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Let $L=\mathbb{Q}(\alpha,\beta)$ such that $\alpha^3 + \alpha + 1 = 0$ and $\beta^2 + \beta -3 = 0$. Find $[L:\mathbb{Q}]$.

I'm having problems with this question because when we think of $L=\mathbb{Q}(\alpha,\beta)$, it should be $\mathbb{Q}(\alpha)(\beta)$, right? And $\mathbb{Q}(\alpha)(\beta)$ is the intersection of all fields $K$ that contain $\mathbb{Q}(\alpha)$ and $\beta$, right? But what is $\beta$? There are multiple solutions for $\alpha$ and \beta$ to begin with.

I also think that the solution to this has something to do with finding the degree of $\mathbb{Q}(\alpha,\beta)$ over $\mathbb{Q}(\alpha)$ or $\mathbb{Q}(\beta)$ and then finding the degree of this over $\mathbb{Q}$, but I don't even know how $\mathbb{Q}(\alpha,\beta)$, $\mathbb{Q}(\alpha)$ and $\mathbb{Q}(\beta)$ looks like

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    $\begingroup$ Since both $x^3+x+1, x^2+x-3$ are irreducible over $\mathbb{Q}$, and their degree are relatively prime, so $[L:\mathbb{Q}] = 6$. $\endgroup$ – pisco Sep 20 '17 at 5:45
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More of a comment: indeed, since the first extension is not normal, it's not clear what is the choice of $\alpha$. As for the second, once $\alpha$ is chosen, any choice of $\beta$ produces the same extension. However, in this case, you have two extensions of relatively prime degrees. So just make a choice, $\alpha$ and $\beta$, and look at the degrees of intermediate extensions.

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