I've read in many books, that $\Bbb Z/3\Bbb Z \times\Bbb Z/5\Bbb Z\cong \Bbb Z/15\Bbb Z$
and $\Bbb Z/3\Bbb Z ⊗ \Bbb Z/5\Bbb Z\cong 0$ .
How can I prove the isomorphism of them?
For $\Bbb Z/3\Bbb Z ⊗ \Bbb Z/5\Bbb Z\cong 0$,
I let B: $\Bbb Z/3\Bbb Z \times\Bbb Z/5\Bbb Z \to A $ , a bilinear form
and $3B(a,b)=B(3a,b)=B(0,b)=0; $
and $5B(a,b)=B(a,5b)=B(a,0)=0.$ Is it right? How can I prove the first one?
Any help would be appreciated!
On
The second isomorphism is a special case of the more general isomorphism, valid in any commutative ring $R$ and any $R$-module $M$:
For any ideal $I$ in the commutative ring and any $R$-module $M$, $$ R/I\otimes_R M\simeq M\otimes_R R/I\simeq M/IM. $$
As a consequence, if $I$ and $J$ are two ideals in $R$, we obtain $$ R/I\otimes_R R/J\simeq (R/J)\Bigm/I\,(R/J)\simeq (R/J)\Bigm/(I+J/J)\simeq R/(I+J)$$ by the 3rd isomorphism theorem.
Now in a P.I.D.,, if $I=(a)$, $J=(b)$, we know that by definition, $(a)+(b)=(\gcd(a,b)).$
For the second isomorphism, one can consider any elementary tensor $x\otimes y$ in $\mathbb{Z}/3\mathbb{Z}\otimes \mathbb{Z}/5\mathbb{Z}$. Then, $3$ is invertible in $\mathbb{Z}/5\mathbb{Z}$ namely $2*3=1$ in $\mathbb{Z}/5\mathbb{Z}$. Hence, we have $$x\otimes y=x\otimes (3\times(2y))=3(x\otimes(2y))=(3x)\otimes(2y)=0\otimes (2y)=0$$ using only computational properties of the tensor product. Because the space $\mathbb{Z}/3\mathbb{Z}\otimes \mathbb{Z}/5\mathbb{Z}$ is generated by elementary tensors, and these are all zero, we deduce that $\mathbb{Z}/3\mathbb{Z}\otimes \mathbb{Z}/5\mathbb{Z}=0$.
For the first isomorphism, I invite you to check the Chinese remainder theorem if you don't already know about it. It exactly states what you want. Click here for the Wikipedia page.