Commutative Algebra – Compute Tensor Product of Z/726Z and Z[1/77]

commutative-algebralocalizationtensor-products

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.

Best Answer

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.

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