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All Graduate Theses, Dissertations, and OtherCapstone Projects Graduate Theses, Dissertations, and Other Capstone Projects 2020
Gröbner Bases and Systems of Polynomial Equations
Gröbner Bases and Systems of Polynomial Equations
Rachel HolmesMinnesota State University, Mankato
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Holmes, R. (2020). Gröbner Bases and Systems of Polynomial Equations [Master’s alternative plan paper, Minnesota State University, Mankato]. Cornerstone: A Collection of Scholarly and Creative Works for Minnesota State University, Mankato. https://cornerstone.lib.mnsu.edu/etds/1079/
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Equations
Rachel Holmes
An Alternative Plan Paper submitted in partial fulfillment of the
requirements for the degree of Masters of Science in Mathematics and
Statistics at Minnesota State University, Mankato
Mankato, Minnesota
committee members:
Dr. Wook Kim, Advisor Dr. Ruijun Zhao Dr. Brandon Rowekamp
Acknowledgments
To my advisor, Dr. Wook Kim, for his unending guidance, knowledge, motivation, and invaluable talent as an educator. To my committee members, Dr. Brandon
Rowekamp and Dr. Ruijun Zhao, for their time and evaluation. To my dear friends, Sarah and Alex, whom I am eternally grateful.
Contents
1 Introduction 1
2 Preliminary Information 2
2.1 Polynomials and Polynomial Rings . . . 2
2.2 Domains . . . 8
2.3 Division in k[x] . . . 11
2.4 Generators and Ideals . . . 13
2.5 Monomial Ordering . . . 21
3 Ideals and Varieties 27 3.1 Varieties . . . 27
3.2 Ideals . . . 29
3.3 Monomial Ideals . . . 33
4 Division in k[x1, . . . , xn] 38 5 Gr¨obner Bases 47 5.1 Gr¨obner Basis and Hilbert Basis Theorem . . . 47
5.2 Buchberger’s Algorithm . . . 59
5.3 History . . . 70
5.4 Applications . . . 72
6 Conclusion 87
Abstract
This paper will explore the use and construction of Gr¨obner bases through
Buchberger’s algorithm. Specifically, applications of such bases for solving
sys-tems of polynomial equations will be discussed. Furthermore, we relate many
1
Introduction
In 1965, Bruno Buchberger’s PhD thesis “An algorithm for finding the basis elements of the residue class ring of a zero-dimensional polynomial ideal” examined an algorithm provided to him by his advisor, Wolfgang Gr¨obner. In his thesis he discusses the termination and computer implementation of this algorithm. Shortly thereafter, he becomes the first person to implement it at the first computer laboratory [2]. This paper will seek to examine this algorithm as well, specifically its termination as well as some applications.
In section 2 we recall some important concepts from abstract algebra, specifically ring and field theory. We also discuss and define some important aspects of polynomi-als, discuss ordering of monomipolynomi-als, and recall the division algorithm for polynomials ink[x].
Section 3 contains many important results and definitions which will be used frequently throughout our discussion of Gr¨obner bases and Buchberger’s Algorithm in Section 5. Here we discuss Buchberger’s Algorithm and Buchberger’s Criterion and examine them in depth to verify their termination. We also consider the Hilbert Basis Theorem, and an equivalent theorem, the Ascending Chain Condition. Prior to this we discuss the division algorithm as it is extended to the polynomial ring k[x1, . . . , xn] in section 4. We conclude by considering a few applications of Gr¨obner bases. Many of the examples throughout this paper will make use of the Groebner package in Maple 2020.
2
Preliminary Information
2.1
Polynomials and Polynomial Rings
Polynomials are key in the study of Gr¨obner bases and while polynomials are familiar to most, some standard definitions need to be reconsidered to hold in the case of multiple variables. Ordering for example is not so clear. One may be familiar with the standard form of a polynomial in one variable, in which we order terms in descending order based on degree of the exponent. Ordering monomials in descending order becomes less obvious when dealing with multivariate polynomials. For example, determining the standard form of a polynomial such as f = 2x2yz3+x6+y5z is not so clear. To determine such a form, a few definitions are in order.
Definition 2.1. A monomial inx1, . . . , xn is a product of the form
xα1
1 ·x
α2
2 · · · · ·x
αn n
where α= (α1, α2, . . . , αn)∈Zn≥0. Then we can write xα =x
α1
1 ·x
α2
2 · · · · ·xαnn.
Definition 2.2. The total degree of a monomial xα1
1 · x
α2
2 · · · · ·xαnn is the sum
α1+· · ·+αn, and we can denote the total degree of xα as |α|=α1+· · ·+an.
Furthermore, we see when α= (0, . . . ,0), xα =x01· · · · ·x0n = 1.
linear combination of monomials whose coefficients are in k. That is, we can write
f =X α
aαxα
where aα ∈k and the sum is over a finite number of n-tuples α= (α1, . . . , αn).
Before discussing polynomial rings, it may first be useful to recall some basic axioms and definitions of rings and fields. Since polynomials have the commutative property, we will be concerned only with commutative rings.
Definition 2.4. Acommutative ringconsists of a setRand two binary operations “ + ” and “·” for which the following axioms are satisfied for all a, b, c∈R:
i. (Associativity) (a+b) +c=a+ (b+c) and (a·b)·c=a·(b·c).
ii. (Commutativity) a+b =b+a and a·b=b·a.
iii. (Distributivity)a·(b+c) =a·b+a·c.
iv. (Existence of identities) There exists 0,1∈R such that a+ 0 =a anda·1 =a.
v. (Additive inverses) For every a there exists −a such thata+ (−a) = 0.
Notice the last axiom fails to require the existence of a multiplicative inverse; if a commutative ring also has multiplicative inverses for all nonzero elements, then it is called a field.
Definition 2.5. Afieldkis a commutative ring with the property that for all nonzero a∈k, there exists a−1 ∈k such thata·a−1 = 1.
Definition 2.6. A polynomial ring k[x1, . . . , xn] is a commutative ring whose el-ements consist of the collection of polynomials whose coefficients belong to a field k
and which has indeterminants x1, . . . , xn.
