# Computing Galois Group of $\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}$

Galois Group of $\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}$ is cumbersome to computing. Is easy to find the all four possible candidates but is cumberstone to show that they are automorphisms:

For multiplication is too long showing that

$f((a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})(a'+b'\sqrt{2}+c'\sqrt{3}+d'\sqrt{6}))=f(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})f(a'+b'\sqrt{2}+c'\sqrt{3}+d'\sqrt{6})$

For some non trivial candidate since the left hand side expands in $16$ terms. What about $\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5}):\mathbb{Q}$? there are $7$ non trivial candidates and multiplication expands in $64$ terms!

How to do it in a tricker form?

Edit(Let me be more precise in my question):

I know that if $f\in\Gamma(\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q})$, then $f(\sqrt{2})=\pm\sqrt{2}$, $f(\sqrt{3})=\pm\sqrt{3}$ so there are $4$ candidates for $\mathbb{Q}$-Automorphisms. Candidates are given by:

$f_1(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})=a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}$

$f_2(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})=a-b\sqrt{2}+c\sqrt{3}-d\sqrt{6}$

$f_3(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})=a+b\sqrt{2}-c\sqrt{3}-d\sqrt{6}$

$f_4(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})=a-b\sqrt{2}-c\sqrt{3}+d\sqrt{6}$

But I need to check that they are $\mathbb{Q}$-Automorphisms in $\mathbb Q(\sqrt{2},\sqrt{3})$. All candidates $f_i$ fixes $\mathbb{Q}$ and is obvious that all preserves sums, $f_1$ obviously preserves product also since is the identity, but is cumberstone to show that $f_i(xy)=f_i(x)_if(y)$ to conclude $f_i\in\Gamma(\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q})$.

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Why would you show that they are automorphisms? The Galois group consists of precisely those automorphisms of the splitting field which leave the base field invariant. Thus by hypothesis your $f$ is an automorphism. However, does it leave the rationals invariant? That's what you need to determine. (Hint: There are only four, and in the second example there are six.) –  Erik Vesterlund Jul 29 '13 at 19:04
@ErikVesterlund: Um, try 8 :) –  Ted Shifrin Jul 29 '13 at 19:17
@Erik I think that there are $8$ in the second example. –  Gastón Burrull Jul 29 '13 at 19:17
Woops, you're right indeed. –  Erik Vesterlund Jul 29 '13 at 19:30

To reiterate a point in @ErikVesterlund's comment, as well as Andrea Mori's and Ayman Hourieh's answers: there is no need to "directly" verify that the indicated maps are $\mathbb Q$ homomorphisms, because we have already proven this indirectly, by invoking the main theorem of Galois theory, in effect. That is, we show that the degree of the extension really is what it appears to be. Then, observe that any Galois automorphism must permute roots of irreducibles over the base, so _at_most_ $\sqrt{2}\rightarrow \pm \sqrt{2}$, and similarly for any other square roots. Thus, in your first example, there are _at_most_ these four. But/and Galois theory says there _at_least_ four, so these four must be "it". Similarly in the second example, after showing that the field extension really is of degree $2^3$, again we see that there are _at_most_ the $8$ automorphisms that flip the signs of the $3$ square roots, and Galois theory say there are _at_least_ $8$, so those must be "it".

No explicit verification that multiplication is preserved is needed!

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Thanks! Your clear explanation helped me a lot! –  Gastón Burrull Jul 29 '13 at 19:45
Good!........ :) –  paul garrett Jul 29 '13 at 19:50

HINT:

Since $\Bbb Q(\sqrt2,\sqrt 3)$ is generated over $\Bbb Q$ by $\sqrt 2$ and $\sqrt 3$, any automorphism will be determined by its action on these two elements.

Let $\sigma$ be such an automorphism. What can $\sigma(\sqrt 2)$ be? Does the fact that $\sqrt 2$ is a root of $X^2-2$ help?

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I don't know how your answer helps, I know the four candidates using your hint. I don't know how to prove that they are homomorphisms. –  Gastón Burrull Jul 29 '13 at 19:12

$\mathbb{Q}(\sqrt 2, \sqrt 3)$ is the splitting field of $(x^2 - 2)(x^2 - 3)$ over $\mathbb{Q}$. Hence, the extension $\mathbb{Q}(\sqrt 2, \sqrt 3) / \mathbb{Q}$ is Galois. Since $\sqrt 2$ and $\sqrt 3$ are not linearly dependent, we have $[\mathbb{Q}(\sqrt 2, \sqrt 3) : \mathbb{Q}] = 4$. Thus, the Galois group $G$ has order $4$.

We know that automorphisms in $G$ permute the roots of $x^2 - 2$ and $x^2 - 3$. It's easy to verify that $$\sigma = \begin{cases} \sqrt 2 \mapsto -\sqrt 2 \\ \sqrt 3 \mapsto \sqrt 3 \end{cases}$$ and $$\tau = \begin{cases} \sqrt 2 \mapsto \sqrt 2 \\ \sqrt 3 \mapsto -\sqrt 3 \end{cases}$$

are members of $G$. Thus, $G = \{1, \sigma, \tau, \sigma \tau\}$. I'll let you find its isomorphism type.

For $\mathbb{Q}(\sqrt 2, \sqrt 3, \sqrt 5)$, follow the same method, as it's the splitting field of $(x^2 - 2)(x^2 - 3)(x^2 - 5)$. Consider the automorphisms: $$\begin{cases} \sqrt 2 \mapsto -\sqrt 2 \\ \sqrt 3 \mapsto \sqrt 3 \\ \sqrt 5 \mapsto \sqrt 5 \end{cases}$$ and $$\begin{cases} \sqrt 2 \mapsto \sqrt 2 \\ \sqrt 3 \mapsto -\sqrt 3 \\ \sqrt 5 \mapsto \sqrt 5 \end{cases}$$ and $$\begin{cases} \sqrt 2 \mapsto \sqrt 2 \\ \sqrt 3 \mapsto \sqrt 3 \\ \sqrt 5 \mapsto -\sqrt 5 \end{cases}.$$

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I think is cumberstone proving that $\sigma$ is an $\mathbb{Q}$-Automorphism (I know that is really easy but is too long proving that is homomorphism) –  Gastón Burrull Jul 29 '13 at 19:21
First you compute $[\mathbb Q(\sqrt{2},\sqrt{3}) : \mathbb Q] = 4$, and you show that the extension is Galois, like you are told in the other answers.
Now you know that the Galois group $G$ has size $4$. We know that elements of Galois groups preserve the minimal polynomials over the base field. So for any $\sigma\in G$, it holds that $\sigma(\sqrt{2})\in\{\pm\sqrt{2}\}$ and $\sigma(\sqrt{3})\in\{\pm\sqrt{3}\}$. Furthermore, $\sigma$ is uniquely determined by $\sigma(\sqrt{2})$ and $\sigma(\sqrt{3})$, since $\sqrt{2}$ and $\sqrt{3}$ generate $\mathbb Q(\sqrt{2},\sqrt{3})$ over $\mathbb Q$. Since $G$ has size $4$, and there are only $4$ combinations of the possible images $\sigma(\sqrt{2})$ and $\sigma(\sqrt{3})$, all these possibilities must indeed occur in $G$. So you have found the elements of $G$ without explicitly checking them for the homomorphism property.
For $\mathbb Q(\sqrt{2},\sqrt{3},\sqrt{5})$, a similar argument applies.