# Square roots of root-expressions: distinguishing quartic Galois groups from copies of $C_4$ or of $D_4$ - exercise M.11, Ch.16 Artin's algebra

$$\newcommand{\gal}{\operatorname{Gal}}$$The setup:

$$F$$ is some field (Artin always assumes characteristic zero to get separability, but I don't like this choice) and $$f(x)\in F[x]$$ is an irreducible monic quartic polynomial $$x^4-a_1x^3+a_2x^2-a_3x+a_4$$. We assume further that it is a separable polynomial, so that its splitting field $$K$$ has $$K/F$$ a Galois extension, and $$f$$ has an ordered set of distinct roots $$\alpha_{1,2,3,4}$$ in $$K$$. I identify $$\gal(K/F)$$ with its permutation representation over these roots, e.g. $$(12)$$ would represent the automorphism (if it is an automorphism) fixing $$F$$ and fixing $$\alpha_3,\alpha_4$$, whilst swapping $$\alpha_1\mapsto\alpha_2\mapsto\alpha_1$$.

Let $$\beta=\alpha_1\alpha_2+\alpha_3\alpha_4$$, define the resolvent cubic $$g(x)\in F[x]$$ which is the monic polynomial with distinct roots $$\beta_i$$, where the $$\beta_i$$ are in the orbit of $$\beta$$ under $$S_4$$. Let $$\delta$$ be the square root of the discriminant: $$\delta=(\alpha_1-\alpha_2)(\alpha_1-\alpha_3)(\alpha_1-\alpha_4)(\alpha_2-\alpha_3)(\alpha_2-\alpha_4)(\alpha_3-\alpha_4)$$

This exercise is about the case where $$\delta\notin F$$ but $$g(x)$$ has one root in $$F$$, $$\beta$$. Artin so far has shown that this forces $$\gal(K/F)$$ to be a copy of $$C_4$$ or of $$D_4$$. The exercise begins by asking us to first identify which permutations feature in these copies. I'm almost certain the correct response to this is: $$\gal(K/F)=\langle(1423)\rangle\cong C_4,\quad\gal(K/F)=\langle(12),(1423)\rangle\cong D_4$$

Continuing, he defines $$\gamma=\alpha_1\alpha_2-\alpha_3\alpha_4$$ and $$\epsilon=\alpha_1+\alpha_2-\alpha_3-\alpha_4$$. He asks us to show, which I have done, that $$\gamma^2,\epsilon^2\in F$$.

The difficulty:

If $$\gamma\neq0$$, show that $$\delta\gamma$$ is a square in $$F$$ iff. $$\gal(K/F)=C_4$$.

Well, I first claim that $$\delta\gamma\in F$$ iff. $$\gal(K/F)=C_4$$:

If $$\gal(K/F)\not\cong C_4$$ it must be the copy of $$D_4$$, containing $$(12)$$. However, $$(12)$$ fixes $$\gamma$$ but negates $$\delta$$, and $$\gamma,\delta\neq0$$, so it follows that $$\delta\gamma$$ is not fixed by $$(12)$$ and thus is not in $$F$$, whereas $$\delta\gamma$$ is fixed by the generator $$\langle(1423)\rangle$$ of the copy of $$C_4$$.

So really it remains to show that $$\delta\gamma\in F,\gal(K/F)\cong C_4$$ imply $$F(\sqrt{\delta\gamma})=F$$. My thoughts on this:

We can factorise $$\gamma=(\alpha_1-\alpha_3)(\alpha_2+\alpha_4)+(\alpha_1-\alpha_4)(\alpha_2+\alpha_3)-(\alpha_1-\alpha_2)(\alpha_3+\alpha_4)$$ in order to get some squared terms going in $$\gamma\delta$$, but it remains horrible:

\begin{align}\delta\gamma&=(\alpha_1-\alpha_3)^2(\alpha_2^2-\alpha_4^2)[(\alpha_1-\alpha_2)(\alpha_1-\alpha_4)(\alpha_2-\alpha_3)(\alpha_3-\alpha_4)]\\&+(\alpha_1-\alpha_4)^2(\alpha_2^2-\alpha_3^2)[(\alpha_1-\alpha_2)(\alpha_1-\alpha_3)(\alpha_2-\alpha_4)(\alpha_3-\alpha_4)]\\&-(\alpha_1-\alpha_2)^2(\alpha_3^2-\alpha_4^2)[(\alpha_1-\alpha_3)(\alpha_1-\alpha_4)(\alpha_2-\alpha_3)(\alpha_2-\alpha_4)]\end{align}

But it's not clear what $$\sqrt{\delta\gamma}$$ would actually look like. So far, Artin's analysed statements like this - "if the discriminant is a square in $$F$$, then..." - since the square root of the discriminant has a very obvious form. But, $$\sqrt{\delta\gamma}$$ has no obvious form, unless my factorisation can be meaningfully improved. In general, I don't know what tools are at my disposal to analyse this element - I can't decide which permutations fix it, since I don't know what it is! What I do know is that it is fixed under some permutation only if that permutation fixes $$\delta\gamma\in F$$, but this is trivial since any $$F$$-automorphism fixes elements of $$F$$.

I have skipped ahead, I should admit - it's possible that the section on Kummer extensions is relevant to help me here, but I don't know how.

Using the main theorem, $$F(\sqrt{\delta\gamma})$$ is either $$F,K$$ or the fixed field of $$H:=\{\mathrm{id},(12)(34)\}\le\gal(K/F)$$. Knowing $$\delta\gamma\in F$$ but $$\delta\notin F$$ shows $$\gamma\notin F$$ so it follows from $$\delta^2,\gamma^2\in F$$ that they are both degree two over $$F$$. $$\delta,\gamma$$ are both fixed under $$H$$ and both not-fixed under $$\gal(K/F)$$, so $$K^H=F(\delta,\gamma)$$, implying $$2=[F(\delta,\gamma):F]=[F(\delta)(\gamma):F(\delta)][F(\delta):F]$$ - so $$F(\delta,\gamma)=F(\delta)=F(\gamma)$$ by similar argument. However, I don't see right now how to leverage that to show $$\sqrt{\delta\gamma}\in F$$.

So far: $$F(\sqrt{\delta\gamma})=\begin{cases}F(\delta)=F(\gamma)\\K\\F\end{cases}$$

I'd really appreciate hints on what to look for, how to proceed (or full answers) - how to attack the square root when the square root has no clear form in terms of the roots $$\alpha_{1,2,3,4}$$. Indeed, I'm now realising that, without a clear form in terms of the roots, I can't guarantee $$\sqrt{\delta\gamma}$$ is even in $$K$$!

N.B. we have $$(\delta-\gamma)^2$$ and $$(\delta+\gamma)^2$$ both elements of $$F$$ too, but I'm not sure if that can be leveraged.

The exercise is wrong. Consider $$x^4-4x^2+2$$. This has resolvent cubic $$x^3 + 4x^2 - 8x - 32$$, which has a single rational root. If one computes $$\delta \gamma$$, one gets $$-128$$, which is not a square in $$\Bbb Q$$.

• I’m going to need a moment to verify these computations myself, but first I must ask - what made you think of this example? Jul 31, 2022 at 15:13
• Thank you for pointing out this error. I might have wasted a lot of time on this. Do you agree that, at least: $$\delta\gamma\in F\iff\operatorname{Gal}(K/F)\cong C_4$$Is correct? Jul 31, 2022 at 15:26
• I just tried some extensions with Galois group $C_4$ that came to my mind. I agree with that equivalence, it‘s even plausible that this is what Artin intended. Jul 31, 2022 at 15:49