Reduction modulo a set

We have studied, back in chapter 1, the idea of reducing an integer modulo another integer. The concept which will be introduced here is similar, except the role of the integers is played by the ring of polynomials over a field $F$.

Definition 2.8.9   Let $f,g\in F[x_1,...,x_n]$ and fix a monomial ordering $<$. We say $f$ reduced to $h$ modulo $g$ in one step if $lt(g)$ divides some term $ax_1^{a_1}...x_n^{a_n}$ of $f$ and

$$h=f-\frac{ax_1^{a_1}...x_n^{a_n}}{lt(g)}g.$$

In this case, we write $f\stackrel{g}{\rightarrow}h$.

Example 2.8.10   Let $<$ be graded lexicographical ordering - terms are listed from highest to lowest degree and $x>y>z$.

Let $f(x,y,z)=2x^2z^4+xz^3+y^3z^3$ and $g(x,y,z)=xz^3+y^2z^2$. Then $lt(f)=2x^2z^4$ and $lt(g)=xz^3$. There are two terms of $f$ which are divisible by $lt(g)$: $2x^2z^4$ and $xz^3$. In particular, $f$ reduced to $h$ modulo $g$ in one step, where

$$\begin{array}{cc} h(x,y,z)&=f-\frac{lt(f)}{lt(g)}g\\ &= f-... ...y^3z^3-2x^2z^4-2xy^2z^3\\ &=-2xy^2z^3+y^3z^3+xz^3. \end{array}$$

Definition 2.8.11   Let $f,<$ be as in the previous definition. Let $G\subset F[x_1,...,x_n]$ be a finite set. We say $f$ reduced to $h$ modulo $G$ if there is a finite sequence of polynomials $h_0=f,h_1,...,h_k=h$ such that $h_i$ reduces to $h_{i+1}$ modulo some $g\in G$ (depending on $h_i$) in one step, $0\leq i\leq k-1$. In this case, we write $f\stackrel{G}{\rightarrow}h$.

We say that $h$ is in normal form modulo $G$ if is no $h$ such that $h\stackrel{G}{\rightarrow}h'$. We say $h$ is the reduced normal form of $f$ modulo $G$ if $f$ reduces to $h$ modulo $G$ and $h$ is in normal form modulo $G$.

The normal form of $f$ modulo $G$, if it exists, is denoted $NF(f,G)$.