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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 $I$?

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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 $N(I)\mid N(m)$.

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  • $\begingroup$ 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

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