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 mona123 November 17th, 2017 12:44 PM

Non trivial subgroup

Let $\alpha=\sqrt{2+\sqrt{3}}$.

Let $L=\mathbb{Q}(\alpha)$.

>For each non trivial $H< \text{Gal}(L/\mathbb{Q})$ determine $\text{Fix}(H)$. Express the answer in form of $\mathbb{Q}(\beta)$ whith $\beta$ given explicitly in term of $\alpha$

$L$ is isomorphic to $\mathbb{Z}_2 \times \mathbb{Z_2}$
(since its generators are $\alpha \to -\alpha,\alpha\to \frac{1}{\alpha}$).

For computing the fix field of $<\alpha \to -\alpha>$ we need to solve the equation $m+n\alpha+n'\alpha^2+n''\alpha^3=m-n\alpha+n'\alpha^2-n''\alpha^3$. Using the fact that $\{1,\alpha,\alpha^2,\alpha^3\}$ is a base for $\mathbb{Z}(\alpha)$ we have $n=n''=0$

For computing the fix field of $<\alpha \to \frac{1}{\alpha}>$ we need to solve the equation $m+n\alpha+n'\alpha^2+n'' \alpha^3=m+\frac{n}{\alpha}+\frac{n'}{\alpha^2}+ \frac{n''}{\alpha^3}$ samely since $\{1,\alpha,\alpha^2,\alpha^3\}$ is a base for $\mathbb{Z}(\alpha)$, we have $m\alpha^3+n+n'\alpha+n''\alpha^2=m\alpha^3+n \alpha^2+n ' \alpha + n''$
so $n=n''$

 ABVictor November 26th, 2017 02:20 AM

The minimal polynomial of $\displaystyle \alpha$ over $\displaystyle \mathbb{Q}$ is
$\displaystyle m(x)=(x^2-2)^2-3.$

The root of $\displaystyle m(x)$ are $\displaystyle \alpha=\sqrt{2+\sqrt{3}},\ -\sqrt{2+\sqrt{3}},\ \sqrt{2-\sqrt{3}},\ -\sqrt{2-\sqrt{3}}.$

$\displaystyle \sqrt{2+\sqrt{3}}\cdot \sqrt{2-\sqrt{3}} = 1.$ It implies $\displaystyle \sqrt{2-\sqrt{3}}=\frac{1}{\alpha}.$

We get expressions of all the roots in terms of $\displaystyle \alpha$:

$\displaystyle \alpha, -\alpha,\ \frac{1}{\alpha},\ -\frac{1}{\alpha}.$

Galois group $\displaystyle \text{Gal}(L/\mathbb{Q})$ of $\displaystyle L=\mathbb{Q}(\alpha)$ can be expressed as the group of following functions:
$\displaystyle \varphi_1(x)=x,\ \varphi_2(x)=-x,\ \varphi_3(x)=\frac{1}{x},\ \varphi_4(x)=-\frac{1}{x}.$

$\displaystyle \text{Gal}(L/\mathbb{Q})$ is isomorphic to $\displaystyle \mathbb{Z}_2 \times \mathbb{Z_2}$.

Let the fix field of the subgroup generated by $\displaystyle <\alpha \to -\alpha>$ be $\displaystyle \mathbb{Q}(\beta)$ where $\displaystyle \beta\in\mathbb{Q}(\alpha)$:

$\displaystyle \beta = m+n\alpha+n'\alpha^2+n''\alpha^3$.

$\displaystyle m+n\alpha+n'\alpha^2+n''\alpha^3=m-n\alpha+n'\alpha^2-n''\alpha^3.$

$\displaystyle 2n\alpha+2n''\alpha^3=0.$

Using the fact that $\displaystyle \{1,\alpha,\alpha^2,\alpha^3\}$ is a base for $\displaystyle \mathbb{Q}(\alpha)$ (a basis of the linear space $\displaystyle \mathbb{Q}(\alpha)$) we have
$\displaystyle n=n''=0$.

And $\displaystyle \beta = m+n'\alpha^2$:
$\displaystyle \beta = \alpha^2$
Let $\displaystyle m=0, \ n'=1$.
$\displaystyle \mathbb{Q}(\alpha^2)$ is the fix field of$\displaystyle <\alpha \to -\alpha>$.

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