As in the question,

I am asked to determine the Galois group of $$f(X)= X^6-2tX^3+1 \in \mathbb{Q}(t)[X] $$ over $ \mathbb{Q}(t) $.

First, I should prove that $ f $ is irreducible over $ \mathbb{Q}(t) $ and to do this I thought of the poynomial $ g(X)=f(X^\frac{1}{3})=X^2-2tX+1 $ since $ g(X-1)=X^2-2(t+1)X+2(t+1) $ is Eisenstein with respect to the prime element $ 2(t+1) $ of $ \mathbb{Q}[t] $ but I am not sure if this is enough to conclude that $ f $ is irreducible.

Furthermore, we have that the splitting field $ L $ of $ f $ over $ \mathbb{Q}(t) $ is $ \mathbb{Q}(\alpha, \omega) $ where $ \alpha $ is a root of $ f $ and $ \omega $ is a primitive third root of unity. Starting from here, I would like to compute the Galois group of $ f $.

I would appreciate any help regarding both questions. Thank you!


You cannot in general deduce the irreducibility of $f(X^n)$ from the irreducibility of $f(X)$.

(Consider $f$ as polynomial over $\mathbb Q[t]$ and reduce modulo the prime element $2t+1$, then we get $\bar f = X^6+X^3+1$ over $\mathbb Q[t] / (2t+1) \cong \mathbb Q$. But the irreducibility of this polynomial over $\mathbb Q$ is well-known: it is in fact the $9$-th cyclotomic polynomial. Thus $f$ is irreducible over $\mathbb Q[t]$ and therefore also over $\mathbb Q(t)$.)

Edit: In fact, let's show something stronger. I claim that $f$ is irreducible over $\mathbb Q(t,\omega)$ where $\omega$ is a third root of unity. To see this, we consider $f$ as polynomial over $\mathbb{Q}(\omega)[t]$ and reduce modulo the prime element $2t+\omega+1$ and get the polynomial $x^6+(1+\omega)x^3+1 \in \mathbb{Z}[\omega][x]$. We apply the automorphism of $\mathbb{Z}[\omega][x]$ given by $x \mapsto x-\omega$ and get the polynomial $x^6-6\omega x^5+15\bar{\omega}x^4+(\omega-19)x^3+(12\omega-3\bar{\omega})x^2+(3-3\bar{\omega})x+(1-\omega) \in \mathbb{Z}[\omega]$. (This could be simplified, but it's not really necessary.)
Now it is a well-known fact that $\mathbb{Z}[\omega]$ is PID and $(1-\omega)$ is a prime element in $\mathbb{Z}[\omega]$. From the equation $\omega^2+\omega+1=0$ we get $(1-\omega)^2=-3\omega$, $-\omega$ is obviously a unit, so this shows that $(1-\omega) | 3$. A simple computation shows that $(1-\omega)(-6\omega-13)=\omega-19$, so that $(1-\omega)|(\omega-19)$, thus the polynomial $x^6-6\omega x^5+15\bar{\omega}x^4+(\omega-19)x^3+(12\omega-3\bar{\omega})x^2+(3-3\bar{\omega})x+(1-\omega)$ is Eisenstein with respect to the prime element $1-\omega$. This completes the irreducibility.

Now let $\alpha$ be a root of $f$. From fact that $f$ is a minimal polynomial of $\alpha$ over $\mathbb Q(t,\omega)$, we easily compute the degree of the extension $\mathbb Q(t,\alpha,\omega) / \mathbb{Q}(t)$ as $12$ using the tower law. (The fact that $\omega \notin \mathbb Q(t)$ and hence $[\mathbb Q(t,\omega):\mathbb Q(t)]=2$ is obvious, as $\mathbb Q(t) / \mathbb Q$ is purely transcendental).

To compute the group $G := Gal(\mathbb{Q}(t,\omega,\alpha)/\mathbb{Q}(t))$, we exhibit some subextensions: First note that $\mathbb{Q}(t,\alpha) | \mathbb{Q}(t)$ is not normal, so $G$ can't be Abelian. Note that the minimal polynomial of $\alpha^3$ over $\mathbb{Q}(t)$ is $x^2-2tx+1$. (Use the tower law and that $[\mathbb Q(t,\alpha):\mathbb Q(t, \alpha^3)] \leq 3$), so that we have at least two distinct quadratic subextensions $\mathbb Q(t,\alpha^3)$ and $\mathbb Q(t,\omega)$. Under the Galois correspondence, the Galois groups of these two quadratic subextensions are quotients of $G$, so that $G$ must have at least two subgroups of order $6$. The rest is group theory: there are not that many non-abelian groups of order $12$, only three in fact: The dicyclic group Dic12, which has a unique subgroup of order $6$, the alternating group A4, which does not have any subgroup of order $6$, so that the only group that remains is the dihedral group D12.

So $G$ is the dihedral group of order 12.


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.