The sum and products of any two polynomials will also result in a polynomial and it can be shown further that k[x1, . . . , xn] satisfies the axioms of a commutative ring. We will often make use of the general commutative ringk[x1, . . . , xn], but occasionally we may specify a specific ring such as R[x, y], which is the ring of polynomials with
indeterminants xand y with coefficients coming from the field of real numbers.
Note that x1, . . . , xn in many texts are referred to as variables; here instead we refer to them asindeterminants. This is to distinguish that in general we are treat-ing x1, . . . , xn as characters or symbols, and not always as variables for polynomial functions. Consider also the following definitions regarding polynomials.
Definition 2.7. Letf =P
αaαx
α be a polynomial in k[x
1, . . . , xn]. Then:
i. aα is called the coefficientof the monomial xα. ii. If aα 6= 0 then we call aαxα aterm of f.
iii. Thetotal degree off 6= 0, which we denote deg(f), is the maximum |α|such that the coefficient aα is nonzero.
As a note, the total degree of the zero polynomial is undefined or in some texts it is considered to have a degree of−∞.
Example 2.8. Consider the polynomial f = −2y3z5 + 13x2 ∈ k[x, y, z]; f has two
terms: −2y3z5 and 13x2, which have total degrees of 8 and 2, and coefficients−2 and 13, respectively.
Alternative notation for the two monomials in the above example would be (0,3,5) and (2,0,0), respectively. These ordered triples (x, y, z) ∈ Z3
≥0 represent
the degree of the individual indeterminants in a monomial and disregards any coef-ficients. This type of notation is convenient for ordering monomials which will be discussed further in section 2.5.
We next define affine spaces to allow us to communicate ideas about polynomials both algebraically and geometrically.
Definition 2.9. Given a fieldkand a positive integern, then define then-dimensional
affine space overk to be the set kn={(a
1, . . . an) :a1, . . . , an∈k}.
Now we discuss the evaluation map of a polynomial f. That is, for a polynomial
f = P
αaαxα ∈ k[x1, . . . , xn] and a = (a1, . . . , an) ∈ kn we have the evaluation map eva :k[x1, . . . , xn]→k defined by the homomorphism
f(x1, . . . , xn)→f(a1, . . . , an).
For (a1, . . . , an)∈kn, eachxi is replaced with ai in the given expression for f. Here
of f are in k, then f(a1, . . . , an) also lies in k. This function takes in some ordered
n-tuple from kn and will output some constant belonging tok, i.e.,
f :kn →k.
A special case of this is whenf(a1, . . . , an) = 0. This has two implications: eitherf is the zeropolynomial orf is the zerofunction. Iff is the zero polynomial, then all of its
coefficients are zero and thus f = 0. If f is the zero function, then f(a1, . . . , an) = 0 for all (a1, . . . , an)∈kn. Fortunately, the latter is only a possibility when k is afinite field.
Proposition 2.10. Let k be a infinite field and f ∈k[x1, . . . , xn]. Then f = 0 if and
only if f :kn →k is the zero function.
The zero polynomial will certainly give the zero function, but to see whyk must
be an infinite field for this proposition to hold, consider the following example.
Example 2.11. Let F2 = {0,1} be the finite field with two elements and consider the polynomial
f =x3−x∈ F2[x].
This is clearly not the zero polynomial, but it is the zero function since
f(0) =f(1) = 0.
We will only be concerned with polynomials rings overinfinite fields such asRor
is the zero polynomial f = 0. When f(a1, . . . , an) = 0, we refer to (a1, . . . , an) as a
zero or a root of f.
Theorem 2.12 (The Fundamental Theorem of Algebra). Every nonconstant polyno-mial f ∈C[x] has a root in C.
Many proofs for this theorem exist and can be found in many texts; while the statement may seem simple, the proof is often not and thus is omitted. Instead, to illustrate how we may fail to have a zero for a polynomial inR, consider the following
example.
Example 2.13. Let f =x2+ 1∈R[x]. To find the zeros of f,
x2+ 1 = 0
x2 =−1
x=±√−1.
Clearly neither zero is a real number, and thus f has no zeros in R. If instead we
take f ∈C[x], then f has two zeros x=±√−1 = ±i.
Furthermore, we can equate two polynomials if we consider their corresponding functions.
Corollary 2.14. Let k be an infinite field, then f =g for f, g ∈k[x1, . . . , xn] if and
Proof. The forward direction follows directly from the definition of the function of a polynomial. For the backward direction, we assume f, g ∈k[x1, . . . , xn] are the same function. Then, for all (a1, . . . , an)∈kn, this implies
f(a1, . . . , an) =g(a1, . . . , an).
It follows that the function f −g = 0 for all (a1, . . . , an)∈ kn, therefore f −g must be the zero function. Hence f −g is the zero polynomial by proposition 2.10, and
thus f =g.
2.2
Domains
Domains, which are special types of rings, are a important topic in the study of ring theory. We define a few here and and discuss their relationship to polynomial rings. One may also recall that these domains are nested proper subsets of each other as follows:
Fields ⊂ Euclidean Domains ⊂ PID’s ⊂ UFD’s ⊂Integral Domains.
An integral domainis a ring R such that R contains no zero divisors. A zero divisor is a nonzero element a ∈R such that a·b = 0 for some b ∈ R. For example,
the ringZ/5Zis an integral domain, butZ/6Zis not an integral domain as it contains
zero divisors. For example, in Z/6Z
so 4 is a zero divisor. Since k[x1, . . . , xn] has no zero divisors, it is a integral domain.
To discuss unique factorization domains, we first recall the following definitions.
Definition 2.15. In a commutative ring R, and for p, x, y ∈ R we call p prime if when p|xy, then p|x or p|y.
Definition 2.16. In a commutative ringR, and fora, b, c∈R, we call airreducible
if when a=bc, b is a unit or cis a unit.
In general, a prime element is also irreducible, though the converse is not always true. In a unique factorization domain, we find that irreducible elements are also prime elements. A unique factorization domain (UFD) is a integral domain in which every element can be written uniquely as a product of prime powers.
