$\def\N{\bf N} \def\S{\sum_{i=1}^n}$ Let ${\bf u} = (u_1, ..., u_n) \in \N^n$

Define the partial order $\geq$: $~~{\bf u} \geq {\bf v}$ if $u_i \geq v_i$ for all $i$.

Define ${ >_{lex}}$ on $\N^n$ by using lexicographical ordering: ${\bf u} >_{lex} {\bf v}$ if the leftmost entry of ${\bf u} - {\bf v}$ is positive.

Ex: $(2, 3, 1) >_{lex} (2, 1, 3)$ since $(2, 3, 1) - (2, 1, 3) = (0, 2, -2)$ and the leftmost term $2 > 0$.

Defn: ${\bf x}^{\bf u} = x_1^{u_1} x_2^{u_2} \cdot \cdot \cdot x_n^{u_n}$ is a monomial

Let $k$ be a field. Then if $f \in k[x_1, ..., x_n]$, $f = \S c_i {\bf x}^{\bf u_i}$ for some $c_i \in k$.

Defn: $c_i {\bf x}^{\bf u_i}$ is a term

If the monomials are ordered such that ${\bf u_i} >_{lex} {\bf u_j}$if $i < j$, then

Defn: $LC(f) =$ leading coefficient of $f = c_1$

Defn: $LM(f) =$ leading monomial of $f = x^{\bf u_i}$

Defn: $LT(f) =$ leading term of $f = c_1 x^{\bf u_i}$

Ex: If $f = -0.2 x_5^2x_6 + \frac{1}{2}x_2^9x_4 - 8x_1x_2^3$

Then $LC(f) = -0.2 ~~~ LM(f) = x_5^2x_6 ~~~ LC(f) = -0.2 x_5^2x_6$

Defn: If ${\bf v} \leq {\bf u}$, then ${\bf x}^{\bf v}$ divides ${\bf x}^{\bf u}$ with quotient ${\bf x}^{\bf u}/{\bf x}^{\bf v} = {\bf x}^{\bf u-v}$

Ex: $x_1^2x_2^3/ x_1x_2 = x_1x_2^2$ while $x_1^2x_2^3/x_1^3x_6$ is not defined in $k[x_1, ..., x_n]$

Defn: $LCM({\bf x}^{\bf u}, {\bf x}^{\bf v})$ = least common multiple = ${\bf x}^{\bf w}$ where $w_i = max(u_i, v_i)$.

Ex: $LCM(x_1^2x_2^3, x_1^3x_6) = x_1^3x_2^3x_6$.

Defn: A Grobner basis is a collection of polynomials $F = \{f_1, ..., f_n\}$ such that for any linear combination of $f_i$'s, $\S c_i f_i$, there exists an $i$ such that $LM(f_i)$ divides $LM(\S c_if_i)$

Note a Grobner basis is a generating set, but linear independence is NOT required.

Let $< F > = \{ \S c_i f_i ~|~ c_i \in k, f_i \in F \}$, the ideal generated by $F$.

Ex: $x_1x_2 = x_1(x_2^3) - x_2(x_1x_2^2 - x_1)$

Thus $x_1x_2 \in < x_2^3, x_1x_2^2 - x_1 >$, but neither $x_2^3$ nor $x_1x_2^2$ divides $x_1x_2$. Thus $\{x_2^3, x_1x_2^2 - x_1\}$ is NOT a Grobner basis.

Finding a Grobner basis:

Let $h = LCM(LM(f), LM(g))$. The syzygy polynomial or S-polynomial is $$S(f, g) = \frac{h}{LT(f)} f - \frac{h}{LT(g)} g$$

Ex: $LCM(x_2^3, x_1x_2^2) = x_1x_2^3$

$S(x_2^3, x_1x_2^2 - x_1) = \frac{x_1x_2^3}{x_2^3}x_2^3 -\frac{x_1x_2^3}{x_1x_2^2}(x_1x_2^2 - x_1)$ $= {x_1}x_2^3 -{x_2}(x_1x_2^2 - x_1)$ $= {x_1}x_2^3 -x_1x_2^3 - x_1x_2$ $= -x_1x_2$

Use division to reduce $p = x_1x_2$ and take remainder: $p = \S c_if_i + r$. Add $r$ to basis. Note $r = x_1x_2$

Division algorithim: If $LT(f_i)$ divides $LT(p)$, replace $p$ with $p - \frac{LT(p)}{LT(f_i}f_i$. Repeat until no $LT(f_i)$ divides the most recent $p$.

$F_1 = \{x_2^3, x_1x_2^2 - x_1, x_1x_2\}$,

$S(x_1x_2, x_1x_2^2 - x_1)$ $= \frac{x_1x_2^2 }{x_1x_2 } x_1x_2 - \frac{x_1x_2^2 }{x_1x_2^2 } (x_1x_2^2 - x_1)$ $= x_2x_1x_2 - x_1x_2^2 + x_1 = x_1$

$F_2 = \{x_2^3, x_1x_2^2 - x_1, x_1x_2, x_1\}$,

$x_1x_2^2 - x_1 = (x_2^2 - 1) x_1$ and $x_1x_2 = x_2(x_1)$.

Thus $S( x_1x_2^2 - x_1, x_1) = 0$ and $S( x_1x_2, x_1) = 0$

$F_3 = \{x_2^3, x_1x_2^2 - x_1, x_1x_2, x_1\}$,

$S(x_2^3, x_1) = \frac{x_2^3x_1}{x_1}x_1 - \frac{x_2^3x_1}{x_2^3}x_2^3$ $= x_2^3x_1 - x_1x_2^3 = 0$, $S(x_2^3, x_1x_2)$ $= \frac{x_1x_2^3}{x_2^3} x_2^3 - \frac{x_1x_2^3}{x_1x_2} x_1x_2 = 0$

Thus $F_3 = \{x_2^3, x_1x_2^2 - x_1, x_1x_2, x_1\}$ is a Grobner basis for $< F >$.

A Grobner basis, $G$, is reduced if for all $g \in G$, (1) $LC(g) = 1$ and (2) no term of $g$ is divisible by $LT(f)$ for any $f \in G$.

Ex: $G = \{x_2^3, x_1\}$ is a reduced Grobner basis for $< F >$.

Note the above can be extended to vectors of functions where ${\bf x}^{\bf u}{\bf e_i}> {\bf x}^{\bf v}{\bf e_j}$ if $i < j$ or ($i = j$ and ${\bf x}^{\bf u} >_{lex}{\bf x}^{\bf v})$

Defn: $LCM({\bf x}^{\bf u}{\bf e_i}, {\bf x}^{\bf v}{\bf e_j})$ $= \cases{ LCM({\bf x}^{\bf u}, {\bf x}^{\bf v}){\bf e_i} & if~ i = j \cr 0 & otherwise}$