A term order (or a term ordering or even a monomial ordering) on K[X] is any relation &lt

1 Preliminary

We denote by K a field and by K[X] the polynomial ring over K in n variables X₁, . . . , X_n. For α = (α₁, . . . , α_n) ∈ Nⁿ we abbreviate X₁^α1X₂^α2· · · X_n^αn by X^α, and we call X^α a power product.

Definition 1.1. A term order (or a term ordering or even a monomial ordering) on K[X] is any relation < on Nⁿ satisfying :

(1) < is a total (or linear) ordering on Nⁿ. (2) If α < β and γ ∈ Nⁿ, then α + γ < β + γ .

(3) < is a well-ordering on Nⁿ. This means that every nonempty subset of Nⁿ has a smallest element under < .

For a simple example of a term order, note that the usual numerical order 0 < 1 < 2 < 3 < · · · < m < m + 1 < · · ·

on the elements of N satisfies the three conditions of Definition 1.1. Hence, the degree ordering 1 < x < x² < · · · < x^m< x^m+1 < · · ·

on the monomials in K[X] is a term order.

Definition 1.2. (Lexicographic Order) For α = (α₁, . . . , α_n), β = (β₁, . . . , β_n) ∈ Nⁿ we define α <_lex β ⇐⇒ there is a j ∈ {1, 2, . . . , n} such that α_k = β_k if k < j and α_j < β_j.

Example 1 : (0, 2, 0) <lex (1, 0, 0) <lex (1, 0, 1).

With P

αcαX^α or P

α∈NⁿcαX^α we always tacitly mean that only finitely many of the coefficients cα are different from zero. If f = P

αcαX^α is a polynomial in K[X] and we have selected a term order <, then we can order the monomials of f in an unambiguous way with respect to <. For example, let f = 4xy²z + 4z²− 5x³+ 7x²z² ∈ K[x, y, z]. Then with respect to the lex order, we would reorder the terms of f in decreasing order as

f = −5x³+ 7x²z²+ 4xy²z + 4z². We will use the following terminology.

Definition 1.3. Let f = P

αc_αX^α be a nonzero polynomial in K[X] and let < be a term order.

(1) The degree of f is

deg(f ) := max { α ∈ Nⁿ| c_α 6= 0 } . (2) The leading coefficient of f is

lc(f ) := c_{deg(f )} . (3) The leading monomial of f is

lm(f ) := X^{deg(f )} . (4) The leading term of f is

lt(f ) := lc(f ) · lm(f ) .

Example 2 : Let f = 4xy²z + 4z²− 5x³+ 7x²z² and let < denote the lex order. Then deg(f ) = (3, 0, 0), lc(f ) = −5, lm(f ) = x³, lt(f ) = −5x³.

For a subset F ⊆ K[X] we define

deg(F ) := { deg(f ) | f ∈ F − {0} } , D(F ) := deg(F ) + Nⁿ and LT (F ) := { lt(f ) | f ∈ F − {0} } .

Let us firstly look at the special case of the division of f by g, where f, g ∈ K[X]. We fix a term order on K[X].

Definition 1.4. Given f, g, h in K[X], with g 6= 0, we say that f reduces to h modulo g in one step, written

f −→ h,^g if and only if lm(g) divides lt(f ) and

h = f − lt(f ) g .

Example 3 : Let f = 6x²y − x + 4y³− 1 and g = 2xy + y³ be polynomials in Q[x, y]. If the term order is lex order with y < x, then f −→ h, where h = −3xy^{g 3}− x + 4y³− 1, since, in this case lt(f ) = 6x²y is the term of f we have canceled using lt(g) = 2xy.

In the multivariable case we may have to divide by more than one polynomial at a time, and so we extend the process of reduction defined above to include this more general setting.

Definition 1.5. Let f , h, and f₁, . . . , f_s be polynomials in K[X], with f_i 6= 0 (1 ≤ i ≤ s), and let F = {f₁, . . . , f_s}. We say that f reduces to h modulo F , denoted

f −→^F ₊ h,

if and only if there exist a sequence of indices i1, i2, . . . , it ∈ {1, 2, . . . , s} and a sequence of polynomials h1, . . . , ht−1 ∈ K[X] such that

f −→ h^fi1 ₁ −→ h^fi2 ₂ −→ · · ·^{fi3 f}−→^it−1 h_t−1 −→ h .^fit

Definition 1.6. A polynomial r is called reduced with respect to a set of non-zero polynomials F = {f1, . . . , fs} if r = 0 or no power product that appears in r is divisible by any one of the lm(fi), i = 1, 2, . . . , s. In other words, r cannot be reduced modulo F .

Definition 1.7. Fix a term order and let I be an ideal in K[X], I 6= {0}. A finite subset G of I − {0} is a Gr¨obner basis of I ⇐⇒ LT (G) generates the ideal < LT (I) > .

Definition 1.8. Let 0 6= f, g ∈ K[X], and let L = lcm lm(f ), lm(g)
. The polynomial S(f, g) = L

lt(f ) f − L lt(g) g is called the S-polynomial of f and g .