To see that irreducible elements are also prime elements, let a ∈ R be a
irre-ducible element. Then, we can writea|bcfor someb, c∈R. If we can showa|b ora|c,
then a is also prime. Since a|bc, by the division algorithm, we can write ad =bc for
some d∈R. Sinceb, c, d∈R, we can write them as a product prime powers. Denote
these unique factorizations by
b=bα1
1 · · · · ·b
αr r ,
c=cβ1
1 · · · · ·c
βs s ,
d=dγ1
1 · · · · ·d
γt t ,
where bαi i , c
βj j , d
γk
a(dγ1
1 · · · · ·d
γt
t ) = (bα11 · · · · ·bαrr)(c β1
1 · · · · ·cβss).
It follows that since these factorizations are unique, then a must be some bi or cj, or
equivalently, a|b or a|c.
Clearlyk[x1, . . . , xn] is a UFD, where an irreducible element is defined as follows.
Definition 2.17. A non-constant polynomial f ∈ k[x1, . . . , xn] is said to be
irre-ducible if when
f =h·g
for h, g ∈k[x1, . . . , xn], then either h org is a constant.
Aprincipal ideal domain(PID) is defined as an integral domain in which every ideal can be generated by a single element. In generalk[x1, . . . , xn] is not a PID, but if n = 1, as we will see in Theorem 2.23, every ideal I ⊆k[x] can be generated by a
single polynomial, and therefore k[x] is PID. We will investigate PID’s further after
defining ideals and generators. Lastly, we have Euclidean domains. A integral domain R is a Euclidean domain if we have some sort of measure ν : R →Z≥0 and
for a, b∈R and b6= 0, then there exists unique q, r∈R such that
a=b·q+r
where r= 0 or ν(r)< ν(b).
We find k[x] is also a Euclidean domain with the usual polynomial division, the
result of which we define in the following section. In this case the measure in question is degree.
2.3
Division in
k
[
x
]
One may recall the usual single indeterminant polynomial division. In k[x] we
can define a polynomial simply as follows.
Definition 2.18. A polynomial f ∈k[x] is of the form f =c0xm+c1xm−1+· · ·+cm−1x+cm
where ci ∈ k and c0 = 0. We call6 c0xm the leading term of f, and denote this by
LT(f) =c0xm. We say f is of degree m and denote this by deg(f) = m.
Theorem 2.19. Let k be a field, and f, g ∈ k[x] such that g 6= 0. Then there exists unique q, r ∈k[x] such that f =gq+r where deg(r)<deg(g).
The proof of this theorem is omitted, not as to dismiss the importance and power of it as its result is used frequently, but to focus on the division algorithm in multiple indeterminants. Proofs for this theorem can be found in many algebra texts. Though instead we will consider the algorithm in pseudo-code, as it relates closely to that of the algorithm in k[x1, . . . , xn]. When performing division in k[x] we work with a single divisor and produce a single quotient. An important result of the division algorithm ink[x] is the quotient and remainder areunique. This may not be the case
when we extend this processes tok[x1, . . . , xn].
Consider the pseudo-code below, presented as it is in [4], of the division of a polynomial f by g to produce unique remainder r and quotientq.
Algorithm 1: Division in k[x]
Input : f, g
Output: q, r
1 q := 0 2 r :=f
3 while r 6= 0 AND LT(g)|LT(r) do 4 q :=q+LT(r)/LT(g)
5 r :=r−(LT(r)/LT(g))g 6 end
7 RETURN q, r;
An important piece of this algorithm is the redefinition of q and r for each
iteration. This gives us a useful identity:
f =q·g+r=
q+
LT(r) LT(g)
·g+
r−
LT(r) LT(g)
·g
.
Consider the following example, which demonstrates how this identity is useful in writing the iterations of division concisely.
Example 2.20. Letf = 2x3+ 8x2−3 and g =x+ 3, then by the division algorithm
2x3+ 8x2−3 = (0)(x+ 3) + (2x3+ 8x2−3)
= (2x2)(x+ 3) + (2x2−3)
= (2x2+ 2x)(x+ 3) + (−6x−3)
= (2x2+ 2x−6)(x+ 3) + (15).
A typical proof of the termination of the division algorithm follows closely to the above example, as note the degree of the remainder is strictly decreasing at
each iteration. Instead of verifying it’s termination formally, we will explore the termination of the division algorithm in k[x1, . . . , xn] in depth in section 4.
2.4
Generators and Ideals
We recall the definition of an ideal, as ideals are central in the discussion of Gr¨obner bases.
Definition 2.21. Let R be a commutative ring. Then I ⊆ R is an ideal if the following conditions hold:
i. 0 ∈I.
ii. a+b∈I for all a, b∈I.
iii. r·a∈I for all a∈I, r∈R.
Polynomial rings also contain ideals, and like any ringk[x1, . . . , xn] has the trivial ideals: the zero ideal {0} and the ring k[x1, . . . , xn] itself. The most common ideals we will encounter are those generated by a collection of polynomials. When we say “generate” this is analogous to “spanning” in the linear algebra sense. In order for a collection of polynomials to generate an ideal I they must span the entirety of I.
Often we will begin with a set of polynomials f1, . . . , fs and we will generate an ideal
I with those polynomials.
To follow with this linear algebra concept, recall the unit vectors ~e1 = (1,0),
~e2 generate all of R2 and we often call {~e1, ~e2} a basis for R2. Since these vectors
form a basis for R2, then we can represent any vector (x, y)∈R2 by a unique linear
combination
(x, y) =c1~e1+c2~e2
with coefficients c1, c2 ∈R.
Example 2.22. Similarly, consider the ring C[x]. If f is some polynomial in C[x], we denote the set of polynomials generated by f ashfi. This set has the form
hfi={h·f :h∈C[x]}.
Take, for example, f = 1; the set of polynomials generated by f is
hfi=h1i={h·1 :h∈C[x]}
which is clearly the entire ring C[x]. As another example, take instead f = x−1;
then the set of polynomials generated by f is
hfi=hx−1i={h·(x−1) : h∈C[x]}.
This set can be also thought of as the collection of polynomials which attain a zero at 1.
