A fact about alternating tensors over rings with enough units

Let $R$ be a commutative ring in which $k!$ is a unit, and let $M$ be an $(R,R)-bimodule such that$latex rm = mr\$. Recall that $\mathcal{A}^k(M)$ is the set of all tensors in $\mathcal{T}^k(M)$ having two consecutive entries equal, and $\mathsf{Alt}(z) = \sum_{\sigma \in S_k} \epsilon(\sigma)\sigma z$, where $S_k$ acts on $\mathcal{T}^k(M)$ by permuting entries.

Prove that $z - (1/k!) \mathsf{Alt}(z) = (1/k!) \sum_{\sigma \in S_n} (z - \epsilon(\sigma)\sigma z)$ for all $z \in \mathcal{T}^k(M)$. Use this to prove that $\mathsf{ker}\ \frac{1}{k!} \mathsf{Alt} = \mathcal{A}^k(M)$.

Let $z \in \mathcal{T}^k(M)$. We have $z - \frac{1}{k!} \mathsf{Alt}(z) = z - \frac{1}{k!} \sum_{\sigma \in S_k} \epsilon(\sigma)\sigma z$ $= \frac{1}{k!}(k!z - \sum_{\sigma \in S_k} \epsilon(\sigma)\sigma z)$ $= \frac{1}{k!}(\sum_{\sigma \in S_k} z - \sum_{\sigma \in S_k} \epsilon(\sigma)\sigma z)$ $= \frac{1}{k!}(\sum z - \epsilon(\sigma)\sigma z)$, as desired.

Suppose $z \in \mathcal{A}^k(M)$; say the $i$ and $i+1$ components of $z$ are equal. Note that $\sigma z = \sigma (1\ i+1) z$, and in particular $\epsilon(\sigma)\sigma z + \epsilon(\sigma(i\ i+1)) \sigma (i\ i+1) z = 0$. In the equation $\mathsf{Alt}(z) = \sum_{\sigma \in S_k} \epsilon(\sigma)\sigma z$, we can break up the right hand side as a summation over the cosets of $\langle (i\ i+1) \rangle$, each of which is 0. Thus $\frac{1}{k!}\mathsf{Alt}(z) = 0$, and we have $\mathcal{A}^k(M) \subseteq \mathsf{ker}\ \frac{1}{k!}\mathsf{Alt}$.

Now suppose $z \in \mathsf{ker}\ \frac{1}{k!} \mathsf{Alt}$. From the equality proved above, we have $\frac{1}{k!} \sum_{\sigma \in S_k} (z - \epsilon(\sigma)\sigma z) = z$. Note that $z - \epsilon(\sigma)\sigma z \in \mathcal{A}^k(M)$ for each $\sigma$. (It suffices to notice this is true for adjacent transpositions- i.e. $\sigma = (i\ i+1)$.) Thus $z \in \mathcal{A}^k(M)$.

So $\mathsf{ker}\ \frac{1}{k!} \mathsf{Alt} = \mathcal{A}^k(M)$.