1

Consider the $\mathbb{Z}$ module $\mathbb{Z}/n\mathbb{Z}$.
What is $\mathbb{Z}/n\mathbb{Z} \otimes_{\mathbb Z} \mathbb{Z}/n\mathbb{Z}$?

pjs36
  • 18,400
AB_IM
  • 6,888
  • 1
    I was just thinking since the left hand side has 1 generator as a $\mathbb{Z}$ module, namely $ 1 \otimes 1$, moreover since any scalar is reduced modulo n (on either side), this should be isomorphic to $\mathbb{Z}/n\mathbb{Z}$. (By mapping $1 \otimes 1 \mapsto 1$) – AB_IM Feb 11 '14 at 20:15

1 Answers1

5

There are different solutions; however, I like this. Consider the following exact sequence:

$$\mathbb Z\xrightarrow{f} \mathbb Z\xrightarrow{g}\mathbb Z_n\xrightarrow{}0$$

where $g(a)=\overline{a}$ and $f(a)=na$. By tensor theorems, the sequence $$\mathbb Z\otimes \mathbb Z_n\xrightarrow{f\otimes id} \mathbb Z\otimes \mathbb Z_n \xrightarrow{g\otimes id}\mathbb Z_n\otimes \mathbb Z_n\xrightarrow{}0\otimes \mathbb Z_n$$ is exact too. Hence, $\frac{\mathbb Z\otimes \mathbb Z_n}{\rm{Im( f\otimes id)}}\cong \mathbb Z_n\otimes_{\mathbb Z} \mathbb Z_n$, but the definition of $f$ implies that $\rm{Im}(f\otimes id)=n(\mathbb Z\otimes\mathbb Z _n)$. Thus, we can conclude that $\frac{\mathbb Z\otimes \mathbb Z_n}{\rm{Im( f\otimes id)}}\cong \frac{\mathbb Z_n}{n\mathbb Z_n}\cong \mathbb Z_n$. By similar argument, you can show that $\mathbb Z_n\otimes_{\mathbb Z} \mathbb Z_m\cong\mathbb Z_{(m,n)}$

Magdiragdag
  • 15,435
Bobby
  • 7,596
  • Nice, that one is alot more elegant than my sketch :) (and more general!! ) – AB_IM Feb 11 '14 at 20:34
  • I'm confused: If $\frac{\mathbb Z\otimes \mathbb Z_n}{\rm{Im( f\otimes id)}}\cong \frac{\mathbb Z_n}{n\mathbb Z_n}\cong \mathbb Z_n$, and $\frac{\mathbb Z\otimes \mathbb Z_n}{\rm{Im( f\otimes id)}}\cong \mathbb Z_n\otimes_{\mathbb Z} \mathbb Z_n$, does this implies $\mathbb Z_n \cong \mathbb Z_n\otimes_{\mathbb Z} \mathbb Z_n$ ? – Peanica Mar 13 '19 at 09:36