I'm trying to figure out generators of subfields given $E$ and $F$, $\text{char}(F) \neq 2$, and $E$ the splitting field of $$f(x) = x^4-2ax^2+a^2-b^2c$$ which has roots

$$\alpha_1 = \sqrt{a+b\sqrt{c}} \quad \alpha_2 = -\sqrt{a+b\sqrt{c}} \quad \alpha_3 = \sqrt{a-b\sqrt{c}} \quad \alpha_4 = -\sqrt{a-b\sqrt{c}}$$

A condition for $E$ being the splitting field I've previously found is that $a^2-b^2c \in E$ is square, since

$$\sqrt{a-b\sqrt{c}} = \frac{\sqrt{a^2-b^2c}}{\sqrt{a+b\sqrt{c}}}$$

Moreover, $\text{Gal}(E/F) \cong V$ (Klein-4 group) if $\Delta(f)$ is square in $F$, so I'm assuming that.

I've thus far figured out that $F[\sqrt{c}]$ is one subfield, corresponding to the cyclic subgroup generated by $\langle (12)(34) \rangle$, however I'm not sure how to determine the other two. I know that

$$(13)(24) \text{ fixes } \sqrt{a+b\sqrt{c}} + \sqrt{a-b\sqrt{c}}$$ $$(14)(23) \text{ fixes } \sqrt{a+b\sqrt{c}} - \sqrt{a-b\sqrt{c}}$$

But I'm unsure how to ascertain the generators of the other two subfields. Since they are quadratic extensions of $F$, I can assert that the other two generators will be of the form $\sqrt{x}$ and $\sqrt{y}$, but aside from that I'm not sure.

Would appreciate any tips.


1 Answer 1


I'm a bit puzzled by your question: you start by defining $E$ to be the splitting field, then later you give a condition for $E$ to be the splitting field. You don't define $F$: is this your base field?

Let's suppose that $F$ is the ground field. Then $F(\alpha_1)$ is the splitting field iff $a^2-b^2c$ is a square in $F$. You want the extension to be $V_4$. The discriminant is $$\begin{align} &(\alpha_1-\alpha_2)^2(\alpha_1-\alpha_3)^2(\alpha_1-\alpha_4)^2(\alpha_2-\alpha_3)^2(\alpha_2-\alpha_4)^2(\alpha_3-\alpha_4)^2\\ =&16\alpha_1^2\alpha_3^2(\alpha_1-\alpha_3)^2(\alpha_1+\alpha_3)^2(-\alpha_1-\alpha_3)^2(-\alpha_1+\alpha_3)^2\\ =&16\alpha_1^2\alpha_3^2(\alpha_1^2-\alpha_3^2)^4\\ =&16(a+b\sqrt c)(a-b\sqrt c)(2b\sqrt c)^4\\ =&256(a^2-b^2c)b^4c^2 \end{align}$$ so it's a square iff $a^2-b^2c$ is a square. Let's assume this.

Write $a^2-b^2c=r^2$ where $r=\alpha_1\alpha_3$. Let $$\beta=\alpha_1+\alpha_3=\sqrt{a+b\sqrt c}+\sqrt{a-b\sqrt c}.$$ Then $$\beta^2=2a+2r.$$ So one of the quadratic extensions you seek is $F(\sqrt{2(a+r)})$. Of course, the other is $F(\sqrt{2(a-r)})$.

  • $\begingroup$ Ah sorry about the confusion, wording should've been the other way around. And yes I was assuming $F$ as the base field. Thanks for the confirmation, I had suspected but wasn't sure if those would be the generators. $\endgroup$ May 21, 2017 at 6:23

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