How to count the number of elements in $(\mathbb{Z}[i]/I^{2014})\otimes_{\mathbb{Z}[i]}(\mathbb{Z}[i]/J^{2014})$?

Question

Let $I,J\unlhd \mathbb{Z}[i]$ be the principal ideals generated by $7-i$ and $6i-7$, respectively. Find the number of elements in the $\mathbb{Z}[i]$-module $A=(\mathbb{Z}[i]/I^{2014})\otimes_{\mathbb{Z}[i]}(\mathbb{Z}[i]/J^{2014})$.

I know that $A\cong \mathbb{Z}[i]/(I^{2014}+J^{2014})$, but I'm not sure where to go from here. I don't think there is anything special about the number 2014 here (the problem is from a past algebra exam question from that year), so I'm inclined to think that there is some trick I'm missing that could be applied for integers other than 2014 as well. My initial thought was to somehow use the fundamental theorem for modules over PID's in order to write $A$ as a direct sum, which would allow us to count the elements, but I can't quite get it to work. Any hints would be much appreciated!

Answer 1

Hint. for principal ideals $(a), (b)$ in a UFD, we have $(a)+(b) = (\gcd(a,b))$.

Further hint: by calculating the norm of a Gaussian integer, one can guess what its prime factorization could be.

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Further hint: Depending on how much algebraic number theory you know, recall or prove using the smith normal form that the cardinality of $\mathbb Z[i]/(a)$ equals the norm $N(a)$.

Solution.

$N(7-i)=50$ so its prime factors can only have norm $2$, $5$ or $25$. That is, they are $1+i$ and $5$, or $1+i$ and two prime divisors of $5$. It is not hard to see that $7-i \neq \pm i \cdot(1+i)\cdot5$, so it has to be of the form $\pm i \cdot (1+i) p_1^{a_1}p_2^{a_2}$, where $p_1,p_2=2\pm i$ and $a_1+a_2=2$. By checking all possibilities, one finds $7-i = (1+i)(2-i)^2$. Similarly, $6i-7=i(2-i)(1+4i)$. Thus $I+J = (2-i)$, and the cardinality we're looking for is $N(2-i)^{2014}=5^{2014}$.