Element Collapsing Between Tensor Products (10.4.2)

Dummit and Foote Abstract Algebra, section 10.4, exercise 2:

Show $2 \otimes 1=0∈\mathbb{Z} \otimes_\mathbb{Z} \mathbb{Z}/2\mathbb{Z}$ but $2 \otimes 1≠0∈\mathbb{2Z} \otimes_\mathbb{Z} \mathbb{Z}/2\mathbb{Z}$.

Proof: For the first, notice $2 \otimes 1 = 1 \cdot 2 \otimes 1 = 1 \otimes 2 \cdot 1 = 1 \otimes 2 = 1 \otimes 0 = 0$.

Now, if $2 \otimes 1$ is zero in the latter tensor product, then by natural extension as a $\mathbb{Z}$-module, for an arbitrary element we have $2a \otimes b = (ab)(2 \otimes 1) = 0$ so that $\mathbb{2Z} \otimes_\mathbb{Z} \mathbb{Z}/2\mathbb{Z}$ is zero. However, we have$$\mathbb{2Z} \otimes_\mathbb{Z} \mathbb{Z}/2\mathbb{Z} ≅ \mathbb{Z}/2\mathbb{Z} \otimes_\mathbb{Z} \mathbb{2Z} ≅ \mathbb{2Z}/((\mathbb{2Z})(\mathbb{2Z})) ≅ \mathbb{2Z}/\mathbb{4Z} \not ≅ 0~\square$$