Vector space basis for a quotient ring

Question

Given the polynomial ring $\mathbb{F}_2[x,y,z]$ with some ideal $I$, what is the method to find a basis (as a vector space over $\mathbb{F}_2$) for the quotient ring $R/I$?
For example, take $$ S = \mathbb{F_2}[x,y,z]/(z^2+zx^{2l}−1+zx+1;y^2+yx^{2l}−1+yx+1; x^{2g}+1) $$ where $l$ and $g$ are integers. We know that $S$ can be expressed with a Groebner basis. How do I find a linear basis or the dimension of $S$?

Answer 1

Suppose that $k$ is a field, and $G$ is a Groebner basis for an ideal $I$ of $R = k[x_1, x_2, \ldots, x_n]$ with respect to some given monomial order. Then a basis for $R / I$ as a $k$-vector space is given by the set of monomials $M$ over $x_1, \ldots, x_n$ such that for every initial monomial $N$ of an element of $G$, $N$ does not divide $M$. (The repeated division process shows that these monomials span the quotient, since any polynomial with initial element not in this set can be reduced by division by the appropriate element of $G$. Conversely, essentially by the definition of Groebner basis, you get that the elements are linearly independent.)

In the example you give, let us consider the monomial order given by lexicographic order on powers of $z$ first, then on powers of $y$, then on powers of $x$. Then since the initial monomials $z^2, y^2, x^{2g}$ are pairwise relatively prime, the given set of generators already forms a Groebner basis with respect to this monomial order. Therefore, a basis for $S$ as an $\mathbb{F}_2$-vector space consists of the monomials $x^a y^b z^c$ which are not divisible by any of $z^2, y^2, x^{2g}$. This condition is equivalent to $0 \le a < 2g, 0 \le b < 2, 0 \le c < 2$. As a corollary, $\dim_{\mathbb{F}_2}(S) = 8g$.