What monic integral polynomial has $V_4$ as the Galois Group?

I am just asked to find a monic integral polynomial of degree 4 whose Galois group over the rationals is $V_4$, however I don't really have an idea of how I can find this?

I understand that I want an irreducible polynomial so that the Galois group acts transitively on the roots, and if it is also separable then as it is integral, we also know that if the discriminant is a square then the Galois group is either $V_4 \text{ or } A_4 \leq S_4$

How though, can I show that $V_4$ is the Galois group and not $A_4$? Should I consider the degree of the splitting field over the rationals? (As this degree should be equal to the order of the Galois group, I believe). Is there anything else I can do?

For example I suspect that either $t^4 + 5t^2 + 1$ or $t^4 + 1$ satisfies the properties that I want, though I have no idea how I can prove this. They are both separable, irreducible over the rationals, and their discriminants are squares in $\mathbb Q$, so that means that their Galois groups are either $V_4$ or $A_4$

For $t^4 + 1,$ the splitting field is $\mathbb Q(i,\sqrt{2}),$ which has degree $4$ over $\mathbb Q$ I believe, so is that enough to justify that $Gal(t^4 + 1) = V_4?$

Also for $t^4 + 5t^2 + 1$, I was reasoning that as the roots are complex conjugate roots, any $\sigma \in Gal(t^4 + 5t^2+1)$ are determined by their actions on a representative from each of the two pairs. I then wanted to say that this meant the permutations could only be double transpositions, but I realised there's no reason why $\sigma$ can swap two roots that are complex conjugates of each other and leave the other two roots untouched, thereby giving us a transposition, though at the same time we already know that $Gal(t^4 + 5t^2 + 1) \in \{V_4, A_4\}$ so I'm very confused now.

If anyone could help me clear this up then that would be great, thank you! I'm really struggling to wrap my head around all of this.

Also, I would like to clarify something about the second half of the question I'm doing to make sure that I understand what's happening.

The question uses this polynomial with Galois group $V_4$ over the rationals and the fact that:

"a separable, monic, integral polynomial of degree n whose Galois group over the rationals does not contain an n-cycle satisfies that all of it's reductions modulo $p$ for a prime $p$ are reducible,"

to arrive at an irreducible, integral polynomial whose reduction$\pmod p$ is reducible for every prime $p$

It seems like if I have the first function so that it is irreducible and separable then the following property that all its reductions are reducible is trivial, as $V_4$ does not contain a 4-cycle. Am I right in thinking this?

• Better to start with a field extension that you know has your desired group as Galois group, and find the minimal polynomial of a primitive element. I would just take two integers that have nothing to do with each other, like two different primes, and adjoin the square roots of both. The field is the compositum of two quadratic extensions, the intersection of these being $\Bbb Q$. So the Galois group is $C_2\times C_2$ (two groups of order two). That’s your $V_4$. Now find the minimal polynomial of a primitive element (sum of the two square roots will always do the trick). – Lubin Nov 23 '17 at 1:51