Galois Group of $x^4+12x^3+44x^2+48x+40\in\mathbb{Q}[x]$

Question

In my script it says that the galois group of $x^4+12x^3+44x^2+48x+40\in\mathbb{Q}[x]$ is given by $V_4$. However, I think this is not correct. The polynomial is irreducible and so are the resolvants $R_2(x)=x^2+1960$ and $R_3(x)=x^3+10x^2-196x-1960$, so the galois group should be $S_4$. Is that correct?

Answer 1

Let $p(x) = x^4 + 12x^3 + 44x^2 + 48 x + 40$. We can look at this question to find a strategy we can try to find the Galois group of $p$. The discriminant of $p$ is $7225344 = 2688^2$ and the resolvent cubic is $$x^3-44 x^2+416 x-1024 = (x-32) (x-8) (x-4),$$ so the resolvent is in fact reducible. Since the resolvent is reducible and the discriminant is a square we see that the Galois group must be $V_4$.

We can also calculate the splitting field of $p$ by letting $$f(x) := p(x-3) = x^4-10 x^2+49$$ and noting that the splitting field of $f$ is that of $p$. The roots of $f$ can be calculated quite easily and they are those $\alpha$ that satisfy $$\alpha^2 = 5-2i\sqrt{6} \quad\text{or}\quad \alpha^2 = 5+2i\sqrt{6}.$$ Now we can check that all these $\alpha$ lie in $\mathbb{Q}[i,\sqrt{6}]$ since for example $(i + \sqrt{6})^2 = 5 + 2i\sqrt{6}$. The automorphism group of $\mathbb{Q}[i,\sqrt{6}]/\mathbb{Q}$ can quite easily be seen to be $V_4$.

Answer 2

The discriminant of $f(x)=x^4+12x^3+44x^2+48x+40$ is a square $2^{14}3^27^2$, so the Galois Group cannot be $S_4$ as it must be contained in $A_4$. In fact your script is correct. The Galois Group is $V$, the Klein $4$-group.