Tensor with Zero Element is Zero in Tensor

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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

$\blacksquare$