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In the ring $$\textbf{Z}[i],$$ if the norm of an element is divisible by $2$, then the element must be divisible by $$1 + i,$$ and vice versa. A similar result holds for $$\textbf{Z}[\sqrt 3]$$ and divisibility by $$1 + \sqrt 3.$$ My question is this:

Consider a field and its ring of integers $$\mathcal{O}_K.$$ If we have an element of norm $2$, does that imply that all elements of even norm are divisible by it?

My question was inspired by this line in David Speyer's proof, namely the part where he concludes for which elements $$\sigma_2(a)$$ is plus or minus $1$.

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This is false since we might have more than one element of norm $2$. For example, consider the field $\mathbb{Q}(\alpha)$, where $\alpha$ is a root of $x^3-3$. Then $a_1=1-\alpha$ and $a_2=1+\alpha+\alpha^2$ both have norm $2$ but neither divides the other.

In general, if we assume $O_K$ is a unique factorisation domain (which it rarely is, but then we should talk about ideals), then one of three things can happen. Firstly, $2$ can remain irreducible in $O_K$, in which case there is no element of norm $2$, but any element with even norm is divisible by $2$. Secondly, we have $2=u \cdot a^k$ for some irreducible element $a$ and unit $u$, in which case your statement also holds.

Lastly though, we can have $2=a_1\cdots a_n$ for some irreducible elements $a_1,\cdots a_n$. Then if the norm of an element is even, then we can say that one of the $a_i$ must divide your element, but we don't know which one.

On the ideals side, these cases correspond to totally inert, totally ramified (in which case you only get one element of norm $2$ so the argument works), and split. In the split case, we can allow ramification but inertia will prevent us from getting an element of order $2$ for that particular prime.

Lastly, I should add that there was nothing special about $2$ here; we can use the same argument for any rational prime $p$.

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