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).

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