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?



Are the p-adics a direct summand of the direct product of the groups $\mathbb{Z}/p^n\mathbb{Z}$?
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.


