# “Prime decomposition” for the profinite integers, adeles, or p-adics?

The strictly positive integers can be decomposed into a product of prime powers. Likewise, the positive rationals can be decomposed into a product of (possibly negative) prime powers.

Another way to state this is that the underlying multiplicative monoids are free and admit the primes as a basis.

If negative integers or rationals are allowed, then we no longer get a free monoid. But, we can still get a suitable prime decomposition if we create an additional "basis" element, representing -1, that takes exponents in $\Bbb Z/ 2 \Bbb Z$.

I'm wondering if a similar sort of decomposition exists for any of the more interesting structures thrown around in number theory, namely

1. The profinite integers, and/or its group of units
2. The ring of adeles, and/or its group of units
3. The field of p-adic numbers for some p

I'm particularly interested in the case where the "basis" is countable, like the primes themselves.

Now, since these are all uncountable, it's clear that a decomposition into a direct sum of countable groups won't work. But perhaps we can allow infinitely many non-zero exponents, or allow the exponents to take values in some interesting group. Does anything like this work?

The supernatural numbers seem related here, but I can't tell exactly what the relationship is, or if they embed into one of the groups mentioned (or vice versa).

## 1 Answer

There is a notion of unique factorization domain, which is a commutative integral domain whose nonzero elements admit unique prime factorizations up to units. (If we don't ignore units then there's no hope of getting uniqueness.)

The $p$-adic integers $\mathbb{Z}_p$ have this property mostly by virtue of having a lot of units: every $p$-adic integer is uniquely of the form $p^n$ times a unit. In fact $p$ is the only prime in the $p$-adics! This reflects the fact that $\mathbb{Z}_p$ is a local ring.

The group of units is well understood. Let me assume that $p$ is odd because something annoying happens when $p = 2$. The group of units is isomorphic to $(\mathbb{Z}/p\mathbb{Z})^{\times}$ times the subgroup of units congruent to $1 \bmod p$. This subgroup is in turn isomorphic to the additive group $\mathbb{Z}_p$, with the isomorphism given by the exponential

$$\mathbb{Z}_p \ni a \mapsto (1 + p)^a \in \mathbb{Z}_p^{\times}.$$

In particular, it is "topologically generated" by $1 + p$.

The profinite integers and the adeles don't fall into this framework because they are not integral domains. In a ring that's not an integral domain it's very unclear what one ought to mean by unique prime factorization of a zero divisor. Nevertheless, since the multiplicative monoid and group of units functors both send products to products, once you've understood what's going on for $\mathbb{Z}_p$ you more or less understand what's going on in these more complicated cases.

• The rationals aren't a UFD, but they admit the sort of structure I'm talking about, right? Likewise with the supernatural numbers. – Mike Battaglia Sep 22 '15 at 23:52
• Sorry, to be technically correct, fields ARE UFD's -- but everything's a unit! Still... – Mike Battaglia Sep 22 '15 at 23:54
• I guess the thing I'm really interested in is finding a decomposition for the group of units, as a starting point. – Mike Battaglia Sep 23 '15 at 2:36
• @Matt: I give such a decomposition above for the $p$-adic integrs, and from there you can infer what's going on for the profinite integers, the adeles, and the $p$-adic numbers. – Qiaochu Yuan Sep 23 '15 at 2:38
• @Mike: yes, it's different. In the usual embedding a natural number $n$ gets sent to (for each prime $p$) $p^k$ times whatever $p$-adic unit $\frac{n}{p^k}$ ends up being (where $p^k$ is the greatest power of $p$ dividing $n$). – Qiaochu Yuan Sep 23 '15 at 5:38