Question

Question: Exhibit the Galois correspondence for the given extension of Q:

(a)$Q\left(\sqrt{2},\sqrt{3}\right)$

(b)$Q\left(i,\sqrt{2}\right)$

Short answer

(a)

$$\begin{gathered} \left[Q\left(\sqrt{2}\right)\left(\sqrt{3}\right):Q\right]=\left[Q\left(\sqrt{2}\right)\left(\sqrt{3}\right):Q\left(\sqrt{2}\right)\right]\left[Q\left(\sqrt{2}\right),\left(\sqrt{3}\right):Q\right] \\ =2\times 2 \\ =4 \end{gathered}$$

(b)

$$\begin{gathered} \left[Q\left(i,\sqrt{2}\right):Q\right]=\left[Q\left(i\right)\left(\sqrt{2}\right):Q\left(i\right)\right]\left[Q\left(i,\sqrt{2}\right):Q\right] \\ =2\times 2 \\ =4 \end{gathered}$$

Step-by-step solution

Automorphism:

The Galols group is denoted by Gal_FK when the set of all F automorphism of K is an extension of F.

It is the Galols group of K over F.

To find Galois correspondence of  (a)

It is given that extension of Q is $Q\left(\sqrt{2},\sqrt{3}\right)$

The two minimal polynomial of the given Galois group function are $x^2-2$and $x^2-3$. It takes a root$\sqrt{2}to\sqrt{2}or-\sqrt{2}andsimilarly\sqrt{3}to\sqrt{3}or-\sqrt{3}$

There are atmost 4 automorphism in given $Ga{l}_{Q}Q\left(\sqrt{2},\sqrt{3}\right)$over Q.

So, possibilities are:

${\sqrt{2}}^i\to \sqrt{2},{\sqrt{2}}^r\to -\sqrt{2},{\sqrt{2}}^{\alpha }\to \sqrt{2},{\sqrt{2}}^{\beta }\to -\sqrt{2}$

Same for${\sqrt{3}}^i\to \sqrt{3},{\sqrt{3}}^r\to \sqrt{3},{\sqrt{3}}^{\alpha }\to \sqrt{3},{\sqrt{3}}^{\beta }\to -\sqrt{3}$

Hence, the given Galois group has a solution

$i\left(\sqrt{3},\sqrt{3}\right),\tau \left(\sqrt{3},-\sqrt{3}\right),\alpha \left(\sqrt{3},\sqrt{3}\right),\beta \left(\sqrt{3},\sqrt{3}\right)$

And$i\left(\sqrt{2},\sqrt{2}\right),\tau \left(\sqrt{2},-\sqrt{2}\right),\alpha \left(\sqrt{2},\sqrt{2}\right),\beta \left(\sqrt{2},-\sqrt{2}\right)$$i\left(\sqrt{2},\sqrt{2}\right),\tau \left(\sqrt{2},-\sqrt{2}\right),\alpha \left(\sqrt{2},\sqrt{2}\right),\beta \left(\sqrt{2},-\sqrt{2}\right)$

Now, the sequence of the simple extension is:

$Q\subseteq Q\left(\sqrt{2}\right)\subseteq Q\left(\sqrt{2}\right)\left(\sqrt{3}\right)=Q\left(\sqrt{2},\sqrt{3}\right)$

Now, if$Q\left(\sqrt{2}\right)\left(\sqrt{3}\right):Q\sqrt{2}=2$

Then,

$$\begin{gathered} \left[Q\left(\sqrt{2}\right)\left(\sqrt{3}\right):Q\right]=\left[Q\left(\sqrt{2}\right)\left(\sqrt{3}\right):Q\left(\sqrt{2}\right)\right]\left[Q\left(\sqrt{2}\right),\left(\sqrt{3}\right):Q\right] \\ =2\times 2 \\ =4 \end{gathered}$$

Thus, the given function is algebraic over Q and ordered is 4.

To find the Galois correspondence of Q(i,2)(b)

The extension field of$Ga{l}_{Q}Q\left(i,\sqrt{2}\right)$ can be obtained by the sequence of simple extension:

$Q\subseteq Q\left(\left(i\right)\right)\subseteq Q\left(i\right)\left(\sqrt{2}\right)=Q\left(i,\sqrt{2}\right)$

Now, the minimal polynomial is$x^2+1$

Then,

$$\begin{gathered} \left[Q\left(i,\sqrt{2}\right):Q\right]=\left[Q\left(i\right),\left(\sqrt{2}\right):Q\left(i\right)\right]\left[Q\left(i,\sqrt{2}\right):Q\right] \\ =2\times 2 \\ =4 \end{gathered}$$

Thus, the given function is algebraic over Q and ordered is 4.