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Given a field extension $L/K$, $\alpha, \beta \in L$ and $f,g \in K[x]$ irreducible polynomials with $f(\alpha)=g(\beta)=0$. Then

$$ \operatorname{dim}_K(K(\alpha,\beta)) = \deg(f) \cdot \operatorname{dim}_{K(\alpha)}(K(\alpha,\beta)) = \deg(g) \cdot \operatorname{dim}_{K(\beta)}(K(\alpha,\beta))$$

holds. I have no idea how to prove that as the degree of the polynomial because the dimension of the splitting field has to only a divisor of the degree but not the exact product.

$f \in K(\beta)[x]$ is ireducible iff $g \in K(\beta)[x]$ is irreducible. Any ideas how to prove that?

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Do you know about the multiplicativity formula $[M:K]=[M:L][L:K]$ where $M|L$ and $L|K$ are field extensions? –  Gregor Bruns Sep 5 '12 at 16:20
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2 Answers

up vote 3 down vote accepted

Hints: the first hint is what Gregor wrote in the comment (Hagen gave a hint on the proof), the second hint that if $K/F$ is a field extension and $K=F(\alpha)$ where $\alpha$ is algebric over $F$ then $[K:F]=deg(m_{\alpha,F})$ also keep in mind my answer from before that this is the degree of any other irreducible polynomial $p$ that $p(\alpha)=0$

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Hint: If $\gamma_1, \ldots, \gamma_m$ are a basis of $K(\alpha,\beta)$ as a vector space over $K(\alpha)$ and $\delta_1, \ldots, \delta_n$ are a basis of $K(\alpha)$ over $K$, then the products $\gamma_i\delta_j$, $1\le i\le m$, $1\le j \le n$, are a basis of $K(\alpha,\beta)$ over $K$.

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