TITLE: What is $\mathbb{Z}[i] \otimes_{\mathbb{Z}[2i]} \mathbb{Z}[i]$?
QUESTION [6 upvotes]: 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?

REPLY [4 votes]: 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 $1 \otimes 1 + i \otimes i$.