Galois theory

**peteryellow** · December 28th 2008, 11:44 AM

Please help me with the following question.

Let $f(x)$ be n-th degree irreducible polynomial in $\mathbb Q[x].$ And let $a_i$ for $i\in [1,n]$ be the roots of $f(x).$ and for each $a_i$ let $d_i(f)$ be the roots of $f(x)$ in $\mathbb Q(a_i)$. Then show that $d_1(f)=...=d_n(f).$

Now let $d(f):=d_1(f)=...=d_n(f).$ Then show that $d(f)$ is the number of automorphisms for the field $\mathbb Q(a)$ where $a$ is a root of $f(x)$.

**ThePerfectHacker** · December 28th 2008, 12:03 PM

Originally Posted by peteryellow

Let $f(x)$ be n-th degree irreducible polynomial in $\mathbb Q[x].$ And let $a_i$ for $i\in [1,n]$ be the roots of $f(x).$ and for each $a_i$ let $d_i(f)$ be the roots of $f(x)$ in $\mathbb Q(a_i)$. Then show that $d_1(f)=...=d_n(f).$

I think you want to say $d_i(f)$ is the number of zeros of $f(x)$ in $\mathbb{Q}(a_i)$ .
Is this what you want to show?

Because otherwise let $f(x) = x^4 - 2$ with $a_i = \zeta^ i \sqrt[4]{2}$ and consider $d_4 (f)$ and $d_1 (f)$ . We have $d_4(f) = \{ \sqrt[4]{2} \}$ and $\{ \zeta \sqrt[4]{2} \} \subseteq d_1 (f)$ which means $d_4(f) \not = d_1(f)$ .

**peteryellow** · December 28th 2008, 12:08 PM

yes it is correct.

**ThePerfectHacker** · December 28th 2008, 12:54 PM

Since $f(x)$ is irreducible it means we can define an isomorphism $\phi_i: \mathbb{Q}(\alpha_i) \to \mathbb{Q}[x]/\left< f(x) \right>$ by $\alpha_i \mapsto x + \left< f(x) \right>$ and $\phi (a) = a$ for $a\in \mathbb{Q}$ . Now for $\alpha_i,\alpha_j$ it means $\phi^{-1}_j\circ\phi_i : \mathbb{Q}(\alpha_i)\to \mathbb{Q}(\alpha_j)$ . Let $\theta_{ij} = \phi^{-1}_j \circ \phi_i$ , from here we see that $\theta_{ij} : \mathbb{Q}(\alpha_i)\to \mathbb{Q}(\alpha_j)$ is an isomorphism with $\theta_{ij} (\alpha_i) = \alpha_j$ and $\theta_{ij}(a) = a$ for $a\in \mathbb{Q}$ .

Consider $\text{Gal}(\mathbb{Q}(\alpha_i)/\mathbb{Q})$ . The order of this Galois group is equal to the number of roots of $f(x)$ in $\mathbb{Q}(\alpha_i)$ i.e. $d_i (f)$ . Therefore, to show that $d_i (f) = d_j(f)$ it sufficies to prove that $|\text{Gal}(\mathbb{Q}(\alpha_i)/\mathbb{Q}) | = |\text{Gal}(\mathbb{Q}(\alpha_j)/\mathbb{Q}|$ . To prove this notice that if $\sigma \in \text{Gal}(\mathbb{Q}(\alpha_i)/\mathbb{Q})$ then $\theta_{ij}\circ\sigma\circ\theta_{ji} \in \text{Gal}(\mathbb{Q}(\alpha_j)/\mathbb{Q})$ . This means, $d_i (f) = | \text{Gal}(\mathbb{Q}(\alpha_i)/\mathbb{Q})| \leq |\text{Gal}(\mathbb{Q}(\alpha_j)/\mathbb{Q})| = d_j(f)$ . By a mirror argument $d_j(f) \leq d_i(f)$ . It follows that $d_i(f) = d_j(f)$ .

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