Theorem 2.23. If k is a field, then every ideal I ⊆ k[x] can be written as hfi for some f ∈ k[x]. Furthermore, f is unique up to multiplication by a nonzero constant
Proof. Assume I ⊆ k[x] is a nonzero ideal, otherwise we are done. We will show
I =hfi by choosing nonzerof ∈I such that f is of minimum degree of the elements
of I. By the division algorithm, for f ∈I we can write any element g ∈I as
g =q·f+r
for unique q, r ∈I, and degr <degf. Then, we write
r=g−q·f.
Ifr6= 0 we arrive at a contradiction, as clearlyr ∈I and degr <degf, but we chose
f to be of minimal degree. Thereforer must be zero, i.e., any element g ∈I can be
generated by f. Uniqueness can be shown similarly by assuming I can be generated
by hfi and hgi. One can then write f = q·g since f ∈ hgi and likewise g = ˜q·f.
Performing appropriate substitution, we obtain
f =q·q˜·f
which implies deg(q) = deg(˜q) = 0, and thusf andg differ only by some constant.
If we are given some idealI ⊆k[x] which is generated by more than one
polyno-mial, say I = hf1, . . . , fsi, how can we find a unique f such that hfi =hf1, . . . , fsi?
This problem is easily answered by use of the greatest common divisor (gcd).
Definition 2.24. A polynomial h ∈ k[x] is said the be the greatest common divisor of the polynomials f, g ∈k[x] when:
ii. If p is another polynomial such that p|f and p|g, thenp|h.
When h satisfies the above conditions, we denote h= gcd(f, g).
The gcd of two polynomials has the following properties.
Proposition 2.25. Let f, g ∈k[x]. Then:
i. gcd(f, g) exists and is unique up to multiplication by a nonzero constant in k.
ii. hgcd(f, g)i=hf, gi.
iii. There is an algorithm for finding gcd(f, g).
Proof. The proof of existence and uniqueness is common and trivial, and the existence will be examined using the Euclidean algorithm. We instead focus on the equivalence of ideals and the algorithm for finding gcd(f, g). To see hgcd(f, g)i=hf, gi, suppose
gcd(f, g) exists and denote it by h = gcd(f, g). It follows from the definition of
gcd(f, g) thath|f and h|g. Therefore, for some ˜f ,˜g ∈k[x], we can write
f =h·f˜
g =h·˜g =⇒
f ∈ hhi
g ∈ hhi
Therefore hf, gi=hhi=hgcd(f, g)i.
The algorithm mentioned in (iii) is the usual Euclidean algorithm for finding the gcd(f, g), and is stated below [4].
Algorithm 2: Euclidean Algorithm
Input : f, g ∈k[x]
Output: h= gcd(f, g)
1 h:=f 2 s :=g
3 while s 6= 0 do
4 r := remainder(h, s) 5 h:=s
6 s :=r 7 end
8 RETURN h;
To see that this algorithm does produce the gcd(f, g), for f =qg+r ∈k[x], we
claim that gcd(f, g) = gcd(f −qg, g) = gcd(r, g). It suffices to show that hf, gi =
hf −qg, gi.
Consider an element ˜f ∈ hf−qg, gi. We can write
˜
f = (f−qg)·h1 +g·h2
=f h1 −qgh1+gh2
=f h1 −g(qh1+h2),
and denoting ˜h2 = qh1 +h2 ∈ k[x], then ˜f is of the form ˜f = f h1 −gh˜2 ∈ hf, gi.
Therefore hf, gi = hf − qg, gi = hr, gi. This shows that the first iteration of the
Euclidean algorithm holds. To see that it holds at every iteration and that it will eventually terminate, consider that after the first iteration deg(g)>deg(r) or r = 0.
Assuming r6= 0, this algorithm continues through another iteration. By the division
deg(r) > deg(r0), or r = 0 and the algorithm terminates. Assuming it does not
terminate, then at each step we continue to have the strict set of inequalities deg(g)>deg(r)>deg(r0)>deg(r00)> . . .
which clearly must terminate since deg(g) is finite. Therefore at some iteration we
will achieve r = 0 which givess= 0, and thus h= gcd(h,0) = gcd(f, g).
Consider the following example, which makes use of both the division and Eu-clidean algorithms.
Example 2.26. Let I = hf1, f2i ⊆ R[x], where f1 = x3 −2x2 +x−2 and f2 =
x5 −2x4 − 10. Since
R[x] is a PID, we must be able to generate this ideal with
a single polynomial. Furthermore, this implies R[x] is UFD, and therefore by the
division and Euclidean algorithms we can write
f1 =x3−2x2+x−2 = (x−2)(x2 + 1)
and
f2 =x5−2x4−10 = (x−2)(x4+ 5).
We see that the gcd(f1, f2) = x−2, implying the ideal can instead be generated by
the single polynomial x−2.
Fortunately, we can extend these properties and definitions of gcd(f, g) to ideals
common divisor for more than two polynomials.
Definition 2.27. A polynomial h ∈ k[x] is said the be the greatest common divisor of the polynomials S ={f1, . . . fs} ⊆k[x] when:
i. h|f for all f ∈S.
ii. If p is another polynomial such that p|f for all f ∈S, then p|h.
When h satisfies these conditions, we denote h= gcd(f1, . . . , fs).
Proposition 2.28. Let f1, . . . fs ∈k[x],where s≥2. Then:
i. gcd(f1, ..., fs) exists and is unique up to multiplication by a nonzero constant.
ii. gcd(f1, ..., fs) is a generator of the ideal hf1, . . . , fsi.
iii. If s ≥3, then gcd(f,. . . , fs) = gcd(f1,gcd(f2, . . . , fs)).
iv. There is an algorithm for finding gcd(f1, . . . , fs).
The proof of this proposition is omitted as it follows similarly to that of proposi-tion 2.25. The algorithm menproposi-tioned in (iv) is the same Euclidean algorithm as before, where the gcd is found in pairs, as mentioned in (iii). Note the gcd(f1, . . . , fs) may be 1 (or some other nonzero constant), in which this would imply the ideal is the entire ring k[x].
