Finding galois group of a $4$ degree polynomial

Question

Find the galois group of $f(x)=x^4+4x-1$

My attempt : I find that it is irreducible over $\mathbb{Q}[X]$ and it's resolvent cubic is $g(x)=x^3+4x-16$ and discriminant is $-7168=-32^2*7$ which is not a square. Also $g(x)=(x-2)(x^2+2x+8)$ that means reducible in $\mathbb{Q}[X]$ and only one linear factor hence possibilites is either $D_4$ or $C_4$ but unable to conclude.

Any help/hint in this regards would be highly appreciated. Thanks in advance!

Answer 1

The equation $x^4+4x-1 = 0$ has two real roots and two complex roots, so complex conjugation is in the Galois group, fixing exactly two roots. Hence the Galois group is $D_4$. Because in $C_4$, you cannot have a permutation that fixes exactly two elements.

Answer 2

Hint: Let $\alpha$ and $\beta$ be the irrational roots of $g(x)$. Is $f(x)$ irreducible in $\mathbb{Q}(\alpha,\beta)$? If it is, then the answer is $D_4$; otherwise, it is $C_4$.

Answer 3

A good text about Galois groups of cubic and quartic polynomials is the text by Keith Conrad. The possible Galois groups are exactly $S_4,A_4,D_4,V_4$ and $C_4$, see section $3$ and $4$. Theorem $3.6$, or Corollary $4.3$ gives you the criterion you want. Section $4$ is all about the distinction between $D_4$ and $C_4$.