What is the tensor product $\mathbb{Z}_{k} \otimes \mathbb{Z}_{(2)}$? I am trying to calculate the tensor product  $\mathbb{Z}_{k} \otimes \mathbb{Z}_{(2)}$ where $\mathbb{Z}_{(2)} = \{\frac{a}{b} | a \in \mathbb{Z} \text{ and } b \text{ is odd}\}$ and $\mathbb{Z}_{k} = \mathbb{Z}/k\mathbb{Z}$. Here is what I have done:
if $k$ is odd then $\mathbb{Z}_{k} \otimes \mathbb{Z}_{(2)} = 0$ as $(x \otimes \frac{a}{b}) = (x \otimes \frac{ak}{bk}) = (0 \otimes \frac{a}{bk})$.
if $k$ is even then $k = 2^{l} \cdot m$ with $m$-odd. This will give us $\mathbb{Z}_{k} \cong \mathbb{Z}_{2^l} \oplus \mathbb{Z}_{m}$ which implies $\mathbb{Z}_{k} \otimes \mathbb{Z}_{(2)} = \mathbb{Z}_{2^{l}} \otimes \mathbb{Z}_{(2)} = \frac {\mathbb{Z}_{(2)}}{2^{l} \cdot \mathbb{Z}_{(2)}}$. I want to know whether this calculation is correct? Also can someone please help me further simplifying the group $\frac {\mathbb{Z}_{(2)}}{2^{l} \cdot \mathbb{Z}_{(2)}}$ for $l \ge 1$.
 A: You can compute the commutative ring $\mathbb{Z}/k\mathbb{Z} \otimes \mathbb{Z}_{(p)}$ for all integers $k$ and primes $p$, where $\mathbb{Z}_{(p)}$ is the localization of $\mathbb{Z}$ at the prime ideal $(p)$.
Since we have $\mathbb{Z}/k\mathbb{Z} \otimes A = A/kA$ in general, we have $\mathbb{Z}/k\mathbb{Z} \otimes  \mathbb{Z}_{(p)} = \mathbb{Z}_{(p)} / k \mathbb{Z}_{(p)}$. For $k = 0$ there is nothing to do, so assume $k \neq 0$. Then $k = p^{\ell} m$ with some $\ell,m \geq 0$ and $p \nmid m$.  Since $m$ is invertible in $\mathbb{Z}_{(p)}$, it follows $k \mathbb{Z}_{(p)} = p^{\ell} \mathbb{Z}_{(p)}$, so that we only need to compute $\mathbb{Z}_{(p)}/p^{\ell} \mathbb{Z}_{(p)}$. There is unique ring homomorphism
$$\mathbb{Z}/p^{\ell} \mathbb{Z}\to \mathbb{Z}_{(p)}/p^{\ell} \mathbb{Z}_{(p)},$$
and I claim that it is an isomorphism. Of course direction calculations are possible, but the most efficient way is to use universal properties:
If $R$ is a commutative ring, then a ring homomorphism $\mathbb{Z}_{(p)}/p^{\ell} \mathbb{Z}_{(p)} \to R$ corresponds 1:1 to a ring homomorphism $\mathbb{Z}_{(p)} \to R$ which kills $p^{\ell}$ (by the universal property of quotient rings, which is usually known as the fundamental theorem on homomorphisms). By the universal property of the localization, this corresponds 1:1 to a ring homomorphism $f : \mathbb{Z} \to R$ (which is necessarily unique) which inverts all elements outside of $(p)$ and kills $p^{\ell}$.  Now we just have to observe that the first condition is implied by the second: If $f : \mathbb{Z} \to R$ satisfies $f(p^{\ell})=0$ and $m \in \mathbb{Z}$ is not divisble by $p$, there are $u,v \in \mathbb{Z}$ with $um + v p^{\ell}=1$, so that $f(u)f(m)=1$ and $f(m)$ is invertible. We deduce that ring homomorphisms $\mathbb{Z}_{(p)}/p^{\ell} \mathbb{Z}_{(p)}  \to R$ correspond 1:1 to ring homomorphisms $\mathbb{Z}/p^{\ell} \mathbb{Z} \to R$, and the correspondence is natural in $R$. The Yoneda Lemma now implies $\mathbb{Z}_{(p)}/p^{\ell} \mathbb{Z}_{(p)} \cong \mathbb{Z}/p^{\ell} \mathbb{Z}$. Since there is actually a unique ring homomorphism, we may even write $\mathbb{Z}_{(p)}/p^{\ell} \mathbb{Z}_{(p)} = \mathbb{Z}/p^{\ell} \mathbb{Z}$.
