1

I just stated tensor product, and I have problem to see how it works. So, why $$\mathbb Z_2\otimes_{\mathbb Z}\mathbb Z_3\cong \{0\}\ \ ?$$

1 Answers1

13

Let $a\otimes b\in \mathbb Z_2\otimes_{\mathbb Z} \mathbb Z_3$. Then $$a\otimes b= (3a)\otimes b=a\otimes (3b)=a\otimes 0=0\otimes 0=0_{\mathbb Z_2\otimes_{\mathbb Z}\mathbb Z_3}.$$

Surb
  • 57,262
  • 11
  • 68
  • 119