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.
Let $M \otimes_R N$ denote their tensor product.
Then:
- $0\otimes_R n = m \otimes_R 0 = 0 \otimes_R 0$
is the zero in $M \otimes_R N$.
Proof
Let $m \in M$ and $n \in N$
Then
\(\ds m \otimes_R n\) | \(=\) | \(\ds \paren {m + 0} \otimes_R n\) | Group Axiom $\text G 2$: Existence of Identity Element | |||||||||||
\(\ds \) | \(=\) | \(\ds m \otimes_R n + 0 \otimes_R n\) | Definition of Tensor Equality | |||||||||||
\(\ds \) | \(=\) | \(\ds m \otimes_R \paren {n + 0}\) | Group Axiom $\text G 2$: Existence of Identity Element | |||||||||||
\(\ds \) | \(=\) | \(\ds m \otimes_R n + m\otimes_R 0\) | Definition of Tensor Equality | |||||||||||
\(\ds \) | \(=\) | \(\ds m \otimes_R n + 0 \otimes_R \paren {n + 0}\) | Group Axiom $\text G 2$: Existence of Identity Element | |||||||||||
\(\ds \) | \(=\) | \(\ds m \otimes_R n + m \otimes_R 0 + 0 \otimes_R n + 0 \otimes_R 0\) | Definition of Tensor Equality | |||||||||||
\(\ds \) | \(=\) | \(\ds m \otimes_R n + 0 \otimes_R 0\) | Definition of Tensor Equality |
Hence $0 \otimes_R n$, $m \otimes_R 0$ and $0 \otimes_R 0$ must all be identity elements for $M \otimes_R N$ as a left module.
$\blacksquare$