Tensor with Zero Element is Zero in Tensor

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.

$\blacksquare$