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Let $K / F$ be a extension of fields and let $x_1 , x_2 \in K$ be two algebraic elements over $F$ and let $p_1 = \deg \min(x_1 , F)$ and $p_2 = \deg \min(x_2 , F)$ be two coprime natural numbers. I want to show that $[F(x_1 , x_2) : F] = p_1 p_2$. By multiplicity of degree formula, we know that $$ [F(x_1 , x_2) : F] = p_1 p\mbox{,} $$ where $p = \deg \min(x_2 , F(x_1))$. Then I need we need to prove that $\min(x_2 , F(x_1))(X) \in F[X]$, using that $\gcd(p_1 , p_2) = 1$. Thank you very much in advance.

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  • $\begingroup$ Use the fact that degree multiplies in a tower of field extensions $E/K/F$. $\endgroup$ Aug 26, 2018 at 22:39

1 Answer 1

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HINT:

$$p_1 = [F(x_1):F]\; \big|\; [F(x_1,x_2):F(x_1)][F(x_1):F] = [F(x_1,x_2):F]$$ $$p_2 = [F(x_2):F]\; \big|\; [F(x_1,x_2):F(x_2)][F(x_2):F] = [F(x_1,x_2):F]$$

Also:

$$[F(x_1,x_2):F] \le p_1p_2$$

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