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This is a natural question which confused me a lot. I think generally it is true but I have no idea how to prove it. Also, can anyone raise any counter-example?

The question follows from a problem from Topics in Algebra by Herstein.

The problem is show that $\sqrt{2}+\sqrt[3]{5}$ is algebraic over $Q$ of degree 6.

If I can prove the proposition $[F(\alpha+\beta):F(\beta)]=[F(\alpha):F(\beta)]$, then I can solve this question. However, now it seems impossible to do so.

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  • $\begingroup$ Are you implying that $\beta \in F(\alpha)$? $\endgroup$ – Arthur Feb 11 '15 at 0:11
  • $\begingroup$ @Arthur No. To let the question be more concrete, you can let $\alpha=\sqrt{2}$ and $\beta=5^$(one over three) $\endgroup$ – scitamehtam Feb 11 '15 at 0:17
  • $\begingroup$ To define the degree of a field extension such as $[K:F]$, you have to have $F$ being a subfield of $K$. $\endgroup$ – Geoff Robinson Feb 11 '15 at 0:25
  • $\begingroup$ @Geoff Robinson Alright. Get your point. $\endgroup$ – scitamehtam Feb 11 '15 at 0:30
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We can show this quickly by just showing that $F(\alpha+\beta)=F(\alpha)$, as field extensions over $F(\beta)$.

We first show that $\alpha\in F(\alpha+\beta)$. As $F(\alpha+\beta)$ contains $\beta$ by being an extension over $F(\beta)$, $\alpha+\beta-\beta=\alpha\in F(\alpha+\beta)$.

A symmetric argument holds for the other direction.

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  • $\begingroup$ Well..Why $F(\alpha+\beta)$ contains $\beta$? $\endgroup$ – scitamehtam Feb 11 '15 at 0:15
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    $\begingroup$ If it doesn't, you can't define $[F(\alpha + \beta):F(\beta)]$. $\endgroup$ – Geoff Robinson Feb 11 '15 at 0:21
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Counter-example: $F=\mathbb{Q}$, $\alpha=\sqrt2$ and $\beta= \sqrt3$. We have $[\mathbb{Q}(\sqrt2+\sqrt3):\mathbb{Q}(\sqrt3)]=2,$ but $[\mathbb{Q}(\sqrt2):\mathbb{Q}(\sqrt3)]$ is not defined.

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