# How to show that this $V_4-$ extension ramifies at the infinite prime?

A result by Boston and Markin states that the minimum number of primes that ramify in a $$G-$$extension of $$\mathbb{Q}$$ is at least $$d(G^{ab})$$, the minimum number of generators of the abelianisation $$G/[G,G]$$.

In the case of $$V_4$$, which is abelian, and has a minimum of two generators, their result states that every $$V_4-$$ extension should be ramified for at least two primes.

I used SageMaths to calculate the discriminant of the number field $$\mathbb{Q}(i+\sqrt{2})$$ and verify that the Galois group is $$V_4$$. The discriminant is a power of 2, and so (2) is the only real prime that ramifies.

sage: R.<X> = PolynomialRing(QQ)
sage: K.<t> = NumberField(X^4-2*X^2+9)
sage: K.integral_basis()
[7/12*t^3 + 3/4*t^2 + 1/12*t + 1/4, 1/6*t^3 + 1/6*t, t^2, t^3]
sage: K.discriminant().factor()
2^8
sage: K.galois_group(type="pari")
Galois group PARI group [4, 1, 2, "E(4) = 2[x]2"] of degree 4 of the Number Field in t with defining polynomial X^4 - 2*X^2 + 9


How would I show that this extension also ramifies at the infinite prime?

• This counterexample is mentioned on p. 146 of the paper you are referring to.
– user23365
Commented Mar 17, 2021 at 16:22
• Looks like yet another instance of two is an odd prime number because it is even (sorry don't remember the exact quote). Anyway, $\Bbb{Z}_{2^n}^*$ is not cyclic, and that leaves this window of opportunity. By the way, the link does not seem to work. I get "failure to connect..." Commented Mar 18, 2021 at 4:39

It ramifies at the infinite prime because $$\Bbb Q$$ is totally real but $$i+\sqrt 2$$ is totally imaginary. (Recall that a real place $$v$$ in $$K$$ is said to ramify over $$L$$ if $$v$$ extends to a complex embedding that is not real.)