The p-adic integers $\mathbb{Z}_p$ can be thought of as a subgroup of the direct product group $P = \prod_{n \geq 1} \mathbb{Z}/p^n\mathbb{Z}$. Are they a direct summand of this group? That is, is the inclusion $\mathbb{Z}_p \hookrightarrow P$ split?

  • 10
    $\begingroup$ Topologically, not: $P$ has dense torsion hence its unique continuous homomorphism to $\mathbb{Z}_p$ is zero. $\endgroup$
    – YCor
    Commented Mar 12, 2015 at 18:27

2 Answers 2


Assuming the axiom of choice, yes. Choose a non-principal ultrafilter on $\mathbb N$. This gives a consistent way to, given a function from $\mathbb N$ to a finite set, choose an element of the finite set, that doesn't depend on any finite subset of $\mathbb N$.

You can take an element of $\prod_{n\geq 1} \mathbb Z/p^n \mathbb Z$ and by modding out by $p^k$ obtain an element of $\prod_{n \geq k} \mathbb Z/p^k$, then by applying the ultrafilter obtain an element of $\mathbb Z/p^k$ (ignroing the first $k-1$ values). Doing this for all $k$ gives an element of $\mathbb Z_p$. Using the standard properties of ultrafilters this is easily seen to be a homomorphism $P \to \mathbb Z_p$ that splits the inclusion.

I'm not sure whether the converse is true, that splitting gives you an ultrafilter.

  • $\begingroup$ Applying the splitting to a vector of all $0$'s and $1$'s will produce either $0$ or $1$. I think that lets you construct an ultrafilter. $\endgroup$
    – user13113
    Commented Mar 13, 2015 at 0:52
  • 1
    $\begingroup$ @Hurkyl Why won't it produce $2$? $\endgroup$
    – Will Sawin
    Commented Mar 13, 2015 at 1:05
  • 1
    $\begingroup$ Hurkyl's suggestion does work if you demand a splitting which is additionally a ring homomorphism; I'm writing an answer which elaborates on this. $\endgroup$ Commented Mar 13, 2015 at 5:06
  • $\begingroup$ Many thanks for this wonderful answer. $\endgroup$
    – M T
    Commented Mar 13, 2015 at 14:51

Let me give a bit of an algebro-geometric perspective on Will's excellent answer, which I note gives not just a splitting of abelian groups but actually a ring homomorphism as well. Let's write $R=\prod \mathbb{Z}/p^n$ and try to understand $\operatorname{Spec} R$. Suppose we have a field $k$ and a homomorphism $f:R\to k$. For any subset $A\subseteq \mathbb{N}$, let $1_A\in R$ be the element that is $1$ on coordinates in $A$ and $0$ on other coordinates. This element is an idempotent, so $f(1_A)$ must be $0$ or $1$. If we write $U$ for the set of all $A$ such that $f(1_A)=1$, then it is easy to see that $U$ is an ultrafilter.

This defines a continuous map $\operatorname{Spec} R\to \beta\mathbb{N}$, where $\beta\mathbb{N}$ is the space of all ultrafilters on $\mathbb{N}$. The fiber over any principal ultrafilter is easily seen to be a single point, corresponding to the projection from $R$ to a single coordinate followed by the quotient map to $\mathbb{F}_p$. Over a nonprincipal ultrafilter $U$, the fiber can be identified with $\operatorname{Spec} R_U$, where $R_U$ is the quotient of $R$ by the elements $1-1_A$ for $A\in U$; alternatively, $R_U$ is the ultraproduct $\prod_U \mathbb{Z}/p^n$. For any $n$, the canonical map $\mathbb{Z}/p^n\to R_U/p^n$ is an isomorphism (because this property can be expressed in the first-order theory of rings and is true on all sufficiently large coordinates of the product), and so there is a unique map $R_U\to \mathbb{Z}_p$. This map is exactly the splitting which Will describes in his answer, and this shows that Will's splittings are the only possible splittings which are ring homomorphisms. In particular, it is surjective, so we obtain a canonical copy of $\operatorname{Spec} \mathbb{Z}_p$ as a closed subset of the fiber over each nonprincipal ultrafilter $U$.

Note, however, that $\operatorname{Spec} R_U$ is a lot larger than $\operatorname{Spec}\mathbb{Z}_p$. In particular, the kernel of the map $R_U\to \mathbb{Z}_p$ includes all elements such that the divisibility of their coordinates by $p$ goes to infinity with respect to $U$. Any such element for which the divisibility goes to infinity slower than $n^{1/k}$ for all $k$ will be non-nilpotent, and in fact it is not hard to construct uncountably many prime ideals in $R_U$ corresponding to different rates of convergence to infinity (compare with my example at the end of this answer). However, we can at least say that every prime ideal of $R_U$ either comes from $\mathbb{Z}_p$ or is contained in the kernel of $R_U\to\mathbb{Z}_p$: any element not in the kernel is of the form $p^nu$ where $u$ is a unit, and $p$ generates the unique maximal ideal of $R_U$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.