More often a generating set is made up of polynomials of multiple indeterminants. Before we formally define a generating set, we consider two examples of ideals in k[x, y].
Example 2.29. Let f1 = x−3y and f2 = 2y2 be polynomials in C[x, y]. Then all
f ∈I =hf1, f2iare of the form
f =h1·f1+h2·f2
=h1·(x−3y) +h2·(2y2),
where h1, h2 ∈C[x, y]. Furthermore, the polynomial 3xy2−3y3−2y2 ∈I, since
3xy2−3y3−2y2 = (y2)(x−3y) + (x−1)(2y2).
Example 2.30. Let I =hx+y2, x3i and J =hx+y2, x2y2i be two ideals in k[x, y].
Notice we can write the second generating polynomial ofJ in terms of the polynomials
fromI; that is,
x2y2 = (x2)(x+y2) + (−1)(x3).
This means J ⊆I. Furthermore, we can representx3 as a linear combination of the
polynomials in J.
x3 = (x2)(x+y2) + (−1)(x2y2).
This showsI ⊆J. Because both ideals are subsets of one another, this impliesI =J.
In general, we can have a generating set of any number of polynomial in the ring k[x1, . . . , xn]. We now formally define such a generating set.
hSi= (
X
α
hαfα :fα ∈S and hα ∈k[x1, . . . , xn]
) ,
for 1 ≤ α ≤ s for some finite s. We call hSi the ideal generated by S where
S = {f1, . . . , fs}. Occasionally, for f ∈ hSi =hf1, . . . , fsi, we may write f explicitly as
f =h1·f1+· · ·+hs·fs.
We occasionally use T to represent some other collection of polynomials.
Lemma 2.32. If S ⊆k[x1, . . . , xn], then hSi is an ideal ofk[x1, . . . , xn].
Clearly by the definition of hSi, the axioms of an ideal are satisfied. Our main
concern lies with ideals which can be finitely generated.
Definition 2.33. An idealI ⊆k[x1, . . . , xn] is said to befinitely generatedif there existsS ⊆k[x1, . . . , xn] such thatI =hSi, then we callS a basisof I.
An amazing fact we will see (by theorem 5.7) isevery ideal ink[x1, . . . , xn] can be
finitely generated. Before proving this, we must first discuss how to order monomials ink[x1, . . . , xn].
2.5
Monomial Ordering
When working in a polynomial ring in one indeterminate, ordering is simple and is based on the degree of the term. In k[x], we saw that the leading term was
an important definition in the division algorithm; this will prove to be true in the algorithm for k[x1, . . . , xn] as well. As such, we need to be able to define what a leading term is when we have multiple indeterminants.
A problem arises when we attempt to order monomials for polynomials in k[x1, . . . , xn]. How, for example, do we order terms in descending order for the poly-nomial
2x3y2z2−13x5y2−2x3z2+ 5y4+ 3z5?
Instinct may lead us to choose to order based on the total degree, but this decision quickly causes a problem: the first two terms in the example above both have a total degree of 7. As such, total degree alone is not sufficient for ordering monomials in k[x1, . . . , xn]. We must then decide how to compare individual indeterminants. This is what we call a monomial ordering. We will use to denote some ordering, for example, xy.
When comparing individual monomials, it is common to use the notation
α = (α1, . . . , αn)∈Zn≥0.
Recall this ordered n-tuple represents the monomial
xα =xα1
1 · · · · ·xαnn ∈k[x1, . . . , xn].
To be able to compare monomials, we need a total ordering. That is, given any two monomials xα, xβ, exactly one of the following three statements must be true:
Any valid ordering should also have transitivity, i.e., if xα xβ and xβ xλ, then
xα xλ. Furthermore, it is required that for xα xβ, and for any monomial xλ, we
havexαxλ xβxλ.
Definition 2.34. A monomial ordering onk[x1, . . . , xn] is a relation onZn≥0,
or equivalently, a relation on the set of monomialsxα, α∈Zn
≥0, satisfying:
i. is a total ordering on Zn≥0.
ii. If α β, and forγ ∈Zn
≥0, then α+γ β+γ.
iii. is a well-ordering on Zn
≥0. That is, if A ⊆ Zn≥0 is nonempty, then there is
α∈A such that β α for every β 6=α inA.
The last condition simply states a smallest element must exists under any order-ing. This well-ordering property becomes important when discussing the termination of the division algorithm in k[x1, . . . , xn]. The following definitions with be useful in the discussion of ordering.
Definition 2.35. Letf =P
αaαx
α be a nonzero polynomial in k[x
1, . . . , xn] and fix a monomial order.
i. The multidegree of f is
deg(f) = max(α∈Zn
(Henceforth, when discussing degree it will be assumed we are discussing the multidegree.)
ii. The leading coefficient of f is
LC(f) =adeg(f) ∈k;
iii. Theleading monomial of f is
LM(f) = xdeg(f) ∈k;
iv. The leading termof f is
LT(f) = LC(f)·LM(f);
An equivalent definition for the degree of a polynomial is
deg(f) = deg(LT(f)).
We will consider three different orderings. Often, when discussing polynomials of three or less variables we often usek[x, y, z], and subscriptsk[x1, . . . , xn] when working
with polynomials of n >3 indeterminants. One of the most common and instinctive
much like alphabetical ordering. In k[x1, . . . , xn], Lex ordering assumes
x1 x2 · · · xn, and xi ≺x2i ≺. . ..
Fork[x, y, z] it is generally assumedxyz.While we can choose any ordering for
individual indeterminants, such as xn · · · x1, we will assume x1 x2 · · · xn
unless another ordering is explicitly stated.
Example 2.36. For an example, let’s consider again the polynomial f = 2x3y2z2−13x5y2−2x3z2+ 5y4+ 3z5
and order it using Lex ordering:
−13x5y2 + 2x3y2z2−2x3z2+ 5y4+ 3z5.
Equivalently, we could compare the individual monomials x5y2 and x3y2z2 using the
notation
(5,2,0)(3,2,2).
