1
$\begingroup$

Suppose $$f_a=x^p-x-\left [\prod_{j=1}^{a-1} \alpha_j\right]^{p-1} \in \mathbb{F}_p(\alpha_1,\ldots,\alpha_{a-1})[x]$$ is irreducible over $\mathbb{F}_p(\alpha_1,\ldots,\alpha_{a-1}) $, where $\alpha_i$ is a root of $f_i$ with $i<a$. And suppose $\alpha_a$ is a root of $f_a$. How do I prove that $$\mathbb{F}_p(\alpha_1,\ldots,\alpha_a)=\mathbb{F}_p(\alpha_a)?$$

It's easy to see that $$\left [\prod_{j=1}^{a-1} \alpha_j\right]^{p-1} \in \mathbb{F}_p(\alpha_a),$$ how do we conclude that $\alpha_i \in \mathbb{F}_p(\alpha_a)$ ?

$\endgroup$
4
  • $\begingroup$ Have you made all necessary assumptions on the $\alpha_i$? This seems false for $p=2,\alpha =1$. $\endgroup$
    – user44191
    Apr 30, 2017 at 2:19
  • $\begingroup$ For all $c \in \mathbf F_p^\times$ the polynomial $x^p - x - c$ is irreducible over $\mathbf F_p$, so is your irreducibility "hypothesis" exactly what you mean to say: irreducibility over $\mathbf F_p$? $\endgroup$
    – KConrad
    Apr 30, 2017 at 3:32
  • $\begingroup$ $\alpha=1$ is not a root of $x^2-x-1$. And each $f_i$ is irreducible, i just put it as an assumption, because im not going to prove it here. $\endgroup$ Apr 30, 2017 at 7:45
  • $\begingroup$ Crossposted at Math.SE $\endgroup$ May 24, 2017 at 23:18

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.