The $i^{th}$ prime in a given ring R

Question

When I say that $p_1=2$, I mean that the first prime in the standard ring of integers $(\mathbb{Z},*,+)$ is $2$. I was wondering whether the notion of ordering the primes like this can be generalized for any ring $R$. First of all, is it even possible to consistently equip a general ring (such as a finite one) with an order? If so, how would the labelling be worked out on primes? (Note that any unit and 0 are not being counted as primes.)

Answer 1

To name the first, second, third, etc. prime will only work under two fairly special conditions. First, your ring $R$ has to be ordered, or maybe have some reasonable measure of “size” of an element. Second, remember that when $p$ is a prime element, there will usually be some other “equally good” primes that look nothing like $p$.

Maybe an example will help. Consider the Gaussian Integers, that is all $z=m+ni$ where $m$ and $n$ are ordinary integers. They can’t be ordered, but you do have the distance from the origin in the Gaussian (Argand, Wessel) plane, so there’s no question, the closest prime to the origin is $p_1=1+i$. But what about $ip_1=-1+i$? It’s equally “good”, except that it isn’t in the first quadrant. But for various reasons, demanding that your prime should be in first quadrant is a little artificial. The “next” prime is $1+2i$, but there’s another entirely different prime in the first quadrant that’s just as far from the origin, namely $2+i$.

Things get messier still, even when you have an ordering, as with the ring $\mathbb Z[\sqrt2\,]$, which is all things of the form $z=m+n\sqrt2$, because there are infinitely many “units” (they behave like the units $i^k$ in the Gaussian integers). There may be a fairly artificial way of specifying which way you want to specify which of the “equally good” primes you want to name, but never very satisfactory.

I’ve only touched on the complications that may arise, and there will undoubtedly be people who see different difficulties than I have here. But the whole question is really interesting, and it does repay study.

Answer 2

If by ''consistently'' you mean "in a canonical way which behaves well with respect to the ordering of $\mathbb Z$", then the obvious answer is "no".

Suppose you have a finite field extension of $\mathbb Q$, call it $K$. Then after doing a bit of algebraic number theory you learn that there is a ring $A$ containing $\mathbb Z$ which has essentially all the nice properties you want concerning prime ideals (called the ring of integers of $\mathbb Q(i)$). In particular, if the class group of $K/\mathbb Q$ is $1$, the ring $A$ is a principal ideal domain, so that the non-zero prime ideals and the prime elements of $A$ are essentially the same thing. Then the ordering you desire is equivalent to look above each prime $p$ in $\mathbb Z$ and fix an ordering in the primes $q$ lying above $p$, and of course there is no canonical way of doing that.

To illustrate what I mean, consider $K = \mathbb Q(i)$, so that the ring $A$ described above is $\mathbb Z[i]$. It is well-known that if $p \in \mathbb Z$ is prime, then $$ (p)_{\mathbb Z[i]} = (a-bi)(a+bi) $$ when $p \equiv 1 \pmod 4$, and when $p \equiv 3 \pmod 4$ then $p$ is also a prime element in $\mathbb Z[i]$ ; furthermore, $(2)_{\mathbb Z[i]} = (1+i)^2$. So an ordering on $\mathrm{Spec}(\mathbb Z[i])$ is essentially equivalent to fixing $a+bi \le a-bi$ or the other way around. The reason why you should believe that there is no canonical way of doing that is because $\mathbb Z[i] \simeq \mathbb Z[-i]$ ; if there was a canonical way of doing it, morally, it should be preserved under isomorphism (or relabelling of the roots if you prefer). But of course, set-theoretically speaking there is nothing wrong in labelling each of the primes lying above each prime $p$ in $\mathbb Z$ and making it consistent, but then it becomes quite ad hoc and practically not so useful in my opinion.

Hope that helps,