What is the correct analogue of $\mathbb N$ in a ring of integers? Question: Let $K$ be a number field. The proper intuitive motivation for the ring of integers $\mathcal O_K$ is that $\mathbb Z$ is to $\mathbb Q$ as $\mathcal O_K$ is to $K$. But what plays the role of $\mathbb N$ in $K$? Surely the analogue of $\mathbb N$ should be important, since $\mathbb N$ is the fundamental object of study in number theory.
My thoughts: It seems like we're looking for a certain semiring contained in $K$. Let's call it $\mathbb N_K$. 
The main difference in my mind between $\mathbb Z$ and $\mathbb N$ is that $\mathbb N$ is an "efficient" version of $\mathbb Z$ containing exactly one element of each orbit of the action of $\mathbb Z^\times$ on $\mathbb Z$ by multiplication. We can't choose elements from the orbits wily-nily, however, since $\mathbb N$ is required to be closed under addition and multiplication.
We would therefore expect $\mathbb N_K$ to satisfy the following properties:


*

*$\mathbb N\subseteq\mathbb N_K$.

*$\mathbb N_K$ is a semiring (closed under addition and multiplication).

*For every $\alpha\in\mathcal O_K$, there exists a unique unit $u\in\mathcal O_K^\times$ such that $u\alpha\in\mathbb N_K$.


But in general, there is no subset $\mathbb N_K$ satisfying properties 1.-3. (See the case study below.) So either there is some other, more fundamental characterization of $\mathbb N$ that I am missing, or my question has an answer in the negative.
Case Study: Let's try to find $\mathbb N_K$ in the Gaussian integers $K=\mathbb Q(i)$. The element $1+i$ or one of its three other associates ($\alpha$ and $\beta$ are associate if $\alpha/\beta$ is a unit) must be contained in $\mathbb N_K$. If $1+i\in\mathbb N_K$ then so is
$$
(1+i)^2 = 2i,
$$
but since $2\in\mathbb N_K$ we've violated condition 3. And if we had chosen a different associate of $1+i$, the same problem would have arisen. These observations are problematic for the construction of $\mathbb N_K$.
Caveat: It's possible that there is no proper analogue $\mathbb N_K$. If this is the case, a good answer should explain why not.
Random: I have heard it said that in the function field analogy for algebraic number theory, the analogue of $\mathbb N$ is the semiring of monic polynomials.
 A: My conviction is also that the correct analogue of $\mathbb{N}$ for an arbitrary ring of integers $\mathcal{O}_K$ is the semiring of nonzero ideals. This construction satisfies versions of your desired properties: there is indeed a copy of $\mathbb{N}$ (with the usual multiplication but a different addition), given by the principal ideals $(n), n \in \mathbb{N}$, and the construction also ignores units. One way in which this conviction is borne out in practice is that this is what you sum over to define the Dedekind zeta function, which reduces to a sum over $\mathbb{N}$ when $K = \mathbb{Q}$, where you get the Riemann zeta function. 
To my mind, the most important property of $\mathbb{N}$ as a subsemiring of $\mathbb{Z}$ hasn't even been pointed out yet: $\mathbb{Z}$ has a natural order with respect to which $\mathbb{N}$ is the subsemiring of positive elements. This is the order that crucially figures in induction, which is arguably the most important thing we use $\mathbb{N}$ for. 
There's no reason for there to be a good analogue of this in an arbitrary number field, since among other things they don't come equipped with natural orderings. (The story of number fields is really about generalizing aspects of $\mathbb{Z}$ and $\mathbb{Q}$, not $\mathbb{N}$.) One way to equip them with orders is to embed $K$ into $\mathbb{R}$, but sometimes (e.g. when $K = \mathbb{Q}(i)$) there are no such embeddings, and sometimes (e.g. when $K = \mathbb{Q}(\sqrt{2})$) there is more than one. 
A: When studying the natural numbers $\mathbb{N}$ in number theory, the key property that they have is not that $\mathbb{N}$ is an "efficient" version of $\mathbb{Z}$ containing exactly one element of each orbit of the action of $\mathbb{Z}^{\times}$ on $\mathbb{N}$ by multiplication. Rather, what is most important about $\mathbb{N}$ in number theory (especially in multiplicative number theory) is that each element of $\mathbb{N}$ factorises into products of powers of primes (i.e. the fundamental theorem of arithmetic).
So as Zhen Lin commented above, the number field analogue of natural numbers are integral ideals, because they satisfy the same factorisation properties. Similarly, integral ideals have a natural ordering given by the absolute norm, which allows one to generalise the notion of summing over the first $n$ natural numbers, by summing over integral ideals of absolute norm at most $n$.
You may be interested to read about Beurling primes, which generalise this even further by thinking of "natural numbers" as the multiplicative semigroup generated by products of powers of an infinite set of "primes", together with an absolute value. It's an active area of study to see what conditions are needed on these generalised primes to still have an analogous form of the prime number theorem.
