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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?

user404920
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    This counterexample is mentioned on p. 146 of the paper you are referring to. –  Mar 17 '21 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..." – Jyrki Lahtonen Mar 18 '21 at 04:39

1 Answers1

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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.)

Kenny Lau
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