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If $K$ is a number field, a result from Bach tells us that the primes in $K$ of norm smaller that $12 (\log |\mathrm{Discriminant}(K)|)^2$ generate the ideal class group $\mathrm{Cl}_K$. Is there any known bound if we require the generating primes to be unramified in $K/\mathbb Q$? (i.e. a result of the form "prime ideals in $K$ above the prime numbers unramified in $K/\mathbb Q$ and smaller than $f(K)$ generate the ideal class group")

In the particular case I study, $[K:\mathbb Q] = 4$. With algebraic methods I can find a bound polynomial in the class number $h_K$ and logarithmic in $|\mathrm{Discriminant}(K)|$, but I am sure we can get rid of the $h_K$, or at least change it for a $\log h_K$, but I'm afraid this would require analytic methods...

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First, do not forget that Bach's result is conditional to GRH.

What Bach proves is that for every nontrivial character $\chi$ of the ray class group of $K$ of conductor $\mathfrak{f}$, there is a prime $\mathfrak{p}$ of norm less or equal to $3\log(\Delta_K^2N\mathfrak{f})^2$ such that $\chi(\mathfrak{p})\neq 1$. In particular, by taking $\mathfrak{f}$ the different of $K$, you get the result you want with a bound $27\log(\Delta_K)^2$.

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  • $\begingroup$ Is there also any way to control the residue degree of the generating primes ? For example if we want all the generators to split completely. It seems to be easier to control the ramification since there are only finitely many ramified primes... $\endgroup$ – Calodeon Mar 24 '14 at 22:36
  • $\begingroup$ You can ask for only degree $1$ primes, the constant is multiplied by $6$. ams.org/journals/mcom/1990-55-191/S0025-5718-1990-1023756-8 $\endgroup$ – Aurel Mar 24 '14 at 22:41
  • $\begingroup$ This is exactly what I need, merci beaucoup ! $\endgroup$ – Calodeon Mar 24 '14 at 22:46
  • $\begingroup$ De rien, my pleasure ! :) $\endgroup$ – Aurel Mar 24 '14 at 23:12
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    $\begingroup$ Ok I have had a closer look (and helpful comments from Eric Bach) : the first bound (3) is with option 1 only, so we don't suppose $(\mathfrak p,\mathfrak f) = 1$. The second bound (12) is with option 1 and 2, i.e. we suppose $(\mathfrak p,\mathfrak f) = 1$. The last bound (18) is with option 1, 2 and 3, i.e. $(\mathfrak p,\mathfrak f) = 1$ and $\deg(\mathfrak p) = 1$. $\endgroup$ – Calodeon Mar 31 '14 at 22:13

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