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My textbook says:

Find dim$_\mathbb{Q}$ $\mathbb{Q}(\alpha, \beta)$, where:

a. $\alpha^3 = 2$ and $\beta^2 = 2$.

b. $\alpha^3 = 2$ and $\beta^2 = 3$.

So by my calculations, $\alpha$ satisfies a polynomial of degree 3, and $\beta$ satisfies a polynomial of degree 2, so the dimension of the field extension is 6 in both cases. This seems "too easy", though. Have I missed something?

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    $\begingroup$ In general, it is not enough to see that the elements satisfy equations of various degrees. It might be that one of the polynomials factors in the other's extension. $\endgroup$ – Michael Burr Mar 1 '15 at 22:46
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Your argument is indeed ``too easy'', but your conclusion is correct.

A better proof would be: First it's clear that since cube roots of 2 don't exist in $\mathbb{Q}$, $X^3-2$ is irreducible, so $\mathbb{Q}(\alpha)$ has dimension 3 over $\mathbb{Q}$. Since $\mathbb{Q}(\alpha,\beta)$ has at most dimension 2 over $\mathbb{Q}(\alpha,\beta)$, $\mathbb{Q}(\alpha,\beta)$ has dimension at most 6 over $\mathbb{Q}$.

On the other hand, again using the fact that dimensions are multiplicative, ie $$[\mathbb{Q}(\alpha,\beta):\mathbb{Q}] = [\mathbb{Q}(\alpha,\beta):\mathbb{Q}(\alpha)][\mathbb{Q}(\alpha):\mathbb{Q}]$$ and also $$[\mathbb{Q}(\alpha,\beta):\mathbb{Q}] = [\mathbb{Q}(\alpha,\beta):\mathbb{Q}(\beta)][\mathbb{Q}(\beta):\mathbb{Q}]$$ Thus, since $[\mathbb{Q}(\beta):\mathbb{Q}] = 2$ and $[\mathbb{Q}(\alpha):\mathbb{Q}] = 3$, (corresponding to the fact that $X^3-2,X^2-2,X^2-3$ are all irreducible over $\mathbb{Q}$, we must have that $[\mathbb{Q}(\alpha,\beta):\mathbb{Q}]$ is divisible by both 2 and 3, and hence is at least 6.

Since from earlier we knew that it's at most 6, it must be 6.

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