Question

(a)Find.$\mathit{G}\mathit{a}\mathit{l}{}_{\mathbf{\mathbb{Q}}}\mathbf{\mathbb{Q}}\mathbf{(}\sqrt{\mathbf{2}}\mathbf{,}\sqrt{\mathbf{3}}\mathbf{)}$

(b) If p,q are distinct positive primes, find $\mathit{G}\mathit{a}\mathit{l}{}_{\mathbf{\mathbb{Q}}}\mathbf{\mathbb{Q}}\mathbf{(}\sqrt{\mathit{p}}\mathbf{,}\sqrt{\mathit{q}}\mathbf{)}$.

Short answer

$$\begin{gathered} \left(a\right) \\ \left[i\left(\sqrt{3},\sqrt{3}\right),\tau \left(\sqrt{3},-\sqrt{3}\right),\alpha \left(\sqrt{3},\sqrt{3}\right),\beta \left(\sqrt{3},-\sqrt{3}\right)\right] \\ \left[i\left(\sqrt{2},\sqrt{2}\right),\tau \left(\sqrt{2},-\sqrt{2}\right),\alpha \left(\sqrt{2},\sqrt{2}\right),\beta \left(\sqrt{2},-\sqrt{2}\right)\right] \\ \left(b\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathrm{\mathbb{Q}}\left(\left(p^{\frac{1}{2}},i\right),\left(q^{\frac{1}{2}},i\right)\right) \end{gathered}$$

Step-by-step solution

Galois group of k over F.

The set of allF automorphisms ofK is an extension of F then Galois group is denoted by $Ga{l}_{F}K$. It is the Galois group of K over F.

Finding the value of Galℚℚ(2,3).

Consider the Galois group of $\mathrm{\mathbb{Q}}(\sqrt{2},\sqrt{3})$over the $\mathrm{\mathbb{Q}}\mathrm{\mathbb{Q}}$. There are two minimal polynomial equations are${\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathrm{and}{\mathbf{x}}^{\mathbf{3}}\mathbf{-}\mathbf{3}$and , It takes a roots$\sqrt{\mathbf{2}}$to $\sqrt{\mathbf{2}}$or$-\sqrt{\mathbf{2}}$and$\sqrt{\mathbf{3}}$to $\sqrt{\mathbf{3}}$or$-\sqrt{\mathbf{3}}$.

Since, there are at most four automorphism in givenlocalid="1658117074322" $Gal{}_{\mathrm{\mathbb{Q}}}\mathrm{\mathbb{Q}}(\sqrt{2},\sqrt{3})$over$\mathrm{\mathbb{Q}}$.

So, the possibilities are:

localid="1657974268358" $\sqrt{\mathbf{2}}\stackrel{\mathbf{i}}{\mathbf{\to }}\sqrt{\mathbf{2}}\mathbf{,}\sqrt{\mathbf{2}}\stackrel{\mathbf{}}{\stackrel{\mathbf{\tau }}{\mathbf{\to }}\mathbf{-}\sqrt{\mathbf{2}}\mathbf{,}\sqrt{\mathbf{2}}\stackrel{\mathbf{}}{\stackrel{\mathbf{\alpha }}{\mathbf{\to }}\sqrt{\mathbf{2}}\mathbf{,}\sqrt{\mathbf{2}}\stackrel{\mathbf{}}{\stackrel{\mathbf{\beta }}{\mathbf{\to }\mathbf{-}\sqrt{\mathbf{2}}}}}}$

Same as:

$\sqrt{\mathbf{3}}\stackrel{\mathbf{i}}{\mathbf{\to }}\sqrt{\mathbf{3}}\mathbf{,}\sqrt{\mathbf{3}}\stackrel{\mathbf{}}{\stackrel{\mathbf{\tau }}{\mathbf{\to }}\mathbf{-}\sqrt{\mathbf{3}}\mathbf{,}\sqrt{\mathbf{3}}\stackrel{\mathbf{}}{\stackrel{\mathbf{\alpha }}{\mathbf{\to }}\sqrt{\mathbf{3}}\mathbf{,}\sqrt{\mathbf{3}}\stackrel{\mathbf{}}{\stackrel{\mathbf{\beta }}{\mathbf{\to }\mathbf{-}\sqrt{\mathbf{3}}}}}}$

Hence, the given Galois group $Gal{}_{\mathrm{\mathbb{Q}}}\mathrm{\mathbb{Q}}(\sqrt{2},\sqrt{3})$has a solution is:

$$\begin{gathered} \left[i\left(\sqrt{3},\sqrt{3}\right),\tau \left(\sqrt{3},\sqrt{3}\right),\alpha \left(\sqrt{3},\sqrt{3}\right),\beta \left(\sqrt{3},\sqrt{3}\right)\right] \\ \left[i\left(\sqrt{2},\sqrt{2}\right),\tau \left(\sqrt{2},\sqrt{2}\right),\alpha \left(\sqrt{2},\sqrt{2}\right),\beta \left(\sqrt{2},\sqrt{2}\right)\right] \end{gathered}$$

Finding the value of .

Given elements$\mathit{p}\mathbf{,}\mathit{q}$ are distinct positive primes. So, Galois group of the element is $Gal{}_{\mathrm{\mathbb{Q}}}\mathrm{\mathbb{Q}}(\sqrt{2},\sqrt{3})$.

If $\mathit{p}\mathbf{,}\mathit{q}$are distinct prime number. So, it is not divisible any number or it is not a square of any number. So, Galois of the given prime elements are consider the complex root of $\mathit{p}\mathbf{,}\mathit{q}$under the multiplication of 2.

So,

role="math" localid="1658117644032" $Gal{}_{\mathrm{\mathbb{Q}}}\mathrm{\mathbb{Q}}(\sqrt{p},\sqrt{q})=\mathrm{\mathbb{Q}}\left(\left(p^{\frac{1}{2},i}\right),\left(p^{\frac{1}{2},i}\right)\right)$

Now, it has a root${\mathit{p}}^{\frac{\mathbf{1}}{\mathbf{2}}}\mathbf{,}{\mathit{p}}^{\frac{\mathbf{1}}{\mathbf{2}}}\mathbf{,}\mathit{p}\mathbf{}{\mathit{i}}^{\frac{\mathbf{1}}{\mathbf{2}}}\mathbf{,}\mathbf{-}{\mathit{p}}^{\frac{\mathbf{1}}{\mathbf{2}}}\mathit{i}\mathbf{,}\mathrm{and}{\mathbf{q}}^{\frac{\mathbf{1}}{\mathbf{2}}}\mathbf{,}{\mathbf{q}}^{\frac{\mathbf{1}}{\mathbf{2}}}\mathbf{,}\mathbf{q}\mathbf{}{\mathbf{i}}^{\frac{\mathbf{1}}{\mathbf{2}}}\mathbf{,}\mathbf{-}{\mathbf{q}}^{\frac{\mathbf{1}}{\mathbf{2}}}\mathbf{i}\mathbf{,}$ and .

Hence, the Galois group is$\mathrm{\mathbb{Q}}\left(\left(p^{\frac{1}{2},i}\right),\left(p^{\frac{1}{2},i}\right)\right)$ .