2

Working on this problem for abstract algebra and I want to check my process:

Let $f = t^6 - 3 \in \mathbb{Q}[t].$ Let $K = \mathbb{Q}$.
(i) Find the splitting field $L$ of $f$.
(ii) Find $G = \Gamma(L : K)$, the Galois group of $L : K$.
(iii) Find all the normal subgroups $N$ of $G$.
(iv) Find the corresponding subfields for each normal subgroup $N$ from part (iii).

(i)The polynomial $f = t^6 - 3$ is irreducible over $\mathbb{Q}[t]$ by Eisenstein's criterion. Thus, the splitting field is $\mathbb{Q}(\sqrt[6]{3}, \zeta_6)$ (where $\zeta_6$ is the sixth root of unity, equal to $\sqrt[3]{-1}$)

(ii) $(\sqrt[6]{3})$ has a minimal polynomial of $t^6 - 3$, with roots $\pm (\sqrt[6]{3}), \pm (\zeta_6\sqrt[6]{3}), \pm \zeta_6^2\sqrt[6]{3}$, and $\zeta_6$ has polynomial $(t^3 + 1)$ of $(t + 1)(t^2 - t + 1)$, where the minimal polynomial is $(t^2 - t + 1)$ with roots $(\pm \zeta_6)$. With that, we can construct our automorphisms. We define $\sigma$ such that \begin{align*} &\sigma(\sqrt[6]{3}) = \zeta_6\sqrt[6]{3}\\ &\sigma(\zeta_6) = \zeta_6 \end{align*} Then, \begin{equation*} \sigma^2(\sqrt[6]{3}) = \sigma(\zeta_6\sqrt[6]{3}) = \zeta_6^2\sqrt[6]{3} \end{equation*} and \begin{equation*} \sigma^3(\sqrt[6]{3}) = \sigma(\sigma^2(\sqrt[6]{3}))) = \sigma(\zeta_6^2\sqrt[6]{3}) = \zeta_6^3\sqrt[6]{3} = -\sqrt[6]{3} \end{equation*} and \begin{equation*} \sigma^4(\sqrt[6]{3}) = \sigma^2(\sigma^2(\sqrt[6]{3})) = \sigma^2(\zeta_6^2\sqrt[6]{3}) = \zeta_6^4\sqrt[6]{3} = -\zeta_6\sqrt[6]{3} \end{equation*} Continuing in this manner, we find that \begin{align*} &\sigma^5(\sqrt[6]{3}) = -\zeta_6^2\sqrt[6]{3}\\ &\sigma^6(\sqrt[6]{3}) = \sqrt[6]{3} \end{align*} Now we define $\tau$ such that: \begin{align*} &\tau(\zeta_6) = -\zeta_6\\ &\tau(\sqrt[6]{3}) = \sqrt[6]{3} \end{align*} Then, \begin{align*} &\tau^2(\zeta_6) = \tau(\tau(\zeta_6)) = \tau(-\zeta_6) = \zeta_6 \end{align*} Thus, our Galois group is $\{e, \sigma, \sigma^2, \sigma^3, \sigma^4, \sigma^5, \tau, \sigma\tau, \sigma^2\tau, \sigma^3\tau, \sigma^4\tau, \sigma^5\tau\}$.
As a check, we know that the degree of an extension of one root is equal to the degree of the minimal polynomial, and we can then apply the Tower Law, so we have that $\mathbb{Q}(\sqrt[6]{3}) : \mathbb{Q} = 6$ and $\mathbb{Q}(\zeta_6) : \mathbb{Q} = 2$, so $|G| = |\mathbb{Q}(\sqrt[6]{3})| \cdot |\mathbb{Q}(\zeta_6)| = 6 \cdot 2 = 12$, the degree of our previously stated Galois group.

(iii) Using Lagrange's Theorem - $H \trianglelefteq G \implies |H| | |G|$ - we know that our subgroups can have order 1, 2, 3, 4, 6, 12. Let's start with $\{e\}$ and $G$, which are subgroups and are normal. We have $\{e, \tau\}, \{e, \sigma, \sigma^2, \sigma^3, \sigma^4, \sigma^5\}$. Subgroups of order 3 and 4 do not exist do not exist. We cannot have any subgroups of 3 or 4, and only 1 subgroup of order 2 and one subgroup of order 6 due to the orders of the elements - $\sigma$ requires a subgroup of at least order 6, $\tau$ requires a subgroup of at least order 2, and $\sigma$ and $\tau$ together requires at least order 12.

Now, we check normalcy. Instead of checking both proper subgroups by hand, note instead that the Kronecker-Weber theorem tells us that we have a Galois group that is abelian, so every subgroup is normal.

