Galois Groups and Minimum Polynomial

In summary, your algebra seems to check out. However, you need to provide more information to verify that you have found the correct minimal polynomial and roots.
  • #1
mathjam0990
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This is what I have done so far. I was wondering if anyone could verify that I found the correct minimum polynomial and roots? If I am incorrect, could someone please help me by explaining how I would find the min polynomial and roots? Thank you.

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  • #2
Your algebra seems to check out.

I would have simplified the working of finding the minimal polynomial of $\alpha$ like so:

$\alpha^3 = \left(\zeta + \dfrac{1}{\zeta}\right)^3 = \zeta^3 + 3\zeta + \dfrac{3}{\zeta} + \dfrac{1}{\zeta^3}$

$= \zeta^3 + \dfrac{1}{\zeta^3} + 3\alpha = \zeta^3 + \zeta^4 + 3\alpha$

and:

$\alpha^2 = \left(\zeta + \dfrac{1}{\zeta}\right)^2 = \zeta^2 + 2 + \dfrac{1}{\zeta^2} = \zeta^2 + \zeta^5 + 2$

so that:

$\alpha^3 + \alpha^2 = \zeta^3 + \zeta^4 + \zeta^2 + \zeta^5 + 3\alpha + 2$

$= \zeta + \zeta^2 + \zeta^3 + \zeta^4 + \zeta^5 + \zeta^6 + 2\alpha + 2$

$= -1 + 2\alpha + 2$, so that:

$\alpha^3 + \alpha^2 - 2\alpha - 1 = 0$.

However, this only shows that $\alpha$ is a root of the cubic $p(x) = x^3 + x^2 - 2x - 1$. To show this is the *minimal* polynomial you have two options:

1) Show it is irreducible over $\Bbb Q[x]$. I would use the rational root theorem (Gauss' lemma).

2) Show that $\sigma_6$ generates a subgroup of order 2 of $\text{Gal}(\Bbb Q(\zeta)/\Bbb Q)$, and that the fixed field of this subgroup is $\Bbb Q(\alpha)$.

You also claim without proof that the roots of $p$ are:

1) $\alpha = \zeta + \zeta^6$ (no need to prove this)
2) $\zeta^2 + \zeta^5$
3) $\zeta^3 + \zeta^4$

I would mention that this is because any automorphism of $\text{Gal}(\Bbb Q(\zeta)/\Bbb Q)$ sends a root of $p$ to another root of $p$.

Since $\Bbb Q(\alpha)$ contains all the roots of $p$, it's a normal extension (thus Galois). Clearly, its Galois group is:

$\dfrac{\text{Gal}(\Bbb Q(\zeta)/\Bbb Q)}{\text{Gal}(\Bbb Q(\zeta)/\Bbb Q(\alpha))} \cong \Bbb Z_6/\langle 3\rangle \cong \Bbb Z_3$

(recall that $3$ is the sole element of order 2 in $\Bbb Z_6$-just as complex-conjugation restricted to $\Bbb Q(\zeta)$ (the map $\sigma_6 : \zeta \mapsto \zeta^6 = \zeta^{-1} = \overline{\zeta}$) is the sole automorphism of the Galois group of $\Phi_7(x)$ of order 2).
 

FAQ: Galois Groups and Minimum Polynomial

What is a Galois group?

A Galois group is a mathematical concept in abstract algebra that describes the symmetries of a field extension. It is named after the French mathematician Évariste Galois and is used to study the properties of polynomials and their roots.

What is the significance of Galois groups?

Galois groups are essential in understanding the solvability of polynomial equations and determining which equations can be solved using radicals. They also have applications in other areas of mathematics, such as number theory and cryptography.

How do you calculate a Galois group?

To calculate a Galois group, you first need to find the minimum polynomial of the field extension. Then, determine the automorphisms of the extension, which are the symmetries that leave the base field fixed. The Galois group is then formed by combining these automorphisms using composition.

What is the minimum polynomial of a field extension?

The minimum polynomial of a field extension is the monic polynomial of lowest degree in the base field that has the extension's elements as its roots. It is used to define the field extension and is crucial in understanding its properties, such as its Galois group.

Can a Galois group be isomorphic to another Galois group?

Yes, two Galois groups can be isomorphic if they have the same structure and number of automorphisms. However, it should be noted that isomorphic Galois groups can arise from different field extensions, so they may not have the same properties or applications.

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