Another common type of ordering is Graded Lexicographic (Grlex) which is similar to Lex ordering but orders monomials by total degree first. Any monomials of a shared total degree are then sorted using Lex ordering. For example, consider the monomialsx3y3 andx5. Using Grlex ordering x3y3 x5, where the contrary was true
with Lex ordering. Let’s order the the polynomialf from Example 2.36 in decreasing
order using Grlex ordering:
The most efficient in terms of computation time, but likely the most difficult conceptually, is Graded Reverse Lexicographic (Revgrlex) ordering. Revgrlex orders first by total degree but sorts monomials of shared degree in “reverse” as compared to Grlex. For example, consider we have to order the two monomialsx4y3z
and x3y2z3, which both have the same total degree of 8. With Revgrlex ordering we break ties by the smallest degree of z (the “smallest” indeterminate), so x4y3z
x3y2z3, or (4,3,1)(3,2,3). Let’s order the polynomialf from Example 2.36 again,
instead using Revgrlex ordering:
−13x5y2 + 2x3y2z2−2x3z2+ 3z5+ 5y4.
To compare how we order in Grlex versus Revgrlex, denote
α= (α1, α2, α3), β = (β1, β2, β3),
where |α| = |β|, and xα, xβ ∈ k[x
1, x2, x3]. Assume x1 x2 x3. If we are using
Grlex ordering then we compare αi and βi starting from the left and are concerned
with which one is largest, i.e.,
(α1, α2, α3) (β1, β2, β3)
if α1 >β1.
If we findα1 =β1, we move to the right and compareα2 andβ2. In general we would
compare by continuing to the right until we find αi 6=βi.
If we are using Revgrlex, then we compare αi and βi starting from the right and
(α1, α2,α3) (β1, β2,β3)
if α3 <β3.
In the same sense, we would continue comparing αi and βi by moving to the left if
we have equality.
3
Ideals and Varieties
In this section we consider the corresponding system of equations for the basis elements of an ideal, and the solutions to this system, or the variety. We also consider the ideal of this variety, and consider ideals generated by monomials.
3.1
Varieties
Definition 3.1. Given S ⊆k[x1, . . . , xn], the variety V(S) is defined as
V(S) ={(a1, . . . an)∈kn:f(a1, . . . an) = 0 for all f ∈S}.
Simply, a variety is the collection of (a1, . . . an)∈knon which all polynomials in
S vanish.
Example 3.2. Consider the polynomialf =x2+x−6∈C[x], which has the variety
V(x2 +x−6) ={a∈C1 :f(a) = 0}={−3,2}.
Example 3.3. Consider the polynomials f1 = −2x+y−3 and f2 = x−y−1 in C[x, y]. The variety V(f1, f2) is the collection of ordered pairs (x, y) ∈ C2 which
satisfy both
−2x+y−3 = 0
x−y−1 = 0 =⇒
−2x+y = 3
x−y = 1.
Using Gaussian elimination, or other algebra techniques, one can find that the only ordered pair which satisfies both equations is (−4,−5). Thus,
V(f1, f2) ={(−4,−5)}.
For S ⊆ k[x1, . . . , xn] we have the variety V = V(S). If f ∈ S, then clearly f vanishes ofV, but consider further the polynomials inhSi. If I =hSiis an ideal, not
only can we consider the variety V(S), but we can also consider the variety of the ideal V(hSi). The following lemma tells us these are equivalent.
Lemma 3.4. If I =hSi ⊆k[x1, . . . , xn], then V(S) = V(hSi).
Proof. By definition, V(S)⊂V(hSi), but to show V(hSi)⊂V(S), let f ∈ hSi=I. Then, if S ={f1, . . . , fs}, we can write
f =h1·f1+· · ·+hs·fs
for hi ∈k[x1, . . . , xn]. Then each fi vanishes for all (a1, . . . , an)∈V(S), i.e,
Then
f(a1, . . . , an) =h1(a1, . . . , an)·f1(a1, . . . , an) +· · ·+hs(a1, . . . , an)·fs(a1, . . . , an)
=h1(a1, . . . , an)·0 +· · ·+hs(a1, . . . , an)·0
= 0.
Therefore V(S) =V(hSi).
The above lemma is useful in proving the following result.
Proposition 3.5. If S and T are bases of the same ideal in k[x1, . . . , xn], so that
hSi=hTi, then V(S) =V(T).
Proof. Let S and T ∈ k[x1, . . . , xn] be bases of the same ideal I such that hSi =
hTi=I. Then consider the variety V(S). By lemma 3.1,V(S) = V(hSi). Therefore
V(S) = V(hSi) =V(hTi) =V(T).
3.2
Ideals
We can also consider the ideal of a variety, which consists of all such polynomials which vanish on the given variety.
Definition 3.6. LetV ⊆kn be an affine variety. Then, we denote
An important observation of the above definition is the following lemma, which tells us I(V) is in fact an ideal.
Lemma 3.7. If V ⊆ kn is an affine variety, then I(V) ⊆ k[x1, . . . , xn] is an ideal.
We call I(V) the ideal of V.
The proof is omitted as I(V) clearly satisfies the requirements of an ideal. We wish to investigate the relationship betweenI(V(S)) and the idealhSiitself. The next lemma states there is a clear relationship if our idealhSiis of the formhx−ai ∈k[x].
Lemma 3.8. Let a ∈ k and let eva : k[x] → k be the evaluation homomorphism.
Then ker(eva) is an ideal of k[x] where ker(eva) = hx−ai. Furthermore, by the First
Isomorphism Theorem, the following diagram commutes
k[x] k
k[x]/kereva eva
π
such that π is the natural homomorphism π(f(x)) =f(x) + ker(eva).
Proof. We claim that ker(eva) =hx−ai. To see this, note that every f ∈ hx−ai is
of the form
f = (x−a)·h
for h ∈ k[x]. Clearly f(a) = (a−a)·h(a) = 0 and therefore ker(eva) ⊂ hx−ai.
To see that ker(eva) ⊂ hx−ai, let f(x) ∈ker(eva) where f(a) = 0. By the division
f(x) = (x−a)q(x) +r(x)
where degr(x) < deg(x−a) and thus r(x) must be a constant r ∈ k. Evaluating
f(a) we find
f(a) = (a−a)q(a) +r
= 0·q(a) +r
=r
but f(a) = 0 therefore r= 0. This implies we can write f(x)∈ker(eva) as
f(x) =q(x)(x−a)
which is the form of all elements in hx−ai.
