Question

Exhibit the Galois correspondence of intermediate fields and subgroups for the given extension of Q:

(a) $\mathit{Q}\left(\sqrt{d}\right)\mathbf{,}\mathbf{}\mathit{d}\mathbf{\in }\mathit{Q}\mathbf{,}\mathbf{}\mathit{b}\mathit{u}\mathit{t}\mathbf{}\sqrt{\mathbf{d}}\mathbf{\notin }\mathit{Q}$.

(b) Q (w),where w is as in Exercise 3.

Short answer

1. The Galois correspondence is$Ga{l}_{Q}Q\left(\sqrt{d}\right)\cong z_2$
2. $Q\left(w\right)=\left\{1+w:a,b\in Q\right\}$
   

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  Q(d)(a)

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Now, the Galois correspondence of intermediate field and subgroup for the extension Q is given by $Q\left(\sqrt{d}\right)$, where $d\in Q,$but $\sqrt{d}\notin Q$

Now, define $Ga{l}_{Q}Q\left(\sqrt{d}\right)$which has order 2 and it is an isomorphism of $Z_2$

The minimal polynomial of the given Galois group function is $x^2-d$and roots are:

$$\begin{gathered} x^2-d=0 \\ x=\pm \sqrt{d} \end{gathered}$$

.

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

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

And, $\tau \left(\sqrt{d}\right)=\sigma \left(\sqrt{d}\right)=-\sqrt{d}$

Similarly, two step arguments of automorphism $\alpha$and $\beta$has order two in group.

So,

$$\begin{gathered} \tau \circ \tau \left(\sqrt{d}\right)=\tau \left(\tau \left(\sqrt{d}\right)\right) \\ =\tau \left(\sigma \left(\sqrt{d}\right)\right) \\ =\tau \left(-\sqrt{d}\right) \\ =-\tau \left(\sqrt{d}\right) \\ \tau \circ \tau \left(\sqrt{d}\right)=-\left(-\sqrt{d}\right) \\ =\sqrt{d} \\ =i\sqrt{d} \end{gathered}$$

Now,

$\tau \circ \tau =i$

Hence, the given galois group is an isomorphism and also an order $Z_2$means $Ga{l}_{Q}Q\left(\sqrt{d}\right)\cong z_2$.

To find the Galois correspondence of Q (w) .(b)

Define the Galois correspondence of Q (w) , where =$\frac{\left(-1+\sqrt{3i}\right)}{2}$

Now, the minimal polynomial of the given function is:

$$\begin{gathered} x=\frac{\left(-1+\sqrt{3i}\right)}{2} \\ 2x+1=\sqrt{3i} \\ {\left(2x+1\right)}^2=-3 \\ {x}^2+x+1=0 \end{gathered}$$

Since, Q (w)=$\{a+wb:a,b\in Q\}$

So, a=1

b=1

Q (w)=$\{1+w:a,b\in Q\}$

Thus, the Galois correspondence of the minimal polynomial $x^2+x+1=0$is $Q\left(w\right)=\left\{1+w:a,b\in Q\right\}.$