# Why are $p$-adic integers integers

When constructing the $$p$$-adic numbers, we proceed for instance as when constructing $$\mathbb{R}$$ for the usual distance. Then the integers are king of natural", we are used to them (are we can see them as the rational algebraic numbers, but this also relies on the specific choice of $$\mathbb{Z}$$ which I cannot motivate, except maybe as the additive group generated by $$1$$?)

In the case of $$p$$-adic numbers, the same process with the $$p$$-adic topology leads to $$\mathbb{Q}_p$$. However, it is often defined right after that $$\mathbb{Z}_p$$ is the ring of $$p$$-adic integers, e.g. as the unit ball or equivalently as the Laurent series that are power series in $$p$$.

Is there a deeper definition of $$\mathbb{Z}_p$$? Can we see them as the algebraic integers of $$\mathbb{Q}_p$$ for instance? Why are they termed $$`$$ìntegers", beyond the analogy of the positive powers appearing only in archimedean integers?

• $\Bbb{Z}_p$ is the completion of $\Bbb{Z}$ for the $p$-adic absolute value. So the choice of notation $\Bbb{Z}_p\subset \Bbb{Q}_p$ is obvious, and once the notation $\Bbb{Z}_p$ is chosen calling it the $p$-adic integers is quite natural. $\Bbb{Z}_p$ is much more analogous to $\Bbb{Z}_{(p)}=(\Bbb{Z}-(p))^{-1}\Bbb{Z}$ than to $\Bbb{Z}$. Commented Feb 12, 2023 at 12:09
• "the specific choice of $\mathbb Z$ which I cannot motivate" -- $\mathbb Z$ is the initial object in the category of unital (commutative) rings. In other words, for every such ring $A$ there is a unique ring homomorphism $\mathbb Z \rightarrow A$. In particular, every ring that contains a subring isomorphic to $\mathbb Z$ contains only one unique such subring, there is no "specific choice" left. Commented Feb 12, 2023 at 16:12

In complex analysis we can speak about meromorphic functions on the plane being holomorphic at a specific point (say, at $$0$$). But we can also speak about the functions meromorphic or holomorphic near a chosen point without those functions being assumed to come from a meromorphic function on the whole complex plane.

That is the kind of analogy Hensel had in mind: meromorphic functions on the plane (or Riemann sphere) are analogous to rational numbers, $$p$$-adic numbers are analogous to meromorphic functions at a point (no assumption they make sense "elsewhere"), and $$p$$-adic integers are analogous to holomorphic functions at a point (no assumptions that they make sense "elsewhere").

By using the label "integer" or "integral" for the valuation ring at a prime (whether it's for the $$p$$-adic valuation on $$\mathbf Q_p$$ or the $$p$$-adic valuation on $$\mathbf Q$$) we can then say that a rational number is an integer (a good old-fashioned ordinary integer: an element of $$\mathbf Z$$) iff it is $$p$$-adically integral for each prime $$p$$. So this sounds like a local-global idea. And this really is how some rational numbers can be proved to be integers: show they are $$p$$-adically integral for each $$p$$. If I want to do this by taking limits in my arguments then I might prefer to be in a complete space when doing this, since limits are better behaved in complete spaces, e.g., I might express my rational number as a $$p$$-adic integral.

Alternatively, the $$p$$-adic integers $$\mathbb{Z}_p$$ can be defined as the projective limit of the sequence $$\cdots\longrightarrow\frac{\mathbb{Z}}{p^3\mathbb{Z}} \longrightarrow\frac{\mathbb{Z}}{p^2\mathbb{Z}} \longrightarrow\frac{\mathbb{Z}}{p\mathbb{Z}},$$ i.e. the ring of sequences $$\{a_n\}_{n\geq1}$$ where $$a_n\in\frac{\mathbb{Z}}{p^{n}\mathbb{Z}}$$ and $$a_{n+1}\equiv a_n\bmod p^n$$.

They are the integers in $$\mathbb{Q}_p$$ in the sense that they are the closure of $$\mathbb{Z}$$ in the $$p$$-adic topology.

In fact, this algebraic definition is independent of that of $$\mathbb{Q}_p$$ in the sense that you could first define $$\mathbb{Z}_p$$ as above, prove that is complete and contains $$\mathbb{Z}$$ as a dense subring, and then obtain $$\mathbb{Q}_p$$ as its field of fractions.

• Thanks for the answer. Are they also the algebraic integers (as roots of polynomials with coefficients in $\mathbb{Z}$ in $\mathbb{Q}_p$? Commented Feb 12, 2023 at 10:13
• @TheStudent: what you can say is the following: suppose $\alpha\in\mathbb{Q}_p$ is the root of a polynomial $P(X)\in\mathbb{Z}[X]$, then $\alpha\in\mathbb{Z}_p$. On the other hand. (1) not all polynomials $P(X)\in\mathbb{Z}[X]$ have roots in $\mathbb{Z}_p$ (e.g. $X^2-p$) and (2) not all $\alpha\in\mathbb{Z}_p$ are algebraic. Commented Feb 12, 2023 at 10:25
• And remember that the algebraic numbers (and a fortiori the algebraic integers) are countable in number, whereas $\Bbb Z_p$ is uncountable. Commented Feb 17, 2023 at 21:58