Find Gal$_\mathbb{Q}(x^4 + 5x^3 + 10x + 5)$ and Gal$_\mathbb{Q}(x^4 - 2)$

I was trying the second one which I think is the easiest case. However I am not able to prove it. Here is what I know

(1) The roots of the polynomial $x^4 - 2$ given by $$\psi_k = 2^{\frac{1}{4}}\exp^{i\displaystyle\frac{\pi k}{2}}, k = 0,...,3$$

(2) The splitting field of $x^4 -2$ is $\mathbb{Q}(2^{\frac{1}{4}},i)$.

(3) $\mathbb{Q}(2^{\frac{1}{4}},i)$ is a Galois extension.

(4) $x^4 - 2$ is an irreducible polynomial (By Einsesteins criterion) over $\mathbb{Q}$

(5) The degree of this extension is the same as the order of the Galois group of this extension.

I need to find the group to which $\mathrm{Gal}_\mathbb{Q}(x^4 - 2)$ is isomorphic. I know it's $D_4$, but I am not sure how to prove it.

How do I construct the isomorphism between the permutation in the galois group and $D_4$?

Also for the other polynomial, I have no Idea how to proceed.

  • 1
    $\begingroup$ The proof for Galois group of $x^4-2$ is given here. $\endgroup$ – Dietrich Burde Apr 23 '18 at 18:02

For $X^4-2$, you have essentially two methods, depending on what you know.

First you need to determine the degree of your extension. Set $M=\mathbb{Q}(\sqrt[4]{2})$, and $L=M(i)=\mathbb{Q}(\sqrt[4]{2},i)$.

You have $[M:\mathbb{Q}]=4$, and $[M(i):M]=2$ (clearly, this degree is atmost $2$ since $i^2+1=0$, but this is exactly $2$, since $M\subset \mathbb{R}$, so $i\notin M$.

Hence $[L:\mathbb{Q}]=8$. Let $G={\rm Gal}(L/\mathbb{Q})$.

First method. This one needs to know the classification of groups of order $8$, and Galois correspondence.

Notice that the subgroup ${\rm Gal}(L/M)$ of $G$ is not normal, since the extension $M/\mathbb{Q}$ is not Galois (this is part of the Galois correspondence).

Now there are 5 groups of order 8: 3 abelian groups, $D_4$ and $Q_8$ (quaternion group). Note that $D_4$ is the only group of order $8$ having a subgroup which is not normal.

Hence your group is necessarily $D_4$.

Second method. You need to compute all the automorphisms of $L$ using the standard methods, then ty to identify the group in the list above, by playing with commutation relations, orders...I assume that you know how to extend a $\mathbb{Q}$-embedding $K\to \mathbb{C}$ to an embedding $K(\alpha)\to \mathbb{C}, $ where $\alpha$ is algebraic over $K$.

The $\mathbb{Q}$-embeddings of $K=\mathbb{Q}(i)$ into $\mathbb{C}$ are $Id_K$ and $\tau$, where $\tau(i)=-i$.

The minimal polynomial of $\alpha=\sqrt[4]{2}$ over $K$ is $X^4-2$ (why ?), so $Id_K$ extends to $4$ automorphisms $\sigma_k$, where $\sigma_k(i)=i, \ \sigma_k(\alpha)=i^k\alpha, k=0,...,3$

Now , $\tau(X^4-2)=X^4-2\in K[X]$, so $\tau$ extends to $4$ automorphisms $\sigma'_k$, where $\sigma'(i)=-i, \ \sigma'_k(\alpha)=i^k\alpha, k=0,...,3$.

No you need to find a element of order 4, and an element of order 2, which satisfy the right commutation relation. I let you write the details.

For the second polynomial, there is a result you should know:

let $P$ be an irreducible polynomial of $\mathbb{Q}[X]$, which is monic and has coefficients in $\mathbb{Z}$. Let $p$ be a prime number. Assume that $P$ mod $p$ may be written as $\pi_1\cdots\pi_r,$ where $\pi_1,\cdots,\pi_r\in\mathbb{F}_p[X]$ are pairwise distinct irreducible polynomials.

If $d_i=\deg(\pi_i)$, then the Galois group of $P$ over $\mathbb{Q}$ (viewed as a subgroup of $S_n,$ where $n=\deg(P)$) contains a permutation of type $(d_1,\ldots, d_r)$.

  • $\begingroup$ Can you elaborate more in the result I should know ? What is the point of the permutation and how that can help me proving that. $\endgroup$ – Richard Clare Apr 23 '18 at 22:38
  • $\begingroup$ For example, factoring mod $2$ and using this result shows that the Galosi group contains a $4$-cycle, so it cannot be $A_4$ or $C_2\times C_2.$ $\endgroup$ – GreginGre Apr 24 '18 at 5:19

The Galois groups of cubics and quartics have been computed here. One looks at the discriminant of $f$, whether or not it is a square, and at the cubic resolvent, whether or not it is irreducible; see section $3$ and $4$. As a summary, from Corollary $4.3$, Table $8$, one can read off the Galois group easily. Altogether there the following $5$ possibilities, for quartics, $S_4,A_4,D_4,C_4,C_2\times C_2$.


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