Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm new user here! I'm stuck on the following and I was wondering if some NT experts can prove some help.

Question (from an entrance exam in my country) 5) What is the density of the rational primes $p$ that factor as $(p)=p_{1}p_{2}$ in $\mathbb{Q}[\sqrt[3]{2}]$. What about $\mathbb{Q}[\sqrt[5]{2}]$?

I think Chebotarev is the answer but I kno only how to use it to get the density of the primes that split completely in those number fields.

share|improve this question
1  
Ignore the ramified primes. If $p$ splits as $p_1p_2$ in $K=\mathbf{Q}(\root 3\of 2)$, then you must have $f(p_i|p)=i$ for $i=1,2$. So when you go the normal closure $L=K(\omega), \omega=(-1+\sqrt{-3})/2$ you must get $f=2, g=3$. It shouldn't be too difficult to prove that all the primes with $f=2,g=3$ in $L/\mathbf{Q}$ behave in this way. Does that help? –  Jyrki Lahtonen Jan 12 '12 at 19:19
1  
Unless I made a mistake in $\mathbf{Q}(\root 5\of 2)$ $(p)=p_1p_2$ implies that the respective inertia degrees must be 1 and 4. As they must agree after extension to $\mathbf{Q}(\root 5\of 2,\zeta_5)$, we can deduce that $f=4$. So the decomposition groups must be cyclic subgroups of order 4 of the Galois group $$F_{20}=C_5\rtimes C_4=\langle s,t\mid s^5=1=t^4, tst^{-1}=s^2\rangle.$$ –  Jyrki Lahtonen Jan 12 '12 at 19:29

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.