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.