$$\text{Prove true or false:}\quad M \bigotimes_{\mathbb{Z}} \mathbb{Z}_{21} \cong (M/3M) \bigoplus (M/7M)$$

$M$ is an abelian group, so a $\mathbb{Z}$ module I tried creating homomorphisms with the universal property of the tensor product for bilinear maps and prove that they are inverse to one another but im not getting the desire isomorphism, yet i dont know if the statement is true.

I've tried creating maps , for example one that sends $(m+3M,m_1 + 7 M)$ to $(m m_1 \bigotimes 1 + 21 \mathbb{Z})$ and one that sends $(m \bigotimes x+21\mathbb{Z})$ to $(xm + 3M, 1 +7M)$.

  • $\begingroup$ what have you tried so far? $\endgroup$ – Praphulla Koushik Jan 2 at 11:52
  • $\begingroup$ ive tried creating maps , for example one that sends $(m+3M,m_1+7M) to (mm_1 \bigotimes 1 +21\mathbb{Z})$ and one that sends $ (m \bigotimes x+21\mathbb{Z}) to (xm + 3M, 1 +7M)$ $\endgroup$ – Pedro Santos Jan 2 at 11:56

See that $\mathbb{Z}_{21}=\mathbb{Z}_3\oplus \mathbb{Z}_7$.

As tensor product distributes with direct sum, $$M\otimes_{\mathbb{Z}}\mathbb{Z}_{21}=M\otimes_{\mathbb{Z}}(\mathbb{Z}_3\oplus \mathbb{Z}_7)=(M\otimes_\mathbb{Z}\mathbb{Z}_3)\oplus (M\otimes_\mathbb{Z}\mathbb{Z}_7)$$

Can you take it from here?

  • $\begingroup$ Oh yeah i forgot about that damn it , Thanks , now i can use the fact that $M \bigotimes \mathbb{Z}{m} \cong M/mM$ $\endgroup$ – Pedro Santos Jan 2 at 11:59
  • $\begingroup$ Enjoy!!!!!!!!!! $\endgroup$ – Praphulla Koushik Jan 2 at 12:01

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