Is this correct? $N(\langle a, b \rangle) = \gcd(N(a), N(b))$

Question

Given a ring $R$, not necessarily a principal ideal domain, and numbers $a, b \in R$, is the norm of the ideal $\langle a, b \rangle$ the greatest common divisor of the norms of the numbers $a$ and $b$?

For example, given that $N(3) = 9$ and $N(1 + \sqrt{-5}) = 6$, is $N(\langle 3, 1 + \sqrt{-5} \rangle) = 3$? Or how about $N(3) = 9$ and $N(1 + \sqrt{10}) = -9$, so $N(\langle 3, 1 + \sqrt{10} \rangle) = 3$ also?

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Related question: Norm of an ideal

Answer 1

In general this is false. A counterezample in $R= \Bbb Z[i]$ follows:

$$a= 2+i \qquad b= 2-i$$ are prime elements since $N(a)=N(b)=5$. Note that $\gcd (N(a), N(b))=5$.

However they are not associate each other, since $$ab^{-1} = \frac{3}{5}- \frac{4}{5}i \notin R$$

In particular they are coprime, i.e. $$\langle a,b \rangle = R$$ Which implies $$N( \langle a, b \rangle)=1 \neq 5$$

The only thing you can say is that

$N( \langle a, b \rangle)$ divides both $N(a)$ and $N(b)$

i.e. it divides their $\gcd$.

Answer 2

My answer at https://mathoverflow.net/questions/26549/is-there-much-difference-between-kroneckers-and-dedekinds-methods-in-algebraic describes a way to compute the norm of a nonzero ideal in the ring of integers of a number field in terms of a given set of generators of the ideal.