(iv)Note that a basis for $\mathbb{Q}(\sqrt[6]{3})$ is $\{1, \alpha, \alpha^2, \alpha^3, \alpha^4, \alpha^5\}$ where $\alpha = \sqrt[6]{3}$, and since $n = 6$ is squarefree, a basis for $\mathbb{Q}(\zeta_6)$ is $\{1, \zeta_6, \zeta_6^2, \zeta_6^3, \zeta_6^4, \zeta_6^5\}$, and also note that a basis for $\mathbb{Q}(\sqrt[6]{3}, \zeta_6)$ can be produced by multiplying the bases of $\mathbb{Q}(\sqrt[6]{3})$ and $\mathbb{Q}(\zeta_6)$. After constructing a table of the elements of the basis and the elements of the Galois group, I found the elements that were fixed by each Galois subgroup and concluded that $G$ corresponds to $\mathbb{Q}(\sqrt[6]{3}, \zeta_6)$, the subgroup $\{e\}$ corresponds to $\mathbb{Q}$, the subgroup $\{e, \tau\}$ corresponds to $\mathbb{Q}(\sqrt[6]{3})$, and the subgroup $\{e, \sigma, \sigma^2, \sigma^3, \sigma^4, \sigma^5\} corresponds to $\mathbb{Q}(\zeta_6)$.

Edited: So I tried to rework the problem according to the comments given below, and I think I'm kind of working myself in circles. Thanks for the help!

Edited (again): I reworked the problem again (all y'all's comment were really helpful, thanks), and it makes a lot more sense now. The work above is the current progress I have made. Note that I still have part (iv) uncompleted - I was attempting to construct a table of the basis of the extension and the Galois group to find the intermediate fields (similar to this video: https://www.youtube.com/watch?v=J9sn5o-pAz8&t=168s&ab_channel=AdamGlesser), and the size seemed completely ridiculous, so I'm currently trying to derive a way to do this algorithmically. Any guidance on that topic would be highly appreciated, thanks!

elpitts
  • 153
  • $t^3+1=(t+1)(t^2-t+1)$ isn't irreducible over $\Bbb Q$ so it's not the minimal polynomial of $\sqrt[3]{-1}$. Note that $(\sigma \tau)^2=\sigma^2\tau^2=1$ so your last sentence of iii is not correct. – Robert Shore Mar 30 '21 at 22:49
  • It's weird to mix $\sqrt[3]{-1}$ and $(-1)^{2/3}$. Also, you're missing those parentheses. – Eric Towers Mar 30 '21 at 22:57
  • In (i), you explain that $t^6 - 3$ has six roots. The very first thing you say in (ii) is that polynomial only has two roots. Then follow that claiming $t^3 + 1$ only has two roots. Are these claims missing a phrase or some other ingredient that might make them correct? – Eric Towers Mar 30 '21 at 23:00
  • Have you plotted the six complex roots of $t^6 - 3$? Do you recognize what geometric figure's vertices are the roots? What is the group of symmetries of this figure? (We should be careful that we fix any points that happen to be in $\Bbb{Q}$, but none of these points are in $\Bbb{Q}$.) – Eric Towers Mar 30 '21 at 23:08
  • This Galois group has been investigated here before, e.g., https://math.stackexchange.com/questions/3919231/what-are-the-generators-of-the-galois-group-of-x6-3-over-mathbbq and https://math.stackexchange.com/questions/52810/degree-of-splitting-field-of-x6-3-over-mathbbq-31-2-and-also-over and https://math.stackexchange.com/questions/1341779/intermediate-prime-extensions-rotman and probably others. – Gerry Myerson Mar 31 '21 at 12:21
  • Have you studied any of those links? There's a lot wrong with what you've written. For example, having no rational roots does not imply irreducibility (see, e.g., $(x^2-2)(x^2-3)$). Also, an irreducible polynomial of degree six will have a splitting field of degree divisible by six and thus a Galois group of order divisible by six, so the order can't possibly be eight. It's best to say the polynomial $x^3+1$ has roots $1,\alpha,\beta$ where $\alpha$ and $\beta$ are complex conjugates of each other and also multiplicative inverses of each other; (continued) – Gerry Myerson Apr 01 '21 at 12:32
  • (continued) it follows that there's no such thing as an automorphism that moves one of them while fixing the other. – Gerry Myerson Apr 01 '21 at 12:32
  • @GerryMyerson I did check out those posts, thank you. On the irreducibility topic, I think t^6 - 3 is actually irreducible, but not for the reasons I had previously stated (I use Eisenstein's criterion now). Is that correct? – elpitts Apr 02 '21 at 05:14
  • 1
    Yes, Eisenstein (with $p=3$) does the job. – Gerry Myerson Apr 02 '21 at 05:15

0 Answers0