What is Gal$_\mathbb{Q}(x^4 + 5x^3 + 10x + 5)$? 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.
 A: 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 $(d_1,\ldots, d_r)$.
A: 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, $S_4,A_4,D_4,C_4,C_2\times C_2$.
