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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!

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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.

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