We've done degree 3 and 4, but not 5.

Find Galois groups of the following polynomials over $\displaystyle \mathbb{Q}$:

$\displaystyle x^5+x^3-3x^2-3$

$\displaystyle (x^2-8)(x^3-7)$

Please help me out, again. :(

Printable View

- May 25th 2009, 03:40 PMmariannegalois group of a polynomial of degree 5
We've done degree 3 and 4, but not 5.

Find Galois groups of the following polynomials over $\displaystyle \mathbb{Q}$:

$\displaystyle x^5+x^3-3x^2-3$

$\displaystyle (x^2-8)(x^3-7)$

Please help me out, again. :( - May 25th 2009, 06:49 PMThePerfectHacker
This polynomial factors as $\displaystyle (x^2 + 1)(x^3 - 3)$.

Over the complex numbers this factors as $\displaystyle (x+i)(x-i)(x - \sqrt[3]{3})(x-\zeta\sqrt[3]{3})(x - \zeta^2 \sqrt[3]{3})$.

Where, $\displaystyle \zeta = -\tfrac{1}{2} + i \tfrac{\sqrt{3}}{2}$.

Thus, the splitting field is given by $\displaystyle \mathbb{Q}(i,\sqrt[3]{3},\zeta) = \mathbb{Q}(i,\sqrt[3]{3},\sqrt{3})$.

Now, $\displaystyle [\mathbb{Q}(i,\sqrt[3]{3},\sqrt{3}) : \mathbb{Q}(\sqrt[3]{3},\sqrt{3})] = 2$ since $\displaystyle i \not \in \mathbb{Q}(\sqrt[3]{3},\sqrt{3})$.

The polynomial $\displaystyle x^2 - 3$ is irreducible over $\displaystyle \mathbb{Q}$ so it must be irreducible over $\displaystyle \mathbb{Q}(\sqrt[3]{3})$ because $\displaystyle [\mathbb{Q}(\sqrt[3]{3}): \mathbb{Q}] = 3$ and $\displaystyle \gcd(2,3) = 1$. This means that $\displaystyle [\mathbb{Q}(\sqrt[3]{3},\sqrt{3}) : \mathbb{Q}(\sqrt[3]{3}) ] = 2$.

And of course, $\displaystyle [\mathbb{Q}(\sqrt[3]{3}) : \mathbb{Q}] = 3$.

Thus, $\displaystyle [\mathbb{Q}(i,\sqrt[3]{3},\sqrt{3}):\mathbb{Q}] = 2\cdot 2\cdot 3 = 12$.

The Galois group therefore must have order $\displaystyle 12$ and that already gives us a few ideas of this group might be.

There are five possibilites: $\displaystyle \mathbb{Z}_{12},\mathbb{Z}_4\times \mathbb{Z}_3, A_4, D_6, T$.

I will let you think about which one out of the five it must be. But before you do that it would be helpful to make the following realizations. First, if $\displaystyle \sigma$ is an automorphism then it is completely determined by the values $\displaystyle \sigma(i),\sigma(\sqrt[3]{3}),\sigma(\sqrt{3})$. Now, $\displaystyle \sigma$ must permute the roots of the minimal polynomial of each element, so, $\displaystyle \sigma(i) = \pm i, \sigma(\sqrt[3]{3}) = \sqrt[3]{3},\zeta\sqrt[3]{3},\zeta^2\sqrt[3]{3},\sigma(\sqrt{3}) = \pm \sqrt{3}$. Thus, there are a total of at most $\displaystyle 12$ such permutations but because the Galois group has order twelve it means for each one of these possibilities we have a permuation. Let $\displaystyle \alpha$ be the permutation: $\displaystyle \alpha(i) = -i,\alpha(\sqrt[3]{3}) = \sqrt[3]{3},\alpha(\sqrt{3}) = \sqrt{3}$ i.e. $\displaystyle \alpha$ is complex conjugation. Let $\displaystyle \beta$ be the permutation: $\displaystyle \beta(i) = i,\beta(\sqrt[3]{3}) = \zeta\sqrt[3]{3},\beta(\sqrt{3}) = \sqrt{3}$. Let $\displaystyle \gamma$ be the permutation: $\displaystyle \gamma(i) = i,\gamma(\sqrt[3]{3}) = \sqrt[3]{3},\gamma(\sqrt{3}) = -\sqrt{3}$. Notice that $\displaystyle \alpha,\beta,\gamma$ are non-trivial automorphisms with $\displaystyle \alpha^2 = \beta^3 = \gamma^2 = \text{id}$.

Remember that $\displaystyle \overline{\zeta} = \zeta^2$, so, $\displaystyle \alpha\beta (\sqrt[3]{3}) = \alpha ( \zeta \sqrt[3]{3}) = \alpha(\zeta)\alpha(\sqrt[3]{3}) = \overline{\zeta}\sqrt[3]{3} = \zeta^2 \sqrt[3]{3}$. While $\displaystyle \beta \alpha(\sqrt[3]{3}) = \beta (\sqrt[3]{3}) = \zeta\sqrt[3]{3}$. We see that $\displaystyle \alpha \circ \beta \not = \beta \circ \alpha$ so the Galois group is not abelian. This eliminates the first two abelian cases and leaves us with only $\displaystyle 3$. Can you argue from here? - May 26th 2009, 02:36 AMgeorgel
Could this be done like this?

$\displaystyle Gal_\mathbb{Q}f=\mathbb{Z}_2\otimes A_3$ or $\displaystyle Gal_\mathbb{Q}f=\mathbb{Z}_2\otimes \mathbb{Z}_3$, since

$\displaystyle Gal_\mathbb{Q}(x^3-3)=\mathbb{Z}_3$ or $\displaystyle A_3$

and then we just check, using discriminant, what is exactly $\displaystyle Gal_\mathbb{Q}(x^3-3)$. - May 26th 2009, 09:05 AMThePerfectHacker
It is not true in general that the Galois group of a product of two polynomials is the direct product of the Galois groups of each of the polynomials. Thus, you do not know if this is one of those cases if it works or does not without further argument.

----

There is something further we can do to determine the Galois group. Notice that $\displaystyle \mathbb{Q}(\sqrt{3}),\mathbb{Q}(i)$ are two distinct subfields of $\displaystyle \mathbb{Q}(i,\sqrt[3]{3},\sqrt{3})$. Now $\displaystyle \mathbb{Q}(\sqrt{3})/\mathbb{Q}$ and $\displaystyle \mathbb{Q}(i)/\mathbb{Q}$ are normal extensions of degree 2. Therefore, by the fundamental theorem of Galois theory it means $\displaystyle \text{Gal}(\mathbb{Q}(i,\sqrt[3]{3},\sqrt{3})/\mathbb{Q}(i))$ and $\displaystyle \text{Gal}(\mathbb{Q}(i,\sqrt[3]{3},\sqrt{3})/\mathbb{Q}(\sqrt{3}))$ are two normal subgroups of index two in $\displaystyle \text{Gal}(\mathbb{Q}(i,\sqrt[3]{3},\sqrt{3})/\mathbb{Q})$. However $\displaystyle A_4$ has no subgroup of order $\displaystyle 6$ while $\displaystyle D_6$ has only one normal subgroup of order $\displaystyle 6$. Thus, it must be the remaining possibility and that is $\displaystyle T$.