Noncommutative Gröbner basis

Remarks

A Gröbner basis $G$ of $I$ generates $I$ as a two-sided ideal. The words that contain no $\operatorname{LW}(g)$, $g \in G$, as a subword are the normal words, and their images form a $k$-basis of $k\langle X \rangle / I$. Example: $X = \set{x, y}$ with the degree-lexicographic order, $x < y$, and $I$ generated by $g = yx - xy$, so $\operatorname{LW}(g) = yx$. The word $yx$ has no overlap with itself, so by the diamond lemma $\set{g}$ is a Gröbner basis; the normal words are $x^a y^b$, and $k\langle x, y \rangle / I \cong k[x, y]$. Non-example: with the same order, let $I$ be generated by $g = xx - y$, so $\operatorname{LW}(g) = xx$. The element $gx - xg = xy - yx$ lies in $I$, and its leading word is $yx$, which does not contain $xx$. So $\set{g}$ is not a Gröbner basis of $I$.