# Tensor with Zero Element is Zero in Tensor It has been suggested that this article or section be renamed: Please make yourself familiar with the house style. You ought to be able to work certain things out for yourself by looking at the naming technique of existing pages. Please note that MediaWiki is case sensitive. One may discuss this suggestion on the talk page.

## Theorem

Let $R$ be a ring

Let $M$ be a right $R$-module

Let $N$ be a left $R$-module

and $M\otimes_R N$ their tensor product

Then

$0\otimes_R n = m\otimes_R 0 = 0\otimes_R 0$

is the $0$ in $M\otimes_R N$.

## Proof

Let $m \in M$ and $n \in N$

Then

 $\displaystyle m \otimes_R n$ $=$ $\displaystyle (m+0) \otimes_R n$ By Definition:Group identity element of $M$ as a group. $\displaystyle$ $=$ $\displaystyle m \otimes_R n + 0 \otimes_R n$ By tensor equality $\displaystyle$ $=$ $\displaystyle m \otimes_R (n+0)$ By Definition:Group identity element of $N$ as a group. $\displaystyle$ $=$ $\displaystyle m \otimes_R n + m\otimes_R 0$ By tensor equality $\displaystyle$ $=$ $\displaystyle m \otimes_R n + 0 \otimes_R (n+0)$ By Definition:Group identity element of $N$ as a group. $\displaystyle$ $=$ $\displaystyle m \otimes_R n + m\otimes_R 0 + 0 \otimes_R n + 0\otimes_R 0$ By tensor equality $\displaystyle$ $=$ $\displaystyle m \otimes_R n + 0\otimes_R 0$ By Previous equality and tensor equality

which implies that $0\otimes_R n$, $m \otimes_R 0$ and $0 \otimes_R 0$ must all be identity elements for $M \otimes_R N$ as a left module

$\blacksquare$