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I would like to know in which conditions the extensions $\mathbb{Q}(\alpha,\beta)$ and $\mathbb{Q}(\alpha\beta)$ over $\mathbb{Q}$ are the same.

Thanks in advance.

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I don't think it's productive to seek a complete characterization. In practice, you could do this by computing the characteristic polynomial of $\alpha \beta$ acting on a basis for $\mathbb{Q}(\alpha, \beta)$ and factor it. In special cases, if $\mathbb{Q}(\alpha, \beta)$ is Galois you might be able to explicitly write down the action of the Galois group on $\alpha \beta$. – Qiaochu Yuan Jun 20 '11 at 17:35
a good thing to remember is that almost all c in Q(a,b) are such that Q(a,b) = Q(c). The fact that c=ab works just means that ab isn't contained in one of the finitely many proper subfields of Q(a,b). c=a+b is another popular choice. On the rare occasion when neither ab nor a+b worked, a+2b worked just fine for me. – Jack Schmidt Jun 20 '11 at 17:37
If $\gamma = \alpha \beta$, you're basically asking whether $\alpha \in {\mathbb Q(\gamma)}$ (if so, then $\beta = \gamma/\alpha$ is too, except in the case $\alpha = 0$ which I'll leave you to figure out). – Robert Israel Jun 20 '11 at 17:54
For a similar question where the product is replaced by the sum, see the MO question…. The article "Observations on primitive, normal, and subnormal elements of field extensions" by Steven Weintraub, appearing in Monatsh. Math. 162 (2011) pp. 239-244 might also be helpful. – KCd Jun 20 '11 at 18:10

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