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Let $k = F_p$, the finite field of $p$ elements and $F = k(t, u)$ where $t$ and $u$ are independent variables. Let $α$ and $β$ be roots of $x^p − t$ and $x^p − u$ respectively. Let $L = F(α, β)$. Consider the intermediate fields $F ⊂ F(α+λβ) ⊂ L$ as $λ$ varies over all elements of $F$. I want to show there are infinitely many intermediate fields between F and L in the following way.

So let's say $λ ≠ µ$ are two elements of $F$ such that $F(α + λβ) = F(α + µβ)$, how to show that $α, β ∈ F(α + λβ)$? If I can do this then I can conclude that $F(α + λβ) = F(α, β)$ which is not possible right? :O

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Assume that $F(\alpha + \lambda\beta) = F(\alpha + \mu\beta)$ for some $\lambda, \mu \in F$. Then we have that $\alpha + \mu\beta \in F(\alpha + \lambda\beta)$ and so:

$$(\mu - \lambda)\beta = \alpha + \mu\beta - (\alpha + \lambda\beta) \in F(\alpha + \lambda\beta)$$

This implies that $\beta \in F(\alpha + \lambda\beta)$, as $(\mu - \lambda)$ is an element of $F$. From here it's not hard to conclude that $\alpha = \alpha + \lambda\beta - \lambda\beta \in F(\alpha + \lambda\beta)$. Therefore as you've noted we get that $F(\alpha + \lambda\beta) = F(\alpha, \beta)$.

To see that this equality is impossible, note that:

$$(\alpha + \lambda\beta)^p = \alpha^p + \lambda^p\beta^p = t + \lambda^pu \in F$$

This means that $[F(\alpha + \lambda\beta):F] = p$, while $[F(\alpha,\beta):F]=p^2$, which is a contadiction.

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  • $\begingroup$ Thank you! Very clear answer, I'm needing help with another qns on galois theory too: math.stackexchange.com/questions/2944604/… $\endgroup$
    – Homaniac
    Oct 6, 2018 at 16:00
  • $\begingroup$ Follow-up question: Are all the intermediate fields necessarily of this form $F(\alpha + \lambda\beta)$? I guess not, but I'm wondering specifically in the case $p = 2$, if there is a way to characterize all the intermediate fields. $\endgroup$
    – Oscar
    Jun 15, 2020 at 23:54

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