Proposition. Let $E/F$ be a finite extension. Assume that for every pair of subfields $E_1$ and $E_2$ of E with $F \subset E_1,E_2$ then $E_1 \subset E_2$ or $E_2 \subset E_1$. Then exists a primitive element.

Dem. As E is a finite extension, we can assume that $E=F(a_1, a_2,...,a_n)$. Then $F(a_1)$ and $F(a_2)$ are finite algebraic extension of F. Using the hypothesis, we get (for instance) $F(a_2) \subset F(a_1)$, so $a_2 \in F(a_1)$ and we can express $a_2$ like $k_1+k_2 a_1 + \dots k_d a_1^d$, with $k_j \in F$. Hence, $E= F(a_1, a_3,...,a_n)$. By induction we get $E = F(a_i)$ for a certain $1\leq i \leq n$. QED.

Could someone check it? Thank you.

  • $\begingroup$ It looks fine to me. $\endgroup$ – DonAntonio Jan 13 '17 at 23:15
  • $\begingroup$ Your proof is right. $\endgroup$ – Xam Jan 14 '17 at 2:34

Your Answer

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

Browse other questions tagged or ask your own question.