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Let $E = F(\alpha, \beta)$ be an extension of the field $F$. We're given that the minimal polynomial of $\alpha$ in $F[x]$ is of degree $d_1$, and the minimal polynomial of $\beta$ in $F[x]$ is of degree $d_2$. I'm supposed to show that if gcd$(d_1,d_2) = 1$, then $[E:F] = d_1d_2$.

This is my attempt:

We have $E = F(\alpha, \beta) \supset F(\alpha) \supset F$ as well as $E = F(\alpha, \beta) \supset F(\beta) \supset F$. So $\alpha$ is the root of $p(x) = a_0 + a_1x + ... + a_{d_1}x^{d_1} \in F[x]$ and $\beta$ is the root of $q(x) = b_0 + b_1x + ... + b_{d_2}x^{d_2} \in F[x]$. We also have $[F(\alpha):F] = d_1$ and $[F(\beta):F] = d_2$.

Assuming that $F(\alpha)$ doesn't contain any roots of $q(x)$ and that $F(\beta)$ doesn't contain any roots of $p(x)$, then we have that $q(x)$ and $p(x)$ are minimal polynomials of $\beta$ and $\alpha$ in $F(\alpha)$ and $F(\beta)$ respectively. From what I can see, this should give us that $[E:F(\alpha)] = d_2$ and $[E:F(\beta)] = d_1$, thus giving $[E:F] = d_1d_2$. Is this last part wrong? Can my assumption be made into certainty with gcd$(d_1,d_2) = 1$?

Any help on what I've done wrong, or if nothing is wrong but it's still an insufficient proof, then help on how to continue are greatly appreciated.

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  • $\begingroup$ This could help you : math.stackexchange.com/questions/1612311/… $\endgroup$ – Arnaud D. Apr 12 '16 at 16:11
  • $\begingroup$ @ArnaudD. Thank you. However, is there an error in my attempt so far, or is it just not complete? $\endgroup$ – Auclair Apr 12 '16 at 16:12
  • $\begingroup$ You can prove that no root of $p$ lies in $F(\beta)$ using the same kind of argument as in the question linked. But as someone else pointed out this is not sufficient to say that $p$ is irreducible in $F(\beta)$. $\endgroup$ – Arnaud D. Apr 12 '16 at 16:17

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