Let $F$ be a number field and $I$ be a nontrivial ideal of the ring of integers. Show that the norm $N_{F/\mathbb Q}(I)$ has the same prime factors as the smallest positive integer in $I$.

We have $N(I) = \prod_{i=1}^n {p_i} ^ {e_i}$ a factorization of rational primes, $I = \prod_{i=1}^n P_i$ of prime ideals. $O_F \cap P_i = (p_i)$. Maybe something can be done with the prime factorization of the smallest positive integer in $I$?


Let $m$ be the smallest positive integer in $I$. Then $I\cap\mathbb Z=m\mathbb Z$. Furthermore, $N(I)\in I$ and $N(I)=[\mathcal O_F:I]$ entail $N(I)\in m\mathbb Z$, so $m\mid N(I)$.

On the other side, since $m\in I$ we have $I\mid (m)$, that is, $(m)=IJ$. Then $N(m)=N(I)N(J)$, so $N(I)\mid N(m)$.

  • $\begingroup$ How do we know there aren't any more prime factors $\endgroup$ – mathdragon Oct 25 '15 at 7:57
  • $\begingroup$ $N(m) = m^a$, where $a= [K: \mathbb{Q}]$, so no new prime factors are being added. $\endgroup$ – Aranya Lahiri Oct 25 '15 at 12:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.