That is, forf =x−a∈k[x] the ker(eva) is precisely the idealI(V(x−a)). Since we have shown ker(eva) = hx−ai, then
I(V(x−a)) = hx−ai. The following example illustrates this.
Example 3.9. Consider the ideal J ⊂C[x] defined by
J =hx−2i={(x−2)·q(x) :q(x)∈C[x]}.
From J, we can consider the variety
Furthermore, we can consider the ideal of the variety
I(V) = {f(x)∈C[x] :f(2) = 0}.
Clearly J ⊆I(V), and to show I(V)⊆ J, let f(x)∈ C[x] be any polynomial in I(V) such that f(2) = 0. Then, by the division algorithm we can write
f(x) = (x−2)·q(x) +r(x)
such that deg(r(x))<1. Then r(x) must be a constant r∈C and
f(2) = (2−2)·q(2) +r
= (0)·q(2) +r
=r,
but f(2) = 0, so therefore r= 0. Therefore J =hx−2i=I(V(x−2)).
The next example further explores this relationship between an idealhSiand an
ideal of the variety I(V(S)), and will show that equality is not always the case.
Example 3.10. Consider f = x2 ∈ k[x]. The variety V(x2) = 0 gives the ideal
I(V(x2)) = I({0}). Notice also that I({0}) = hxi. Therefore, since hx2i
( hxi, we
have
hx2i(I(V(x2)).
Lemma 3.11. Let S ⊆k[x1, . . . , xn]. Then, hSi ⊆I(V(f1, . . . , fs)).
The proof is trivial and is illustrated by the preceding example. Given two varietiesV andW we can discuss the relationship between their corresponding ideals.
Proposition 3.12. Let V and W be two affine varieties in kn.
i. V ⊆W if and only if I(W)⊆I(V)
ii. V =W if and only if I(V) =I(W).
Let us examine the following example in place of a proof.
Example 3.13. Consider the following varieties in k:
V1 =V(x−1) and V2 =V((x−1)(x+ 2)).
These varieties have the following ideals, respectively:
I(V1) ={f(x)∈k[x] :f(1) = 0}
I(V2) ={f(x)∈k[x] :f(1) = 0 and f(2) = 0}.
Then we have V1 ⊆V2 and I(V2)⊆I(V1).
3.3
Monomial Ideals
Here we consider ideals generated by monomials as they are the building blocks of polynomials.
Definition 3.14. An ideal I ⊆ k[x1, . . . , xn] is a monomial ideal if there is a (possibly infinite) subset A ⊆ Zn
≥0 such that I consists of all polynomials which are
finite sums of the form P
α∈Ahαxα, where hα ∈k[x1, . . . , xn].
Since an element of Zn
≥0 is of the form α = (α1, . . . , αn), which represents the monomial xα =xα1
1 · · · · ·xαnn, A is simply a collection of monomials xαi for αi ∈ A. Then the ideal I consists of linear combinations of these monomials with coefficients
hα ∈ k[x1, . . . , xn]. Hence we can write I = hxα : α ∈ Ai ⊆ k[x1, . . . , xn]. For
example, I =hx2y4, x3y3, x6y2i is a monomial ideal.
Lemma 3.15. Let I =hxα :α∈Ai be a monomial ideal. Then, a monomial xβ ∈I
if and only if xβ is divisible by xα for some α∈A.
The proof is trivial and instead we consider the following example.
Example 3.16. Let I = hx5y3, x3y2i. Then clearly x6y6 ∈ I since x5y3|x6y6, and likewise since x5y3 6 |x2y and x3y2 6 |x2y, thenx2y6∈I.
In general, for α∈A, if xβ ∈ hxαi, then we must be able to write
xβ =xα·xγ for some γ ∈Zn
≥0
which implies β = α+γ. That is, we can write the elements of the ideal I = hxαi
equivalently as
α+Zn≥0 ={α+γ :γ ∈Z
n
Example 3.17. Consider again the monomial ideal I = hx2y4, x3y3, x6y2i ⊆ k[x, y].
Elements in this ideal then take the form
xβ =h1x2y4+h2x3y3+h3x6y2
or equivalently
xβ =X α
hαxα
whereα1 = (2,4), α2 = (3,3), α3 = (6,2), andhαi ∈k[x1, . . . , xn]. Furthermore, these elements could be represented in the form
((2,4) +Z2≥0)∪((3,3) +Z2≥0)∪((6,2) +Z2≥0).
This can also be visualized by the following graph:
m n
(2,4)
(3,3)
(6,2)
of all possible monomials (which occur at the lattice points) of the elements of the ideal I.
Lemma 3.18. Let I be a monomial ideal, and let f ∈ k[x, y]. Then, the following are equivalent:
i. f ∈I.
ii. Every term of f lies inI.
iii. f is a k-linear combination of the monomials inI.
The proof is trivial and the third result gives the following corollary.
Corollary 3.19. Two monomial ideals are the same if and only if they contain the same monomials.
Recall by the definition of a monomial ideal, the generating set of monomials may be infinite. The following theorem tells us even if we have an infinite generating set we can always reduce this set to a finite basis.
Theorem 3.20 (Dickson’s Lemma). Let I = hxα : α ∈ Ai ⊆ k[x1, . . . , xn] be
a monomial ideal. Then, I can be written in the form I = hxα1, . . . , xαsi, where α1, . . . , αs∈A. That is, I has a finite basis.
