# $\mathbb{Q}_p\otimes_{\mathbb{Q}} \mathbb{Q}_q$ and $\mathbb{Z}_p\otimes_{\mathbb{Z}} \mathbb{Z}_q$

Let $p, q$ be prime numbers which may or may not be distinct. Let $\mathbb{Q}_p$ be the field of $p$-adic numbers. Let $\mathbb{Z}_p$ be the ring of $p$-adic integers. We define similarly $\mathbb{Q}_q$ and $\mathbb{Z}_q$.

Let $A = \mathbb{Q}_p\otimes_{\mathbb{Q}} \mathbb{Q}_q$. Let $\lambda\colon \mathbb{Z}_p \rightarrow A$ and $\mu\colon \mathbb{Z}_q \rightarrow A$ be the canonical ring homomorphisms. Let $B = \mathbb{Z}_p\otimes_{\mathbb{Z}} \mathbb{Z}_q$. Let $\psi\colon B \rightarrow A$ be the ring homomorphism induced by $\lambda$ and $\mu$.

Is $\psi$ injective?

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This is an interesting question (I'll give it some thought). Any motivation for where it came from, besides the obvious "it's a semi-obvious question to ask"? –  Alex Youcis Feb 27 '14 at 10:05
@AlexYoucis math.stackexchange.com/questions/684146/… –  Makoto Kato Feb 27 '14 at 10:24
@Makato: You don't injectivity in order to prove the (trivial) claim about the localization in the other question. –  Martin Brandenburg Feb 27 '14 at 22:03

Since $A$ is the localization of $B$ at the elements $p$ and $q$ (by formal nonsense), the question is if $p$ and $q$ are regular elements of $B=\mathbb{Z}_p \otimes \mathbb{Z}_q$. But this is because $p$ is a regular element of $\mathbb{Z}_p$ and $\mathbb{Z}_q$ is torsion-free and hence flat over $\mathbb{Z}$; similarly for $q$.

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All rings will be assumed to be commutative with unity.

Lemma 1 Let $A$ be a ring, $M, N$ $A$-modules. Suppose $N$ is flat over $A$. Let $a$ be an element of $A$. Suppose the map $f\colon M \rightarrow M$ defined by $f(x) = ax$ is injective. Then the map $g\colon M\otimes_A N \rightarrow M\otimes_A N$ defined by $g(z) = az$ is also injective.

Proof: Let $1_M\colon M \rightarrow M$ be the identity map. Since $ax\otimes y = a(x\otimes y)$ for $x \in M, y \in N$, $g = f\otimes 1_M$. Hence $g$ is injective.

Lemma 2 Let $A$ be a ring, $N$ an $A$-module. Let $(M_i)_{i\in I}$ a direct system of $A$-modules. Then colim$_i (M_i\otimes_A N) =$ (colim$_i M_i)\otimes_A N$.

Proof: This is well known. See, for example, Matsumura's Commutative Ring Theory, Appendix A1.

Remark Lemma 2 can be proved by using category theory as follows. $-\otimes_A N$ is a left adjoint functor of Hom$_A(N, -$)(see my answer to this question). Hence it commutes with colim(see MacLane: Categories for the working mathematician, Chapter V, Section 5, Theorem 1, p.114).

Lemma 3 Let $A$ be a ring. Let $(M_i)_{i\in I} \rightarrow (N_i)_{i\in I}$ and $(N_i)_{i\in I} \rightarrow (L_i)_{i\in I}$ be maps of direct sytems of $A$-modules. Suppose each sequence $M_i \rightarrow N_i \rightarrow L_i$ is exact. Then colim $M_i \rightarrow$ colim $N_i \rightarrow$ colim $L_i$ is exact.

This is well known. See, for example, Matsumura's Commutative Ring Theory, Appendix A2.

Lemma 4 Let $A$ be a ring. Let $(M_i)_{i\in I}$ a direct system of flat $A$-modules. Then colim $M_i$ is flat.

Proof: This follows immediately from Lemma 2 and Lemma 3.

Lemma 5 Let $G$ be a torsion-free abelian group. In other words, the order of every nonzero element of $G$ is infinite. Then $G$ is flat. In other words, the functor $-\otimes_{\mathbb{Z}} G$ is exact.

Proof: Let $(G_i)_{i \in I}$ be the family of finitely generated subgroups of $G$. Since each $G_i$ is torsion-free, it is free. Hence it is flat. Hence $G =$ colim $G_i$ is flat by Lemma 4.

Proposition Let $p, q$ be prime numbers which may or may not be distinct. Let $A = \mathbb{Q}_p\otimes_{\mathbb{Q}} \mathbb{Q}_q$. Let $\lambda\colon \mathbb{Z}_p \rightarrow A$ and $\mu\colon \mathbb{Z}_q \rightarrow A$ be the canonical ring homomorphisms. Let $B = \mathbb{Z}_p\otimes_{\mathbb{Z}} \mathbb{Z}_q$. Let $\psi\colon B \rightarrow A$ be the ring homomorphism induced by $\lambda$ and $\mu$. Then $\psi$ injective.

Proof: Let $S = \mathbb{Z} - \{0\}$. Then $S^{-1} (\mathbb{Z}_p \otimes_{\mathbb{Z}} \mathbb{Z}_q) \cong \mathbb{Q}_p \otimes_{\mathbb{Q}} \mathbb{Q}_q$ by Lemma 3 of my answer to this question. Since $\mathbb{Z}_q$ is torsion-free, it is flat by Lemma 5. Hence $\mathbb{Z}_p \otimes_{\mathbb{Z}} \mathbb{Z}_q$ is torsion free by Lemma 1. Hence $\psi$ injective.

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