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I am trying to calculate when the extension $\mathbb Q\subset \mathbb Q(\sqrt n,\sqrt{a+b\sqrt n})$ is cyclic Galois. To determine when it's Galois it suffices to verify the minimal polynomials are separable and split in the extension field. The first minimal polynomial is just $x^2-n\in \mathbb Q[x]$ which is irreducible when $n\in \mathbb Z$ is squarefree. For the second, I think the minimal polynomial is $(x^2-a)^2-b^2n\in \mathbb Q[x]$, which is irreducible if $a^2-b^2n$ is not divisible by four by Eisenstein.

So I am left with calculating the automorphism group of the extension. I thought I should find a primitive element and use its minimal polynomial to calculate, but a minimal polynomial of a linear combination of $\sqrt n,\sqrt{a+b\sqrt n}$ seems unpleasant to find. On second inspection it seems $\mathbb Q(\sqrt n)\subset \mathbb Q(\sqrt{a+b\sqrt n})$ whence it suffices to deal only with $\mu=(x^2-a)^2-b^2n$. So the extension is $\mathbb Q[x]/ \langle \mu \rangle$. An automorphism of the extension $\varphi$ must satisfy $$(\varphi(x)^2-a)^2-b^2n=f(x)((x^2-a)^2-b^2n)$$for some $f\in \mathbb Q[x]$. Comparing free coefficients forces $\deg f=0$, so $f=\lambda\in \mathbb Q$ is a scalar. Comparing coefficients gives $\lambda=1$. Hence we are left with $(\varphi(x)^2-a)^2=(x^2-a)^2$. Two solutions are $\varphi (x)=\pm x$, but the other equation $\varphi (x)^2=-x^2+2a$ seems to have no solutions. What am I missing?

I'm sure there's an easier way using the roots $\pm \sqrt{a\pm b\sqrt n}$ and the fact $\varphi$ must permute them, but I'm not sure how to proceed.

Arrow
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1 Answers1

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You are correct that you only need to consider $\mathbb{Q}(\sqrt{a+b\sqrt{n}})$. If $n$ is a rational square, the extension is either trivial or quadratic depending on whether $a+b\sqrt{n}$ is a rational square. In either case, the extension is cyclic.

Otherwise, Theorem 1.2.1 from Serre's book ``Topics in Galois Theory'' shows your extension is cyclic if and only if $n(a^2-nb^2)$ is a rational square.

lm6612
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  • Dear lm6612, the proof in Serre is unintelligible to me. I have tried to outline my elementary approach and would appreciate helps along these lines. Thanks! – Arrow Aug 18 '21 at 22:07
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    You can try consulting the American Math. Monthly paper ``An Elementary Test for the Galois Group of a Quartic Polynomial'' by Kappen and Warren. In particular, Theorem 3 is concerned with polynomials of the form $x^4+Ax^2+B$; your polynomial $\mu$ is of this form. The approach in this article is more elementary, and the exposition might be more to your liking. – lm6612 Aug 18 '21 at 22:20