I'm reading the proof of the existence of the tensor product. If $M,N$ are two $R$-modules then we can construct the tensor product $T$ as the quotient $C/D$ where $C$ is the free module over $M \times N$ and $D$ is the submodule generated by the set of all elements in $C$ of the form $(m+m^\prime, n) - (m,n) - (m^\prime, n)$ $ (m, n+n^\prime) -(m,n) - (m,n^\prime) $ $ (am, n) - a(m,n)$ $ (m,an) - a(m,n)$
I use $(m,n)$ to denote the element $e_{(m,n)} \in F(M\times N)$. Since $F(S) \cong \bigoplus_{s \in S} R$ I picture these elements as $e_{(m,0)} = (0, \dots, 0,1, 0, \dots)$ where the $1$ here is at position $m$ and $e_{(m,n)}$ the sequence with $1$ at position $m \cdot |M| + n$ and so forth.
Is this correct so far?
Now I wanted to see what this looks like. So I computed the tensor product of $M = N = \mathbb Z / 2 \mathbb Z$ over $R=\mathbb Z$. For $C$ I get that $C \cong \mathbb Z^4$. Then I computed all the elements above and noticed that $D \cong \langle \{(1,0,0,0), (0,1,0,0), (0,0,1,0)\} \rangle$. Hence $M \otimes N = \mathbb Z / 2 \mathbb Z \otimes_{\mathbb Z} \mathbb Z / 2 \mathbb Z = \langle (0,0,0,1) \rangle \cong \mathbb Z$
Is this correct?