The symmetric algebra of an $R$-module $M$ is defined to be the quotient of the tensor algebra $\mathcal{T}(M)$ by the ideal $C(M)$ generated by all elements of the form $m_1\otimes m_2-m_2\otimes m_1$ for all $m_1,m_2\in M$, and it is denoted by $S(M)$.
A theorem in Dummit and Foote claims that the $k$th symmetric power $S^k(M)$ of $M$ is equal to $M\otimes\cdots\otimes M$ (k factors) modulo the submodule generated by all elements of the form $(m_1\otimes\cdots\otimes m_k)-(m_{\sigma(1)}\otimes\cdots\otimes m_{\sigma(k)})$ for all $m_i\in M$ and all $\sigma\in S_k$.
Now the authors claim that it is "easy to see" that since $\sigma$ can be written as a product of transpositions, every element of the form above can be written as a sum of elements of the form $m_1\otimes\cdots\otimes m_{i-1}\otimes(m_i\otimes m_{i+1}-m_{i+1}\otimes m_i)\otimes m_{i+2}\otimes\cdots\otimes m_k$.
Why is this so?
Let's do an example. $$m_1\otimes m_2\otimes m_3-m_3\otimes m_1\otimes m_2=(m_1\otimes m_2\otimes m_3-m_1\otimes m_3\otimes m_2)+(m_1\otimes m_3\otimes m_2-m_3\otimes m_1\otimes m_2)$$ Remember the tensor product is bilinear, so this is the same as $$m_1\otimes (m_2\otimes m_3-m_3\otimes m_2)+(m_1\otimes m_3-m_3\otimes m_1)\otimes m_2$$ You fill in the sum with terms that differ by an adjacent transposition that telescope, and you use the bilinearity to get adjacent differences into an element of the desired form.