Question

4. Consider the field i) Compute the Galois group Aut(L/Q). Explicitly specify each automorphisa σ E Aut(L/Q) in terms of the permutation of induced by σ. (ii) Compute L: Q (without using the Fundamental Theorem of Galois Theory) Include justifications for your assertions in (i) n

Answer 1

Part-(I):

The Galois Group $Aut(L/\Bbb Q)$ consists of all those auto-morphism which fixes elements of $\Bbb Q$ .

Also if $\alpha\in Aut(L/\Bbb Q)$ then $\alpha(\sqrt 2)=\pm \sqrt 2$ ,

$\alpha(\sqrt 5)=\pm \sqrt 5\\ \alpha(\sqrt 7)=\pm \sqrt 7$

and $\alpha(x)=x \forall x\in \Bbb Q$ .

So the possible choices of $\alpha(x)$ are the following:

(i):

$\alpha(\sqrt 2)=\sqrt 2,\alpha (\sqrt 5)=\sqrt 5,\alpha(\sqrt 7)= \sqrt 7\\ \alpha(x)=x \forall x\in \Bbb Q$

(ii):

$\alpha(\sqrt 2)=-\sqrt 2,\alpha (\sqrt 5)=\sqrt 5,\alpha(\sqrt 7)= \sqrt 7\\ \alpha(x)=x \forall x\in \Bbb Q$

(iii):

$\alpha(\sqrt 2)=\sqrt 2,\alpha (\sqrt 5)=- \sqrt 5,\alpha(\sqrt 7)= \sqrt 7\\ \alpha(x)=x \forall x\in \Bbb Q$

(iv):

$\alpha(\sqrt 2)=\sqrt 2,\alpha (\sqrt 5)=\sqrt 5,\alpha(\sqrt 7)=- \sqrt 7\\ \alpha(x)=x \forall x\in \Bbb Q$

(v):

$\alpha(\sqrt 2)=\sqrt 2,\alpha (\sqrt 5)=- \sqrt 5,\alpha(\sqrt 7)=- \sqrt 7\\ \alpha(x)=x \forall x\in \Bbb Q$

(vi):

$\alpha(\sqrt 2)=- \sqrt 2,\alpha (\sqrt 5)= \sqrt 5,\alpha(\sqrt 7)=- \sqrt 7\\ \alpha(x)=x \forall x\in \Bbb Q$

(vii):

$\alpha(\sqrt 2)=- \sqrt 2,\alpha (\sqrt 5)=- \sqrt 5,\alpha(\sqrt 7)= \sqrt 7\\ \alpha(x)=x \forall x\in \Bbb Q$

and

(viii):

$\alpha(\sqrt 2)=- \sqrt 2,\alpha (\sqrt 5)=- \sqrt 5,\alpha(\sqrt 7)=- \sqrt 7\\ \alpha(x)=x \forall x\in \Bbb Q$

Thus

$Aut(L/\Bbb Q)$ has 8 elements as listed above.

Part-ii:

Note that :

The minimal polynomial of $\sqrt 7$ over $\Bbb Q(\sqrt 2,\sqrt 5)$    is $x^2-7$ and has degree 2.

Thus

$[\Bbb Q(\sqrt 2,\sqrt 5,\sqrt 7):\Bbb Q(\sqrt 2,\sqrt 5)]=2$

Similarly

The minimal polynomial of $\sqrt 2$ over $\Bbb Q(\sqrt 2,\sqrt 5)$    is $x^2-2$ and has degree 2.

Thus

$[\Bbb Q(\sqrt 2,\sqrt 5):\Bbb (\sqrt 5)]=2$

Thus we have

$[\Bbb Q(\sqrt 2,\sqrt 5,\sqrt 7):\Bbb Q]\\ =[\Bbb Q(\sqrt 2,\sqrt 5,\sqrt 7):\Bbb Q(\sqrt 2,\sqrt 5)][ \Bbb Q(\sqrt 2,\sqrt 5):\Bbb Q(\sqrt 5)][\Bbb Q(\sqrt 5):\Bbb Q]\\ =2\times 2\times 2\\ =8$