I'm self studying some Algebraic Number Theory, looking at norms of ideals within rings of integers for some number field.
I know that if we have a principal ideal $I= (a)$, then the norm of the ideal is equal to the absolute value of the norm of the generator $a$.
I am curious about whether the converse is true. That is, if the norm of an ideal is equal to the norm of an element, must it be a generator for that ideal?
I have seen results that say that every ideal in a ring of integers has at most two generators, which makes me think the converse may be false but I can't think how to prove this is the case.
Suppose that you have an ideal $I\subseteq \mathcal O_K$ of norm $N$ where $\mathcal O_K$ is a number ring. Suppose also that $a\in I$ has norm $N$ as well. Then $(a)\subseteq (I)$, which gives you a canonical surjection $\mathcal O_K/(a)\to \mathcal O_K/I$. Now both these quotient rings are finite, and their cardinality are precisely norm of $a$ and norm of $I$. But then this means that the surjection has to be an isomorphism, and this amounts to say that $I=(a)$. So the answer is yes in the case you ask $a\in I$.
On the other hand, if you don't require $a\in I$ then the answer is clearly no, as for example the norm of $1+i\sqrt{5}$ in $\mathbb Z[i\sqrt{5}]$ equals the norm of $1-i\sqrt{5}$, but $1+i\sqrt{5}\notin (1-\sqrt{5})$