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


1 Answer 1


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.