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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 $\hat{\mathbb{Z}}$?

Mike Battaglia
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2 Answers2

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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.

Martin Brandenburg
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    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). –  Mar 19 '13 at 13:26
  • Yes, I agree (for $p \neq q$ of course). – Martin Brandenburg Mar 19 '13 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. – Mike Battaglia Mar 20 '13 at 09:11
  • 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? – Mike Battaglia Mar 20 '13 at 09:14
  • As an abelian group, $\mathbb{Z}[[X]]$ is an infinite product of copies of $\mathbb{Z}$, and tensor products do not commute with infinite products (or rather this happens only rarely). I have never heard about the isomorphism $\mathbb{Z}[[X]]/(X-p) \cong \mathbb{Z}p$ so far. This is really cute. Can you name a reference for this? For the tensor product I can only remark that $\mathbb{Z}_p \otimes{\mathbb{Z}} \mathbb{R} \cong \mathbb{Q}p \otimes{\mathbb{Q}} \mathbb{R}$, which is an incredibly large $\mathbb{Q}$-algebra. – Martin Brandenburg Mar 20 '13 at 13:58
  • Interesting. Sounds like the ring I was really curious about, $\mathbb{R} \otimes_\mathbb{Z} \mathbb{Z}2 \otimes\mathbb{Z} \mathbb{Z}3 \otimes\mathbb{Z} \mathbb{Z}_5 ...$ for all p-adic integer rings, is going to be rather messy then. – Mike Battaglia Mar 20 '13 at 21:14
  • I'm not sure where I first came across the theorem regarding formal power series, but a bunch of stronger results are proven in this thesis here. Page 38, example 3.2.1, lists the case of $\mathbb{Z}[[X]]/(X-p)$ as an implication of some of these stronger results. – Mike Battaglia Mar 20 '13 at 21:14
  • Actually, I did some searching up on infinite tensor products and found one of your MathOverflow posts here which seems to apply to the ring I've described, so maybe the stuff you worked out might give me a clue. (I have no idea if the tensor product I'm describing works out the same as $\mathbb{R} \otimes_\mathbb{Q} \mathbb{Q}2 \otimes\mathbb{Q} \mathbb{Q}3 \otimes\mathbb{Q} \mathbb{Q}_5 ...$ though, as I don't have much intuition for these sorts of infinite tensor products). – Mike Battaglia Mar 20 '13 at 21:59
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    Dear Martin, In case you are still interested, this answer describes the isomorphism you asked about. (See also here.) Cheers, – Matt E Dec 29 '13 at 16:45
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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.

Matt E
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