1
$\begingroup$

I have this multi-part problem I have worked on in Galois Theory. I am particularly unsure abut finding all roots of our polynomial and the action of the Galois group. Also, I cannot see how we can have a subfield of order $4$


Find the splitting field $L$ of the polynomial $X^3-5 \in \mathbb{Q}[x]$

Let $\mathbb{Q}$ be a field and $f(X) \in \mathbb{Q}[X]$. A field extension $L$ where $\mathbb{Q} \subset L $ is a splitting field for $f(X)$ over $\mathbb{Q}$ if $L=\mathbb{Q}(\alpha_1,..., \alpha_n) $ with $f(X)=c(X-\alpha_1)...(X-\alpha_n)$. $c$ is the leading coefficient of $f(X)$

Here we have $f(X)=X^3-5$ so we must find the roots of $f(X)$. Clearly $\alpha_1=\sqrt[3]{5}$ is a root which means $L=\mathbb{Q}(\sqrt[3]{5})$. Are there any others?

Prove that $Gal(L/\mathbb{Q}) \simeq S_3$ and describe the action of its elements on $L$

The polynomial is a reduced cubic, so the discriminant is given by $-4a^3-27b^2=-27(25)$ which is not square so the Galois group is $S_3$

Is the action given by $\tau: \sqrt[3]{5} \rightarrow -\sqrt[3]{5}$?

Find all subfields of degree $2$ over $\mathbb{Q}$ in $L$

The Galois group has order $6$ so all subgroups will have orders that divide $6$, namely $1, 2, 3$ and $6$.

So the subfield of degree $2$ will correspond to the subgroups of degree $4$ in $S_3$ such as $\{e, (12)\}$

What is the corresponding subfield?

Find all subfields of degree $4$ over $\mathbb{Q}$ in $L$

I am confused here since I thought that degrees of subfields had a 1:1 correspondence with the order of the subgroups. But $S_3$ has no subgroup of order $4$. Where I am going wrong?

$\endgroup$
  • 2
    $\begingroup$ $L$ is not equal to $\mathbb Q(\sqrt[3]5)$. This is obvious because $5$ has two complex cube roots, but $\mathbb Q(\sqrt[3]5)\subset\mathbb R$. $\endgroup$ – user336735 Jun 2 '16 at 22:10
  • 1
    $\begingroup$ What are these two complex cubed roots? $\endgroup$ – maths2332 Jun 2 '16 at 22:42
1
$\begingroup$

Step by Step:

  1. $L\neq \mathbb{Q}(\sqrt[3]{5})$. To see this, does your polynomial split? It can't: this is a real extension, but the roots are $\sqrt[3]{5}\zeta_3^i$, with $i=0,1,2$, some of which are complex. So our splitting field requires we adjoin all three.

  2. Show that $L=\mathbb{Q}(\sqrt[3]{5},\zeta_3)$. We already know that $L=\mathbb{Q}(\sqrt[3]{5},\sqrt[3]{5}\zeta_3,\sqrt[3]{5}\zeta_3^2)$ as these are the roots. So show these are equal.

  3. Show that $[L:\mathbb{Q}]=6$. You should really do this yourself: I recommend using the Tower Law by first adjoining $\sqrt[3]{5}$, then $\zeta_3$.

  4. Because this is a splitting field over $\mathbb{Q}$, the extension is normal and separable, hence Galois, so the Galois group has order 6. Automorphisms need to send an element to another element that satisfies its minimal polynomial. So we can specify automorphisms by where they could send our generators to: $\sqrt[3]{5}$ must be sent to $\sqrt[3]{5}\zeta_3^i$ for $i=0,1,2$ as these are the roots to the minimal polynomial, and $\zeta_3$ could be sent to $\zeta_3^j$, $j=1,2$ (show this). Combinations of these images give a total of 6 possible automorphisms, and because we know that the Galois group has order 6, these are all valid and are exhaustive.

  5. You know it is $S_3$ for a lot of reasons: for one, the splitting field of a degree $n$ polynomial is a subgroup of $S_n$, and $\vert S_3\vert=6$, so has to be it. Otherwise, see these maps don't commute, and $S_3$ is the only group of order 6 that is not abelian (can be shown via semi direct products). Or you could even use the discriminant.

  6. By the Galois Correspondence, subfields of degree 2 correspond to subgroups of index 2 in $S_3$. I'll leave it to you to find how many there are and what the correspondence is. In the same way, subfields of degree 4 correspond to subgroups of index 4. Are there any?

$\endgroup$
  • 1
    $\begingroup$ Thank you so much, this is very helpful. So to fill in the extras: $\mathbb{Q}(\sqrt[3]{5}, \zeta \sqrt[3]{5} , \zeta^2 \sqrt[3]{5})=\mathbb{Q}(\sqrt[3]{5}, \zeta \sqrt[3]{5})$ since $\zeta^2 \sqrt[3]{5}=\frac{(\zeta \sqrt[3]{5})^2}{\sqrt[3]{5}}$ i.e. it can be written as a rational combination of the other elements so can be removed. Also, $[\mathbb{Q}(\sqrt[3]{5}) : \mathbb{Q}]=[\mathbb{Q}(\sqrt[3]{5}) : \mathbb{Q}(\sqrt[5]{3})] \times [\mathbb{Q}(\sqrt[5]{3}) : \mathbb{Q}]=2 \times 3 = 6$. $\endgroup$ – maths2332 Jun 2 '16 at 22:58
  • 1
    $\begingroup$ I am struggling with finding indexes of a given value in $S_3$ $\endgroup$ – maths2332 Jun 2 '16 at 22:58
  • 1
    $\begingroup$ I was under the impression the index must divide the group order $\endgroup$ – maths2332 Jun 2 '16 at 23:01
  • 1
    $\begingroup$ That's true, by Lagrange. So the possible indices are 1, 2, 3, 6, which correspond to subgroups of order 6 (trivial), 3 (are there any?), 2 (same?), 1 (trivial). $\endgroup$ – user293121 Jun 3 '16 at 0:16
  • 1
    $\begingroup$ So no subfields of index 4? $\endgroup$ – thinker Jun 3 '16 at 6:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.