Suppose for the $R$ module $M$, we consider its exterior algebra $\bigwedge(M)$, which is the
$$\bigwedge(M)=\frac{T(M)}{A(M)}$$ where the two sided ideal $A(M)$ is generated by all the elements of the form $m \otimes m $ for $m \in M$.
Further suppose that the element of form $m_1 \otimes m_2 \otimes \dots \otimes m_k \in T(M)$ has the image $m_1 \wedge m_2 \wedge \dots \wedge m_k \in T(M)$
How do I show that $m_1 \wedge m_2 \wedge \dots \wedge m_k=0 $ whenever $m_i=m_j$ for some $ i \neq j$?
I think that using the fact $x \wedge y= -y \wedge x $ and the fact that any permutation is product of transpositions, I can rearrange $m_i$'s to get $m_i$ and $m_j$ together along with $(-1)^{\operatorname{sgn}\{\sigma\}}$ for some appropriately choosen permuatation $\sigma \in S_k$. Then because it is generated by elements of the form $m \otimes m$, we conclude that $m_1 \otimes m_2 \otimes \cdots \otimes m_k \in A(M)$ proving our result.
I am doubtful because I am trying to develop all the properties without assuming any other result, apart from what we used to define it. So is the proof alright? or did it assume something non-trivial?
