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$\newcommand{\gal}{\operatorname{Gal}}$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.

Another slightly dead-end attempt:

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.

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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 $\Bbb Q$.

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  • $\begingroup$ I’m going to need a moment to verify these computations myself, but first I must ask - what made you think of this example? $\endgroup$
    – FShrike
    Jul 31, 2022 at 15:13
  • $\begingroup$ 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? $\endgroup$
    – FShrike
    Jul 31, 2022 at 15:26
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    $\begingroup$ 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. $\endgroup$ Jul 31, 2022 at 15:49

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