Tensor product $Z[1/2]\otimes Z/3$ 
Compute  $\mathbb{Z}[1/2]\otimes_{\mathbb{Z}} \mathbb{Z}/3$

My initial instinct was that it is equal to $\mathbb{Z}/3[1/2]$, but I couldn't show this is correct by the universal propert. Hence, I think that my instinct misled me. However, I would like to know what is the right way to think about it, without unnecessary guesses.
 Thank you!
 A: Your guess is correct, except $2$ is a unit modulo $3$, so that $\;\mathbf Z[1/2]\otimes\mathbf Z/3\mathbf Z\simeq \mathbf Z/3\mathbf Z$.
To see it in a rigourous way, note that
$$\mathbf Z[1/2]\otimes\mathbf Z/3\mathbf Z\simeq\mathbf Z[X]/(2X-1)\otimes\mathbf Z/3\mathbf \simeq\mathbf Z/3\mathbf Z[X]/(-X-1)\simeq\mathbf Z/3\mathbf Z.$$
A: Since $2 \cdot 2 = 1 \mod \mathbb{Z}/3\mathbb{Z}$ we obtain
$$\frac{n}{2^k} \otimes x = \frac{1}{2^{k'}} \otimes nx$$
with $k' = 0,1$. If $k' = 1$ we obtain
$$\frac{n}{2^k} \otimes x = 1 \otimes 2nx$$
and otherwise we have
$$\frac{n}{2^k} \otimes x = 1 \otimes nx.$$
Thus every element is of the form $1 \otimes x$. Therefore $\mathbb{Z}[1/2] \otimes \mathbb{Z}/3\mathbb{Z} \simeq \mathbb{Z} / 3 \mathbb{Z}$.
Edit: Just to mention one thing: If $M$ is a $R$-Module and $S \subseteq R$ a multiplicative subset, then $M \otimes S^{-1}R \simeq S^{-1}M$, where $S^{-1}$ denotes the localisation. In this case $\mathbb{Z}[1/2] = S^{-1}\mathbb{Z}$ with $S := \{2^k \mid k \in \mathbb{Z}\}$. This yields
$$\mathbb{Z}[1/2] \otimes \mathbb{Z}/3\mathbb{Z} \simeq S^{-1}\mathbb{Z} \otimes \mathbb{Z}/3\mathbb{Z} \simeq S^{-1}\mathbb{Z}/3\mathbb{Z} \simeq \mathbb{Z}/3\mathbb{Z}.$$
A: You want to use some facts. First, note that $\mathbb Z[2^{-1}] = \mathbb Z[X]/(2X-1)$. Second, for any polynomial ring $A[X]$ and any morphism $A\longrightarrow B$ of rings that makes $B$ into an $A$-module, it follows that $B\otimes_A A[X] = B[X]$. Third, quotients and tensor products commute. Thus
$$\mathbb Z/3\otimes \mathbb Z[2^{-1}] = (\mathbb Z/3\otimes \mathbb Z[X])/(2X-1)$$
Now note that adjoining an inverse of $2$ to $\mathbb Z/3$ is without effect, since $2$ is invertible in $\mathbb Z/3$. So the tensor product is $\mathbb Z/3$. 
