sqrt{3})\in \mathbb{Q}(\sqrt{2},\sqrt{3})\subset L$ then $\alpha\beta\in L$ therefore $\beta\in L$ etc.


Affirmation 4. $\text{Gal}(L/\mathbb{Q})\simeq Q_8$.

Indeed, let $\varphi_{\beta},\varphi_{\gamma}\in \text{Gal}(L/\mathbb{Q})$ given by $\varphi_{\beta}(\alpha)=\beta$ and $\varphi_{\gamma}(\alpha)=\gamma$ with $\beta:=\sqrt{ (2-\sqrt{2})(3+\sqrt{3})},\gamma:=\sqrt{(2+\sqrt{2})(3-\sqrt{3})}$ then $\varphi_{\beta}\circ \varphi_{\gamma}\neq \varphi_{\gamma}\circ \varphi_{\beta}$ therefore $\text{Gal}{L/\mathbb{Q}}$ is non abelian, this implies is not cyclic.

Moreover, $\varphi_{-\alpha}(\alpha):=-\alpha$ is the unique $\mathbb{Q}$-automorphism with order 2.

Because any group not cyclic and with a unique element of order $2$ is isomorphism to $Q_8$, then $\text{Gal}(L/\mathbb{Q})\simeq Q_8$.

To proves that iv) I am somewhat lost. I put below what I know about the minimal polynomial of $\alpha$.

Because $Irr_{L,\alpha}(x)=x-\alpha$ then

$Irr_{\mathbb{Q},\alpha}$ $=\prod_{\sigma\in \text{Gal}(L/\mathbb{Q})}(x-\sigma(\alpha))$ $=(x-\alpha)(x+\alpha)(x-\beta)(x+\beta)(x-\gamma)(x+\gamma)(x-\delta)(x+\delta)$

with $\pm\alpha=\pm\sqrt{ (2+\sqrt{2})(3+\sqrt{3})},\, \pm \beta=\pm\sqrt{ (2-\sqrt{2})(3+\sqrt{3})}$, $ \pm\gamma=\pm\sqrt{ (2+\sqrt{2})(3-\sqrt{3})},\, \pm\delta=\pm\sqrt{ (2-\sqrt{2})(3-\sqrt{3})}$ (conjugates of $\alpha$) (this is a knows results)

By contradiction, if $p(x)=x^4-ax^2+b\in\mathbb{Q}[x]$ has a root in $L$ then I should get a contradiction but I can't think of how.

How can I demonstrate this last part iv)? thank you

Actualization 1. I fix part iv)

iv) If $p(x):=x^4-b$ (irreducible add) with $b\in\mathbb{Q}$ is not a square show that $p(x)$ has not root in $L$.

  • 1
    $\begingroup$ Maybe you can show the Galois group of $p$ is Klein-4, which is not a quotient of the quaternion group. By the way, double-f in "affirmation". $\endgroup$ Nov 30, 2020 at 0:55
  • 1
    $\begingroup$ @GerryMyerson isn't $Q_8/\langle -1\rangle$ Klein-4? $\endgroup$ Nov 30, 2020 at 1:19
  • $\begingroup$ Assuming I show that the Galois group of p (x) is the Klein group. What would this be for? $\endgroup$
    – eraldcoil
    Nov 30, 2020 at 2:47
  • 1
    $\begingroup$ Never mind, @Lukas is right. [But you should still correct the spelling of "affirmation".] $\endgroup$ Nov 30, 2020 at 6:06
  • $\begingroup$ Now i fix part iv) with $p(x)=x^4-b$ and $b$ integers not square. $\endgroup$
    – eraldcoil
    Nov 30, 2020 at 14:03

1 Answer 1


answer to the original question: $\sqrt 2+\sqrt 3 \in L$ has minimal polynomial $x^4-10x^2+1$, so iv) is simply not true.

answer to the revised question: There are two cases: either $x^4-b$ is irreducible over $\Bbb Q$ or $b=-4c^4$ for some $c \in \Bbb Q$, see here. Let us treat the second case first, so assume that $b=-4c^4$, then $x^4-b=x^4+4c^4=(x^2+2cx+2c^2)(x^2-2cx+2c^2)$. This has the four roots $x=(\pm1\pm i)c$. So if $x^4-b$ has root in $L$, then $\Bbb Q(i) \subset L$. But we also have $\Bbb Q(\sqrt{2}) \subset L$ and $\Bbb Q(\sqrt{3}) \subset L$ and $\Bbb Q(\sqrt{6}) \subset L$. By the Galois correspondence, having these four degree 2 subextensions implies that the Galois group $\operatorname{Gal}(L/\Bbb Q)$ has four subgroups of order $4$. But $Q_8$ only has three subgroups of order $4$.

For the second case, we may assume that $x^4-b$ is irreducible. Then because $L$ is a normal extension of $\Bbb Q$, if there is one root of $x^4-b$ in $L$, then all roots are in $L$. This implies that $L$ contains $\frac{i\sqrt[4]{b}i }{\sqrt[4]{b}}=i$, as both $i\sqrt[4]{b}$ and $\sqrt[4]{b}$ are roots of $x^4-b$. This implies that $\Bbb Q(i) \subset L$ which is impossible, as we have already seen.

  • $\begingroup$ but the polynomial must be irreducible $\endgroup$
    – eraldcoil
    Nov 30, 2020 at 1:04
  • 1
    $\begingroup$ @eraldcoil $x^4-10x^2+1$ is irreducible over the rationals. $\endgroup$ Nov 30, 2020 at 1:06
  • $\begingroup$ oh! you got right... $\endgroup$
    – eraldcoil
    Nov 30, 2020 at 1:36
  • $\begingroup$ A query, to understand your argument. Why all roots of $x^4-b$ are in $L$? Is it because of the following? If $x^4-b$ is irreducible and $\xi$ is a root in $L$ of this polynomial then $x^4-b= \text{Irr}_{\mathbb{Q},\xi}(x)$ right?. Besides $\mathbb{Q}(\xi)/\mathbb{Q}$ is the splitting field of $x^4-b$. Therefore all roots of $x^4-b$ are contained in $\mathbb{Q}(\xi)$. Now, $\mathbb{Q}(\xi)\subset L$ therefore all roots of $x^4-b$ are in $L$. This is right? Thanks $\endgroup$
    – eraldcoil
    Dec 1, 2020 at 1:04
  • 1
    $\begingroup$ @eraldcoil okay, there are a couple of equivalent definitions for normal extensions, being the splitting field of a collection of polynomials is just one possible definition. An equivalent definition is that $L/K$ is normal if for any irreducible polynomial over $K$ that has a root in $L$, $L$ has all roots of that polynomial. $\endgroup$ Dec 1, 2020 at 1:16

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