Is there an algorithm that finds, given a subgroup $G$ of a finite permutation group $S_n$, a quadratic number field $K=Q(\sqrt a), a \in Q$ and a Polynom $f \in K[x]$ such that the Galois Group of the splitting field $L$ of $f$ has $Gal(L/K)=G$ or fails if no such field exists? Are anyhow any finite groups known that cant be Galois groups of quadratic number fields?
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7$\begingroup$ I am pretty sure that for fixed base fields questions like these are wide open. See en.wikipedia.org/wiki/Inverse_Galois_problem . $\endgroup$– Qiaochu YuanJul 27, 2011 at 15:49
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1$\begingroup$ [All the "editing" I did was to correct the spelling of "Galois" in the title.] $\endgroup$– Noam D. ElkiesJul 27, 2011 at 16:48
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1$\begingroup$ If by `fails' you allow it to never stop, then yes. Everything in sight is countable, and so you just list all possible polynomials, and check (one by one) whether they give you $G$ as the Galois group. Conjecturally, this algorithm always terminates. Of course, it is terrible! Given a polynomial, it is difficult to compute the Galois group. Conversely, it is occasionally easy to find polynomials for certain groups, but currently not for all groups (even over $\mathbb{Q}$ much less a quadratic field). $\endgroup$– Pace NielsenJul 27, 2011 at 16:54
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2$\begingroup$ Also, the title is a little misleading: of course it is not very hard to compute the Galois groups of quadratic number fields... $\endgroup$– Qiaochu YuanJul 27, 2011 at 16:58
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1$\begingroup$ by fail I meant fail after a finite time, since quadratic number fields have a close connection wiht CM elliptic curves I think, (or naivly hope) that it might be easier to find such extension fields than over Q, since more structure is their? Why is it nevertheless easier over Q to calculate Galois gropups? $\endgroup$– Markus UlkeJul 27, 2011 at 17:14
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