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Our professor gave an example in class: compute the Galois group of $ x^4-2x^2-1$ over $\mathbb{ Q}$. So the main steps are: $\alpha:=\sqrt{1+\sqrt{2}}$, and observe that the splitting field is $\mathbb{ Q}(\alpha,i)$. Now the degree of the extension is 8, since $ i$ isn't real. Then after computing the relations of the Galois elements we see that the group is exactly dihedral.

O.K. So then he left us an exercise with some more delicate technicalities.

Let $ K=\mathbb{C}(t)$, the function field of one variable. Let $u$ be some nonzero complex number, and consider the polynomial $x^4+2utx^2+ t\in K[x]$ ($t$ is the variable of the function field). Compute its Galois group.

I try to immitate the example in class. Then we let $\alpha$ be a root of the polynomial ( it's irreducible thanks to Eisenstein criterion). We can similarly obtain that the splitting field is $ K(\sqrt{t}, \alpha)$. But this time I found it hard to explain why $\sqrt{t}\notin K(\alpha)$. I'm even not sure whether this depends on the value of $ u$.

Could anyone give me some hint? Thanks a lot in advance!

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youknowwho
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    +1 certainly not too weak in asking good questions in mathematics! (In fact, very strong at it). Keep it up. I hope and pray you get a great answer. – Sarvesh Ravichandran Iyer Jul 02 '22 at 08:38
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    @SarveshRavichandranIyer Ah ha, thanks for your encouragement! – youknowwho Jul 02 '22 at 10:17
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    Why is the splitting field $K(\alpha,\sqrt t)$? I don't see where the $\sqrt t$ comes from, or how to write the other roots in terms of $\alpha$ and $\sqrt t$? Also, the first example quartic is biquadratic, which is a well studied special case, but the second quartic is not. – Jyrki Lahtonen Jul 02 '22 at 10:17
  • @JyrkiLahtonen $\alpha:=\sqrt{-ut+\sqrt{u^2t^2-t}},\beta:=\sqrt{-ut-\sqrt{u^2t^2-t}}$, then $\alpha \beta=\sqrt{t}$. – youknowwho Jul 02 '22 at 10:27
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    @JyrkiLahtonen oh no, I’m very sorry. I left a serious typo. The second term of the equation should be a square… – youknowwho Jul 02 '22 at 10:33
  • Ok, so it is a biquadratic. That makes more sense. – Jyrki Lahtonen Jul 02 '22 at 10:50
  • But in that case I encourage you to check out this older thread. – Jyrki Lahtonen Jul 02 '22 at 10:51
  • @JyrkiLahtonen thank you very much! It never occurred to me that I should try to think of the biquadratic cases in full generality? ( is there any difficulty in char. 2 case?) – youknowwho Jul 02 '22 at 10:54
  • In characteristic two the derivative of a biquadratic polynomial is constant zero. Therefore the splitting field of an irreducible biquadratic cannot be a separable extension (all the zeros have multiplicity two or four). Hence no Galois group. – Jyrki Lahtonen Jul 02 '22 at 10:59
  • @JyrkiLahtonen ah yeah, that’s right. – youknowwho Jul 02 '22 at 11:11

1 Answers1

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Some elements, not a solution

If $\sqrt t \in K(\alpha)$, it exists $p_0, p_1, p_2, p_3 \in K = \mathbb C(t)$ such that

$$\sqrt t= p_0 + p_1 \alpha + p_2 \alpha^2 +p_3 \alpha^3$$ with $$\alpha^4 + 2 ut \alpha^2 + t=0.$$

Hence $\alpha$ is a root of $q(x) = (p_0 + p_1 x + p_2 x^2 +p_3 x^3)^2 -t \in K[x]$ and also of $p(x) = x^4 + 2 ut x^2 + t \in K[x]$. Performing the long polynomial division of $q$ by $p$ in $K[x]$ you get

$$q(x) = p(x) b(x) + r(x)$$ where the degree of $r$ is equal to 3 at most and $r(\alpha) = 0$. You can conclude of that that the 4 coefficients of $r$, which belong to $K(t)$ are all equal to zero.

In the end, the question is: "are those 4 equations possible or not?".

Note: sympy polynomial may help to perform the long polynomial division if you know it and are lazy at performing it by hand.

Doing so, you'll get that $\sqrt t \in K(\alpha)$ if and only if the following equations in $p_0, p_1, p_2, p_3 \in K$ have a solution

$$\begin{cases} p_{0}^{2} - 2p_{1} p_{3} t - p_{2}^{2} t + 2 p_{3}^{2} t^{2} u - t &= 0\\ p_{0} p_{1} - p_{2} p_{3} t &= 0\\ 2p_{0} p_{2} + p_{1}^{2} - 4 p_{1} p_{3} t u - 2 p_{2}^{2} t u + 4 p_{3}^{2} t^{2} u^{2} - p_{3}^{2} t &= 0\\ p_{0} p_{3} + p_{1} p_{2} - 2 p_{2} p_{3} t u &= 0 \end{cases}$$

... and I don't see an immediate argument to state that those equations are not compatible.

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