Compute the tensor product $\mathbb{Z}/726\mathbb{Z} \otimes_{\mathbb{Z}} \mathbb{Z}[\frac{1}{77}]$

Question

I am computing the cardinality of the tensor product $\mathbb{Z}/726\mathbb{Z} \otimes_{\mathbb{Z}} \mathbb{Z}[\frac{1}{77}]$. Following are my several attempts:

1. Let $R=\mathbb{Z}$, $I=726\mathbb{Z}$ which is an ideal of $R$, and $M=\mathbb{Z}[\frac{1}{77}]$. Then $(R/I) \otimes_R M \cong M/IM$ as $R$-modules. After this I can re-consider $M/IM$ as an $R/I$ module to compute the cardinality. But I got stuck since $M/IM$ is not a free $R/I$-module.

2. Let $R=\mathbb{Z}$, $S=\{77^n:n \in \mathbb{Z}_{\geq 1}\}$ is multiplicatively closed in $R$ and $A=\mathbb{Z}/726\mathbb{Z}$. Then $\mathbb{Z}[\frac{1}{77}]=S^{-1}\mathbb{Z}$ and $A \otimes_{\mathbb{Z}} S^{-1}\mathbb{Z} \cong S^{-1}A$ as $S^{-1}\mathbb{Z}$-modules. Again at this point I am stuck.

3. Treat the tensor product as usual, i.e. using the pure tensors. Actually I can prove that any elements of the tensor product can always be written as $a(\overline{1} \otimes \frac{1}{77^n})$ for some integer $a$ and some non-negative integer $n$. Again I got stuck with what to do next because the pure tensors with different $n$'s can be equal, for instance $\overline{1} \otimes \frac{1}{77}=\overline{77} \otimes \frac{1}{77^2}$ etc...

I think I got stuck because both $726$ and $77$ are not prime numbers. It makes the actual computation become more complicated. Also I guess that the possibilities for the cardinality are $726$, $77$, $11$ (the $\gcd$), $5082$ (the ${\rm lcm}$) and even infinite. However I still can not predict the structure of the tensor product, at least the most convenient form to compute the cardinality.

Any helps or hints are appreciated.

Answer 1

Your first approach works: since $726=11^2\cdot6$ and $11$ is invertible in $M,$ $$M/IM=M/6M.$$ Let us now prove that the group $M/6M$ is isomorphic to $\Bbb Z/6\Bbb Z,$ or more generally, that $$\gcd(m,n)=1\implies\Bbb Z[1/m]/n\Bbb Z[1/m]\cong\Bbb Z/n\Bbb Z$$ Consider the canonical homomorphism $$f:\Bbb Z\to\Bbb Z[1/m]/n\Bbb Z[1/m].$$

• Its kernel is $n\Bbb Z,$ because $f$ sends $n\Bbb Z$ to $0$ and conversely, any integer $k\in\ker f$ is such that $k=n\frac a{m^b}$ for some $a\in\Bbb Z,b\in\Bbb Z_{\ge0},$ which implies $n\mid m^bk$ and therefore $n\mid k$ (since $m^b$ and $n$ are coprime).
• $f$ is surjective, since every $\frac a{m^b}\in\Bbb Z[1/m]$ ($a\in\Bbb Z,b\in\Bbb Z_{\ge0}$) is equal to $c+n\frac d{m^b}$ for some $c,d\in\Bbb Z,$ i.e. $a\in m^b\Bbb Z+n\Bbb Z,$ by Bézout.

By the first isomorphism theorem, the conclusion follows.