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I am working on this problem:

Let $K$ be an extension field of a field $F$, and let $\alpha \in K$ be algebraic over $F$, with minimal polynomial $p(x)$. Let $\beta \in F(\alpha)$ be algebraic over $F$, with minimal polynomial $q(x)$. Prove that $\deg(q)\mid \deg(p)$.

I'm stuck at starting this problem so I would appreciate some hints for it.

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    $\begingroup$ $F \subseteq F(\beta) \subseteq F(\alpha)$ $\endgroup$
    – Daniel Fischer
    Aug 28, 2017 at 15:15

1 Answer 1

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Hint:

$$\dim_F F(\alpha)=\dim_{F(\beta)} F(\alpha)\cdot \dim_F F(\beta).$$

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  • $\begingroup$ do you mean this: $[F(\alpha):F] = [F(\alpha):F(\beta)] [F(\beta):F]$ $\endgroup$
    – Quy Xuan Cao
    Aug 28, 2017 at 15:28
  • $\begingroup$ Yes, it's just another notation. $\endgroup$
    – Bernard
    Aug 28, 2017 at 15:32
  • $\begingroup$ From the problem, I have $[F(\alpha):F] $ = deg $q$, and $ [F(\alpha):F(\beta)] = 1$, hence $[F(\beta):F]$ = deg $q$. Do I understand it correctly? $\endgroup$
    – Quy Xuan Cao
    Aug 28, 2017 at 15:36
  • $\begingroup$ Not quite: How do you know $F(\alpha)=F(\beta)$? $\endgroup$
    – Bernard
    Aug 28, 2017 at 15:45
  • $\begingroup$ From $[F(\alpha):F] = [F(\alpha):F(\beta)] [F(\beta):F]$ one gets deg $p = [F(\alpha):F(\beta)]$ deg $q$. $\endgroup$
    – Marc Bogaerts
    Aug 31, 2017 at 9:22

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