Galois group of $f(X) = X^4 + 3X^2 + 1$

Question

I'm trying to compute the Galois group $G$ of $$f=X^4 + 3X^2 + 1$$ over $\mathbb{Q}$. This is what I worked out so far:

1. The discriminant is $2^4\cdot 5^2$, a square, so $G\subset A_4$

2. $f$ is irreducible, as one can see by the Cohn's criterion with the prime $10301$, hence $G$ is a transitive subgroup of $S_4$

3. $f$ can be rewritten as $Y^2+3Y+1$, so the roots are easily found by the quadratic formula. It turns out that they are all pure imaginary, namely $$ \pm\sqrt{\frac{-3\pm\sqrt{5}}{2}} $$ From this I deduce that (denote by $K$ the splitting field of $f$) $[K:\mathbb{Q}] = 4$, and so also $|G|=4$. The only candidates for $G$ are therefore the Klein group $V_4$ or the cyclic $C_4$.

4. With Mathematica I computed the factorisations of $f$ modulo $p$ for the first $1000$ primes, and $f$ is never irreducible. This strongly suggests that $G=V_4$, because it looks like it doesn't contain any $4$-cycle.

If you agree with my reasoning, please can you help me to prove that the Galois group is actually $V_4$? Otherwise shame on me! ;-)

Answer 1

Let $\alpha = \sqrt{(-3 + \sqrt{5}) / 2} $ and $\beta = \sqrt{ (- 3 - \sqrt{5} )/ 2}$. The key point is that $\alpha \beta = 1$.

So if you have any $g \in G$, it's determined by where it sends $\alpha$. In particular $g(\alpha)$ is one of $\alpha, -\alpha, \beta, -\beta$. Since $|G| = 4$, you know that each of these options determines a unique element of the Galois group, and you can check that each is of order 2.

Answer 2

From $$\pm\sqrt{\frac{-3\pm\sqrt{5}}{2}}$$ I would guess intuitively that the Galois group has two generators, one flipping the first $\pm$ sign, and one flipping the second. If this is the case, then $V_4$ is the obvious answer. I'm not 100% certain, though.

Answer 3

But C4 is not in A4, right? So once you know that |G|=4 you know that G=V4.