# Tensor products of p-adic integers

These are relatively simple questions, but I can't seem to find anything on tensor products and p-adic integers/numbers anywhere, so I thought I'd ask.

My first question is: given some $\mathbb{Z}_p$, $\mathbb{Q}_p$ can be constructed as its field of fractions. Is the tensor product $\mathbb{Z}_p \otimes \mathbb{Q}$ also equal to $\mathbb{Q}_p$?

My second question is: given some composite p, you can still construct a ring (now with zero divisors) that can be constructed as the inverse limit of $\mathbb{Z}/p^n\mathbb{Z}$. Any such p has a finite prime factorization, so is there a way to construct this non-integral-domain ring of composite p-adics by taking some sort of product of the rings of p-adics of its prime factors? For instance, can you construct the 10-adics by taking the direct product of the 2-adics and the 5-adics, or perhaps is it the tensor product, or...?

My last question is: I haven't seen much about taking tensor products about p-adic integers in general; I've only seen stuff about taking direct products, as in the case of the profinite completion $\hat{\mathbb{Z}}$ of the integers. However, I find the tensor product to be of particular interest, since it's a coproduct in the category of commutative rings. So what, in general, do you get if you take the tensor product of two rings of p-adic integers? And I'm especially curious to know, what do you get if you take the tensor product of all of the rings of p-adic integers, rather than the direct product in the case of $\hat{\mathbb{Z}}$?

• In the last question, do you mean tensor product of $\mathbb Z_p$ for various $p$ ?
– user18119
Mar 19, 2013 at 12:07
• Short answer to the first question: $\mathbb{Z}_p[p^{-1}] = \mathbb{Q}_p$.
– user14972
Mar 19, 2013 at 14:35
• QiL'8: Yes, that's exactly right. Mar 20, 2013 at 5:09
• Hurkyl: Ah, quite interesting! Thanks for that information. Mar 20, 2013 at 5:48

1) Yes. $\mathbb{Z}_p \otimes_{\mathbb{Z}} \mathbb{Q}$ is the localization $(\mathbb{Z}\setminus \{0\})^{-1} \mathbb{Z}_p$. Thus, elements have the form $a/b$ with $a \in \mathbb{Z}_p$ and $b \in \mathbb{Z} \setminus \{0\}$. Clearly this is a subring of $\mathbb{Q}_p$. In order to show that it is the whole of $\mathbb{Q}_p$, it suffices to prove that it contains all $1/u$ for $u \in \mathbb{Z}_p \setminus \{0\}$, i.e. that there are $a,b$ as above satisfying $b=ua$, i.e. that $u$ divides some positive integer in $\mathbb{Z}_p$. But actually $u$ is associated to some positive integer, namely to $p^n$ where $n$ is the $p$-adic valuation of $u$.

Actually this shows that already the localization at the element $p$ gives $\mathbb{Q}_p$. More generally, if $R$ is a DVR with uniformizer $\pi$, then $R_{\pi}=Q(R)$.

2) Yes, If $n$ is any positive integer, you can define $\mathbb{Z}_n := \varprojlim_k~ \mathbb{Z}/n^k$, the $n$-adic completion of $\mathbb{Z}$. The Chinese Remainder Theorem gives $\mathbb{Z}_{nm} \cong \mathbb{Z}_n \times \mathbb{Z}_m$ for coprime $n,m$, and we have $\mathbb{Z}_{n^v}=\mathbb{Z}_{n}$ for $v>0$ since limits of cofinal subsystems agree. Thus, if $n = p_1^{v_1} \cdot \dotsc \cdot p_n^{v_n}$ is the prime decomposition of $n$ with $v_i > 0$, then $\mathbb{Z}_n \cong \mathbb{Z}_{p_1} \times \dotsc \times \mathbb{Z}_{p_n}$. In principle one gets nothing new.

3) I don't think that there is a nice description of $\mathbb{Z}_p \otimes_{\mathbb{Z}} \mathbb{Z}_q$. The tensor product behaves well for finite products and all colimits, but $\mathbb{Z}_p$ is an infinite projective limit. So you should better consider $\mathbb{Z}_p \widehat{\otimes} \mathbb{Z}_q$, some completed tensor product, having in mind that the $p$-adics form a (very nice) topological ring. I suspect that this is a ring which has not been considered in the literature, but I am not sure ...

Perhaps someone else can add a reference on the tensor product of topological rings, because I could only find this for topological $\mathbb{C}$-algebras.

• I think the completed tensor product will be zero (because an arbitrary simple tensor $a \otimes b = p^n a \otimes p^{-n}b$ is arbitrarily small).
– user29743
Mar 19, 2013 at 13:26
• Yes, I agree (for $p \neq q$ of course). Mar 19, 2013 at 13:35
• Thanks, great answer. Some questions - first, just curious, do you know if the ring $\mathbb{Z}_p \otimes \mathbb{R}$ has any sort of nice interpretation as well? I can't seem to find much information on p-adics and tensor products anywhere I look. Mar 20, 2013 at 9:11
• Secondly, I'm not sure I understand your answer in #3. One way to construct $\mathbb{Z}_p$ is as an inverse limit, but another way is to just start with the ring of formal power series $\mathbb{Z}[[X]]$ and quotient out by the ideal generated by $X-p$. This ring of formal power series can be constructed as taking the set of functions $\mathbb{Z^N}$ and equipping it with pointwise addition and discrete convolution, and now once you take the quotient you've got $\mathbb{Z}_p$ without having used limits at all in the construction. So why should the tensor product not behave well? Mar 20, 2013 at 9:14
Since $\mathbb Z$ embeds into $\mathbb Z_q$, we see that $\mathbb Z_p$ embeds into $\mathbb Z_p\otimes\mathbb Z_q$. (Here we use that $\mathbb Z_p$ is flat over $\mathbb Z$, being torsion free.)
Now multiplication by any number coprime to $p$ induces a bijection n $\mathbb Z_p$, and hence a bijection on $\mathbb Z_p\otimes\mathbb Z_q$. On the other hand, multiplication by $p$ induces a bijection on $\mathbb Z_q$, and hence a bijection on $\mathbb Z_p\otimes \mathbb Z_q$.
Thus mult. by any non-zero integers induces a bijection on $\mathbb Z_p\otimes\mathbb Z_q$, and so the $\mathbb Z$-module structure on $\mathbb Z_p\otimes \mathbb Z_q$ can be promoted to a $\mathbb Q$-v.s. structure.
In other words, $\mathbb Z_p\otimes\mathbb Z_q$ is naturally a $\mathbb Q$-algebra. Since it contains $\mathbb Z_p$ as a subring, it contains lots of transcendtal elements, as well as many algebraic elements. I'm not sure how much more precise one can be.