Example 4 : Let f = 2yx − y, g = 3y²− x ∈ Q[x, y]. If the term order is lex order with x < y, then L = y²x, and S(f, g) = ^y_2yx^2x f − ^y_3y^2x2 g = ¹₂yf − ¹₃xg = −¹₂y²+¹₃x².

Algorithm 1.9. ( Buchberger’s Algorithm for Computing Gr¨obner Bases ) INPUT : F = {f₁, . . . , f_s} ⊆ K[X] with f_i 6= 0 (1 ≤ i ≤ s) OUTPUT : G = {g₁, . . . , g_t}, a Gr¨obner basis for < f₁, . . . , f_s >

INITIALIZATION : G := F , G :={f_i, f_j} | f_i 6= f_j ∈ G WHILE G 6= Ø DO

Choose any {f, g} ∈ G G := G −{f, g}

S(f, g) −→^G + h, where h is reduced with respect to G IF h 6= 0 THEN

G := G ∪{u, h} | for all u ∈ G G := G ∪ {h}

Example 5 : Let f1 = xy − x, f2 = −y + x² ∈ Q[x, y] ordered by the lex term ordering with x < y.

INITIALIZATION : G := {f1, f2}, G :={f1, f2} First pass through the WHILE loop

G := Ø

S(f1, f2) −→^G + x³ − x = h ( reduced with respect to G ) Since h 6= 0, let f3 := x³− x

G :={f1, f3}, {f2, f3} G := f1, f2, f3

Second pass through the WHILE loop

G :={f2, f3}

S(f₁, f₃) −→^G ₊ 0 = h Third pass through the WHILE loop

G := Ø

S(f₂, f₃) −→^G ₊ 0 = h The WHILE loop stops, since G := Ø .

Thus, {f₁, f₂, f₃} is a Gr¨obner basis for the ideal < f₁, f₂ > .

Definition 1.10. Let I be an ideal of K[X]. The initial ideal of I for the term order < is the ideal

in_<(I) =X^α | for all α ∈ D(I) .

The ideal in_<(I) is highly dependent on the chosen ordering ; once < is fixed, the ideal is

Let us fix, for our discussion, a term order which we denote simply by <. Let S = {X^α+ I | α /∈ D(I)}, where I is a non-zero ideal in K[X]. We are going to show that S is a basis for the K-vector space K[X] / I.

We first show that every element f + I in K[X] / I is a linear combination of S. Let f =Pr

i=0c_βiX^βi ∈ K[X], and we may assume that β₀ < β₁ < β₂ < · · · < β_r := deg(f ).

Let l₁ = max { i | β_i ∈ D(I), 0 ≤ i ≤ r }.

If l₁ = 0, then obviously, f + I is a linear combination of S.

Suppose l₁ > 0. Since β_l1 ∈ D(I), there exists a non-zero polynomial g₁ ∈ I such that β_l1 = deg(g₁) and lt(g₁) = c_βl1X^βl1 .

Let h₁ = f − g₁. Then f + I = h₁+ I.

Write h₁ =Ps

i=0d_γiX^γi ∈ K[X].

Again, let l₂ = max { i | γ_i ∈ D(I), 0 ≤ i ≤ s }.

If l₂ = 0, then h₁+ I is a linear combination of S, so is f + I.

Suppose l₂ > 0. Since γ_l2 ∈ D(I), there exists a non-zero polynomial g₂ ∈ I such that γ_l2 = deg(g₂) and lt(g₂) = d_γl2X^γl2 .

Let h₂ = h₁− g₂. Then f + I = h₁+ I = h₂+ I.

Doing the same thing on h₂ and keeping going, we will get that f + I is a linear combination of S.

Next, we show that S is a linearly independent subset of K[X] / I. Let Pm

i=0a_αi(X^αi+I) = I, where X^αi+ I ∈ S and α₀ < α₁ < · · · < α_m. Then Pm

i=0a_αiX^αi ∈ I.

If a_αm 6= 0, then α_m ∈ D(I), a contradiction. So, a_αm = 0 and Pm−1

i=0 a_αiX^αi ∈ I.

If a_αm−1 6= 0, then α_m−1 ∈ D(I), a contradiction. So, a_αm−1 = 0 and Pm−2

i=0 a_αiX^αi ∈ I.

Continuing in the same way, we get that a_αm = a_αm−1 = · · · = a_α0 = 0.

Therefore, S is a basis for the K-vector space K[X] / I.

Definition 1.11. For a given f ∈ K[X], the unique polynomial N ormalF orm(f ) =X

c_αX^α, where each α /∈ D(I), such that

f − N ormalF orm(f ) ∈ I, is called the normal form of f with respect to the chosen ordering.

Lemma 1.1. Let I be an ideal of K[X]. Then the mapping

N ormalF orm : K[X] / I −→ K[X] / in(I) is an isomorphism of K-vector spaces.

Proof : Define

N ormalF orm : K[X] / I −→ K[X] / in(I)

f + I −→ N ormalF orm(f ) + in(I)

Since {X^α+ I | α /∈ D(I)} is a basis of the K-vector space K[X] / I, clearly, NormalForm is well-defined, epimorphism, and ker(N ormalF orm) = { I }.

Therefore, K[X] / I ' K[X] / in(I) .