What is $\mathbb{Z}_{p}\otimes_{\mathbb{Z}}\mathbb{R}$ equivalent to?
where $\mathbb{Z}_p$ are the p-adic integers.
I am specially interested in the case $p=2$.
Do know that $\mathbb{Z}_p\otimes_{\mathbb{Z}}\mathbb{Q} \approx \mathbb{Q}_p$ but I think that with $\mathbb{R}$ it's not analogous. Made an extensive search but haven't found anything that even mentions the above.
Thanks for your time.
Since $\mathbb R$ is a $\mathbb Q$-vector space, we have $\mathbb Z_p\otimes_{\mathbb Z} \mathbb R= \mathbb Q_p \otimes_{\mathbb Q} \mathbb R$. So this is the special case of the general phenomenon of a tensor product of two fields.
This is a very large ring about which not much can be said about its structure. First note that for any field $F$ containing subfields isomorphic to both $\mathbb Q_p$ and $\mathbb R$, we get a natural map $\mathbb Q_p \otimes_{\mathbb Q} \mathbb R$ to $F$, whose image is the subring generated by $\mathbb Q_p$ and $\mathbb R$, which will often, but not always, be a field. So this ring has a bunch of different residue fields - a lot isomorphic to $\mathbb C$, but many others.
Next note that we can always find independent transcendentals $\tau_1,\dots, \tau_n$ in $\mathbb Q_p$ and $\alpha_1,\dots,\alpha_m$ in $\mathbb R$, and this tensor product will contain as a subring $\mathbb Q(\tau_1,\dots, \tau_n) \otimes \mathbb Q(\alpha_1,\dots, \alpha_m)$ which is $\mathbb Q[\tau_1,\dots,\tau_n,\alpha_1,\dots,\alpha_m]$ inverting all nonzero polynomials in just the $\tau_i$ and all nonzero polynomials in just the $\alpha_i$. Geometrically, this corresponds to a high-dimensional affine space with all proper "vertical" Zariski-closed sets removed and all proper "horizontal" Zariski-closed set removed. Thinking about this ring, and its geometry, is a good clue to the broader structure of rings like $\mathbb Q_p \otimes_{\mathbb Q} \mathbb R$.
As noticed, the tensor product ${\mathbb Z}_p\otimes_{\mathbb Z}{\mathbb Q}$ is isomorphic to the injective envelope of ${\mathbb Z}_p$, namely ${\mathbb Q}_p$.
The abelian (additive) group $\mathbb R$ is divisible, hence injective, and torsion free. Therefore it is a direct sum of copies of $\mathbb Q$. By cardinality reasons, the number of copies of $\mathbb Q$ is $c=|{\mathbb R}|$. But the tensor product commutes with the direct sums, therefore ${\mathbb Z}_p\otimes_{\mathbb Z}{\mathbb R}$ is a direct sum of $c$ many copies of ${\mathbb Q}_p$.
