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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 mathoverflow.net/questions/26832/…. 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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