# Basis of the symmetric algebra $S(M)$ given $R$-module basis of $M$ using the diamond lemma?

Over the past week, I read this secret blogging seminar post concerning the diamond lemma, which got me to reading about Bergman's paper on the diamond lemma.

Now suppose you have a free $A$-module $M$ over some commutative ring $A$, with basis $\{e_i\}_{i\in I}$. Standard methods using the tensor product yield the fact that the basis of the symmetric algebra $S(M)$ is given by monomials $e_{i_1}\dots e_{i_k}$ for $i_1\leq\cdots\leq i_k$ for some linear ordering on $I$, and similarly for the exterior algebra $\Lambda(M)$.

How can the same results be reached using Bergman's diamond lemma, and avoiding the tensor product arguments? I'm hoping to see how it can be applied in these situations. How can I see what the possible ambiguities are to check? Thank you.

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The symmetric algebra is the quotient of the free algebra on the basis by the ideal generated by the relations $e_ie_j-e_je_i$ with $i$, $j\in I$. Pick a linear order $\prec$ on $I$, extend lexicographically to the non-commutative monomials on the generators, find the ambiguities (there are no containment ambiguities, and all overlap ambiguities are of length $3$) You have to check that each of these can be resolved. Do that. Now see what it means in this particular case for a monomial to be in normal form, and observe that the conclusion of the Diamond Lemma is that the set of monomials of the form $$e_{i_1}e_{i_2}\cdots e_{i_k}$$ with $i_1\preceq i_2\preceq\cdots\preceq i_k$ a non-decreasing sequence of indices with respect to $\prec$. This is the usual basis.
Later. For simplicity, let us suppose that we have finitely many generators $e_1$, $\dots$, $e_n$, and that we order them so that $e_1\prec\cdots\prec e_n$. The relations which define the symmetric algebra are $$e_ie_j-e_je_i, \qquad 1\leq j<i\leq n.$$ Now, if $1\leq j<i\leq n$, the leading term of relation $e_ie_j-e_je_i$ is $e_ie_j$, so the rewriting rules (in Bergman's paper these the pairs $(\sigma,f_\sigma)$...) are $$e_ie_j\leadsto e_je_i, \qquad 1\leq j<i\leq n.$$ To find the ambiguities, we need to see in what ways the monomials in the set $M=\{e_ie_j:1\leq j<i\leq n\}$ interact. There are no containment ambiguities, because they all have length $2$, so we just need to find the overlaps. An overlap of two monomials of length $2$ can only happen in a monomial of length $3$. If $e_ie_je_k$ is such an overlap, we must have $e_ie_j\in M$ and $e_je_k\in M$. We thus conclude that the ambiguitites that we have to resolve are $$e_ie_je_k, \qquad 1\leq k<j<i\leq n.$$