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This question might be stupid for you.I ask because I have no clue about it.

I don't really understand what is tensor product,although I know its definition. I have search what is tensor product,so I know it's a coset of a quotient group and its motivation.

But I still don't figure out why 2x become 0 in this paragraph.

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Remember that if $N = \Bbb Z / 2 \Bbb Z$ that $2n = 0$ for every $n \in N$. –  T. Bongers Mar 8 at 2:16
    
if N=Z/mZ that mn=0 for every n∈N.right?@T.Bongers –  amateur Mar 8 at 2:19
    
Yes. ${}{}{}{}$ –  T. Bongers Mar 8 at 2:19
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There is nothing difficult about tensor products, it is simply a formal way of making a bilinear product –  user40276 Mar 8 at 2:20

1 Answer 1

up vote 1 down vote accepted

One of the properties of the tensor product is that if $r \in \mathbb{Z}$ is an integer, and we're tensoring over $\mathbb{Z}$, then

$$mr \otimes n = m \otimes rn$$

for all $m \in M$ and $n \in N$. Since $N = \mathbb{Z} / 2 \mathbb{Z}$ is the group with two elements, twice any element is the identity; that is, $2n = 0$ for every $n \in N$. As a result, we have

\begin{align*} 2 \otimes x &= 1 \cdot 2 \otimes x =1 \otimes 2x = 1 \otimes 0 = 0 \end{align*}


More generally, if $R$ is a ring with a right module $M$ and left module $N$, we have that

$$mr \otimes n = m \otimes rn$$

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