What is $\mathbb{Z}[i] \otimes_{\mathbb{Z}[2i]} \mathbb{Z}[i]$? What is $\mathbb{Z}[i] \otimes_{\mathbb{Z}[2i]} \mathbb{Z}[i]$? Also, since $\mathbb{Z}[i]$ is a PID, we should be able to write this $\mathbb{Z}[i]$-module as a direct sum of cyclic $\mathbb{Z}[i]$-modules. Can we?
 A: The simpler tensor product $\mathbb{Z}[i] \otimes \mathbb{Z}[i]$ is free abelian on the generators $1 \otimes 1, 1 \otimes i, i \otimes 1, i \otimes i$. As either a left or a right $\mathbb{Z}[i]$-module it is free on two generators $1 \otimes 1, i \otimes i$.
This tensor product is the quotient of the above tensor product by the additional relation that $a \otimes 2i b = 2ia \otimes b$ for all $a, b \in \mathbb{Z}[i]$. This amounts to quotienting by the four elements


*

*$1 \otimes 2i - 2i \otimes 1 = 2 (1 \otimes i) - 2 (i \otimes 1)$

*$1 \otimes (-2) - 2i \otimes i = -2 (1 \otimes 1) - 2 (i \otimes i)$

*$i \otimes 2i - (-2) \otimes 1 = 2 (i \otimes i) + 2 (1 \otimes 1)$

*$i \otimes (-2) - (-2) \otimes i = -2 (i \otimes 1) + 2 (1 \otimes i)$.


The third and fourth relations are redundant. The first relation implies that $1 \otimes i$ and $i \otimes 1$ generate a subgroup isomorphic to $\mathbb{Z} \oplus \mathbb{Z}_2$, with $1 \otimes i$ a generator of the $\mathbb{Z}$ part and $1 \otimes i - i \otimes 1$ a generator of the $\mathbb{Z}_2$ part. Similarly, the second relation implies that $1 \otimes 1$ and $i \otimes i$ generate a subgroup isomorphic to $\mathbb{Z} \oplus \mathbb{Z}_2$, with $1 \otimes 1$ a generator of the $\mathbb{Z}$ part and $1 \otimes 1 - i \otimes i$ a generator of the $\mathbb{Z}_2$ part.  Altogether, as an abelian group this tensor product is
$$\mathbb{Z} \oplus \mathbb{Z} \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_2.$$
As either a left or a right $\mathbb{Z}[i]$-module it is
$$\mathbb{Z}[i] \oplus \mathbb{Z}[i]/(2)$$
where the first part is generated by $1 \otimes 1$ and the second part is generated by $1 \otimes 1 + i \otimes i$. 
A: If $R=\mathbb{Z}[2i]$, then $\mathbb{Z}[i] \cong R[X]/I$, where $I=(X^2+1,2X-2i)$.
So we have $\mathbb{Z}[i] \otimes_R R[X]/I \cong \mathbb{Z}[i][X]/I$.
Substituting $Y=X-i$, this is $\mathbb{Z}[i][Y]/(Y^2+2iY,2Y) \cong \mathbb{Z}[i][Y]/(Y^2,2Y) \cong \mathbb{Z}[i] \oplus \mathbb{Z}[i]/2\mathbb{Z}[i]$.
We can also see this isomorphism more directly by noting that $\mathbb{Z}[i]\otimes_R \mathbb{Z}[i]$ is generated as a $\mathbb{Z}[i]$-module by $1\otimes 1$ and $1\otimes i - i\otimes 1$, the latter of which is annihilated by $2$.
