# Showing that $\mathbb{Q}(\sqrt{2},\sqrt{3},\alpha)$ with $\alpha=\sqrt{(2+\sqrt{2})(3+\sqrt{3})}$ is a Galois extension.

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

i)

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 $\mathbb{Q}$

ii)

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.

iii)

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 $$L$$.

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

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.
• @eraldcoil $x^4-10x^2+1$ is irreducible over the rationals. Nov 30, 2020 at 1:06
• 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 Dec 1, 2020 at 1:04
• @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. Dec 1, 2020 at 1:16