Let $A$ be a finite dimensional algebra generated by $x_1, \ldots, x_n$ subject to certain relations $I_1, \ldots, I_m$. Could we obtained a linear basis $B$ consisting of monomials in $x_1, \ldots, x_n$ from a Groebner basis of $A$? For example, let $A$ be the $\mathbb{C}$-algebra generated by $x_{ij}$, $i \in I=\{1,2,3\}$, $j \in J=\{1,2\}$, subject to the relations: \begin{align*} & x_{ij}^2=x_{ij}, i \in I, j \in J, \\ & x_{21}-x_{11}x_{21}-x_{21} x_{31}=0. \end{align*} The Hilbert series of $A$ is $2t^5+9t^4+16t^3+14t^2+6t+1$. The Groebner basis of $A$ with respect to the graded reverse lexicographic order $tdeg$ is $$x_{32}^2-x_{32}, x_{31}^2-x_{31}, x_{22}^2-x_{22}, x_{21}^2-x_{21}, x_{11} x_{21}+x_{21} x_{31}-x_{21}, x_{12}^2-x_{12}, x_{11}^2-x_{11}.$$ Is there some general method to derive a linear monomial basis (consists of $\dim A$ monomials) from a Groebner basis? Let $B'$ be the Groebner basis with respect to some order. Denote by $B''$ the set of monomials which are initial terms of elements in $B'$. Is $B''$ linearly independent? How to extend it to a basis of $A$? Thank you very much.
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asked Dec 12, 2019 at 11:13
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1$\begingroup$ I assume $A$ is commutative? When you say "Gröbner basis of $A$", do you mean a Gröbner basis of the ideal generated by $I_1,\ldots,I_m$ ? In that case, the answer is "yes" and it's just the Macaulay-Buchberger basis theorem (see, e.g., Proposition 3.10 in my $t$-unique reductions for Mészáros's subdivision algebra (detailed version), where I give a proof because everyone else leaves it as an exercise). $\endgroup$– darij grinbergCommented Dec 12, 2019 at 13:22
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2$\begingroup$ To say it in one sentence: Look at all possible monomials and keep those, which are not divisible by any lead monomial of a member of a fixed gröbner base. In your case above only those with exponent smaller than 2 come into consideration. A little calculation with Macaulay2 shows, that $\dim A = 48$ (in your notation, with Macaulay2 the command is "degree A" where A=R/J and R=QQ[x_(1,1)..x_(3,2)] and J is the ideal of R comprising the 7 polynomials you give above. To find the monomials explicitly, use basis(d, A) where d is the degree of the monomials sought for. $\endgroup$– Jürgen BöhmCommented Dec 12, 2019 at 13:54
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$\begingroup$ @darij grinberg, thank you very much. Yes, $A$ is commutative. Yes, I mean a Groebner basis of the ideal generated by $I_1, \ldots, I_m$. $\endgroup$– Jianrong LiCommented Dec 12, 2019 at 16:00
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$\begingroup$ @Jürgen Böhm, thank you very much. $\endgroup$– Jianrong LiCommented Dec 12, 2019 at 16:00
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