Let $E_1,E_2$ be finite extensions of a field $F$, and assume $E_1,E_2$ contained in some field. If $[E_1:F]$ and $[E_2:F]$ are relatively prime, show that $[E_1E_2:F]=[E_1:F][E_2:F]$ and $[E_1E_2:E_2]=[E_1:F]$.

Would anyone know how to approach this problem/solve it?

  • $\begingroup$ Perhaps I should say that the rules of the game suggest strongly that you explain how much you already know in Galois Theory/Field Theory, what you’ve tried, and how far you’ve gotten so far. $\endgroup$ – Lubin Mar 9 '17 at 23:14

Two theorems apply here: first, if $k\subset E\subset\Omega$ and $k\subset F\subset\Omega$, where all the letters denote fields, then $[EF:E]\le[F:k]$, and in particular if both $[E:k]$ and $[F:k]$ are finite, so is $[EF:k]$. The other is that if $k\subset E\subset K$, then $[K:k]=[K:E][E:k]$.
The rest follows when you realize that the relatively prime numbers $[E_1:F]$ and $[E_2:F]$ both divide $[E_1E_2:F]$, and you then apply the inequality from the first quoted theorem.

  • 1
    $\begingroup$ You beat me to it! (+1) By the way, do you think my suggested method for showing $[EF:E]\leq [F:k]$ is efficient (or even correct)? I have to confess I didn't even know this result before! $\endgroup$ – Kenny Wong Mar 9 '17 at 23:06
  • $\begingroup$ Yes, @KennyWong I believe you need to do a proof that mentions degree(s) of minimal polynomials, and apply the tower theorem. For this reason, the multiplicativity of degree is logically prior to this theorem. But you may know this theorem in a form that’s stronger, if one of the fields is Galois over the smallest field: it’s called the Theorem on Natural Irrationalities. $\endgroup$ – Lubin Mar 9 '17 at 23:12
  • $\begingroup$ I'll certainly look up the Theorem on Natural Irrationalities. Also, I presume the theorem $[EF:E] \leq [F:k]$ works fine even if $[E:k]$ is infinite? (i.e. we only need to assume $[F:k]$ is finite?) Or am I mistaken? $\endgroup$ – Kenny Wong Mar 9 '17 at 23:14
  • 1
    $\begingroup$ Yes, absolutely, @KennyWong, as I believe your sketch of a proof shows. $\endgroup$ – Lubin Mar 9 '17 at 23:16
  • $\begingroup$ Ah yes, the Theorem on Natural Irrationalities does indeed look somewhat familiar, and I now see how it is similar to this problem. Thank you very much! $\endgroup$ – Kenny Wong Mar 9 '17 at 23:23

(i) Suppose $F \subset E_2 \subset L$ and suppose $\alpha \in L$. Can you show that $[E_2(\alpha):E_2] \leq [F(\alpha):F]$?

[Hint: $[F(\alpha):F]$ is the degree of the minimal polynomial of $\alpha $ over $F$, and $[E_2(\alpha):E_2]$ is the degree of the minimal polynomial of $\alpha$ over $E_2$. How are the degrees of these minimal polynomials related?]

(ii) Let $E_1, E_2$ be finite extensions of $F$, all contained inside $L$. Convince yourself that $E_1 = F(\alpha_1, \dots, \alpha_n)$ for certain $\alpha_1, \dots, \alpha_n \in L$. Convince yourself that $E_1E_2 = E_2(\alpha_1, \dots, \alpha_n)$. Can you use the result of (i) inductively to show that $[E_1 E_2 : E_2] \leq [E_1 : F]$?

[Of course, a similar argument gives $[E_1 E_2 : E_1] \leq [E_2 : F]$.]

(iii) Apply the tower law to the inclusions $F \subset E_1 \subset E_1 E_2$ and $F \subset E_2 \subset E_1 E_2$. Use the fact that $[E_1 : F]$ and $[E_2 : F]$ are relatively prime, together with the result from (ii), to show that $[E_1 E_2 : E_2] = [E_1 : F]$ and $[E_1 E_2 : F] = [E_1 : F][E_2 : F]$.


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.