We consider an informal sketch of the proof which follows from the proof given in [2]. The proof of this theorem is generally done by induction on the number of indeterminants n. One can see this is true for n = 1, and we assume the theorem is
true for n−1 as our hypothesis. To see that it holds forn indeterminants, we write
k[x1, . . . , xn] as k[x1, . . . , xn−1, y] so monomials in k[x1, . . . , xn−1, y] take the form
xαym, forα = (α
1, . . . , αn−1) andm= (αn). Then to find the generators for an ideal
I ⊆ k[x1, . . . , xn−1, y], we look at the generators for J ⊆ k[x1, . . . xn−1]. Specifically,
we let J be generated by the elements xα such that xαym ∈ I. By the induction
hypothesis, we know this generating set is finite and we write J = hxα1, . . . , xαsi. Then, we choose m such that m is the maximum mi for xαiymi ∈ I. From here we create a “slice” for each monomial containing yl for 0 ≤ l ≤ m−1, and use these
slices to generate ideals. This process will achieve a finite set of generators. Let’s consider an example to see how these “slices” work.
Example 3.21. Take the idealI =hx2y4, x3y3, x5y3, x6y2i ⊆k[x, y]. We first find the
idealJ ⊆k[x] whereJ =hxαisuch thatxαym ∈I. We findJ =hx2, x3, x5, x6i=hx2i, which is clearly finitely generated. Then note x2y4 contains the largest such m = 4.
Then we create “slices”. Consider the following ideals defined byJl =hxα :xαyl ∈Ii
for 0≤l≤m−1 = 3, and the ideal J:
J0 =hxα :xαy0 ∈Ii=∅
J1 =hxα :xαy1 ∈Ii=∅
J2 =hxα :xαy2 ∈Ii=hx6i
J3 =hxα :xαy3 ∈Ii=hx3, x5i=hx3i
If we use the corresponding monomials xαyl ∈ I which come from J
l 6= ∅, these monomials will be the finite generating set for the ideal I. That is, for this example
we see J2, J3, and J are nonzero ideals and thus correspond to the monomials for a
finite generating set. Therefore {x6y2, x3y3, x2y4} is a finite basis for the monomial
ideal I =hx2y4, x3y3, x5y3, x6y2i. Though this was clearly already finite this process helps illustrate how we can find a minimal basis for a monomial ideal. We see that the slice made with J3 allowed us to remove the monomial x5y3 from the generating
set and maintain the same ideal.
Proposition 3.22. A monomial ideal I ⊆k[x1, . . . , xn] has a basis xα1, . . . , xαs with
the property that xαi does not divide xαj for i6=j. Furthermore, this basis is unique
and is called the minimal basis of I.
The proof follows closely based on the example below.
Example 3.23. Continuing with the ideal I = hx2y4, x3y3, x5y3, x6y2i, we see the
basis x6y2, x3y3, x2y4 is in fact a unique minimal basis ofI since none of the elements
of the basis divide another element. This was to be expected and is clear based on the graphical representation of this ideal in example 3.17.
4
Division in
k
[
x
1, . . . , x
n]
“The algebra behind the division algorithm is very simple...which makes it sur-prising that this form of the algorithm was first isolated and exploited only within the
past 50 years” [4]. Examples of such division will quickly show why computational advancements were the turning point in the development of this division algorithm and other lengthy algorithms in the field of computational algebra.
Theorem 4.1. Let be a monomial ordering on Zn
≥0 and let F = (f1, . . . , fs) be an
ordered s-tuple of polynomials in k[x1, . . . , xn]. Then, every f ∈ k[x1, . . . , xn] can be
written as
f =q1f1+· · ·+qsfs+r
where qi, r ∈k[x1, . . . , xn], and either r = 0 or r is a linear combination, with
coeffi-cients in k, of monomials, none of which is divisible by any of the LT(f1), . . . ,LT(fs).
We call r a remainder of f on division by F, and often denote it as fF.
Further-more, if qifi 6= 0, then
deg(f)≥deg(qifi).
We prove this theorem by means of examining the pseudo-code for this algorithm and through use of examples. The pseudo-code, as provided by [4], establishes the existence of qi and r and can be found in Algorithm 3. Furthermore, we will discuss
the termination of the algorithm within the section.
Division ink[x1, . . . , xn] and its algorithm follow closely to that of division ink[x].
We will denote the polynomial in the dividend by f, divisors byfi, and quotients by
qi. First we must fix a monomial order , if one is not specified. Since the order
Algorithm 3: Division in k[x1, . . . , xn]
Input : f1, . . . , fs, f
Output: q1, . . . , qs, r
1 qi := 0 2 r := 0 3 p:=f
4 while p6= 0 do 5 i:= 1
6 divisionoccured:=f alse 7 s :=r
8 while i≤s AND divisionoccurred=f alse do 9 if LT(fi) divides LT(p)then
10 qi :=qi+LT(p)/LT(fi) 11 p:=p−(LT(p)/LT(fi))∗fi 12 divisionoccurred:=true
13 else
14 i:=i+ 1
15 end
16 end
17 if divisionoccurred= false then 18 r :=r+LT(p)
19 p:=p−LT(p)
20 end
21 end
consistency in which the order of the polynomials as they appear in the basis. To the right of this division, we track any remainder terms we obtain along the way. We set this division up to appear in the following fashion:
q1:
. . . qs:
f1
. . . fs
f
r
As to maintain consistency with the pseudo-code, we initialize p=f. If there is
more than one divisor we begin by finding the firstfi in which LT(fi)|LT(p). We then
perform division of these leading terms LT(p)/LT(fi) and place this value with the
corresponding quotient qi. We then multiply this value by the corresponding fi, and
subtract the result from p. We then redefine p using the result of this subtraction.
Next we consider the redefined LT(p) and continue the process mentioned until we
encounter a LT(p) which is not divisible by any of the LT(fi). If this happens we
remove this term and place it with our remainder. We proceed with either a “division step” or a “remainder step” until there are no terms remaining forp. Once completed,
this process will achieveqi andr; it is important to noteqi andr are unique only for
the specified order in which the divisors are listed. In k[x], a wonderful property of
division is the uniqueness of the quotient and remainder. If we decide to change the order of the polynomials in our basis, we find we are not guaranteed to achieve the
same quotients and remainder.
As with division ink[x], the division algorithm ink[x1, . . . , xn] will also terminate. Consider at each iteration either a division step or a remainder step is performed. Regardless of which, every time we redefine the dividend p the multidegree drops or
p becomes zero. Assuming a division step is performed, observe that we redefine p
(denoted here as p0) as
p0 =p− LT(p)
LT(fi)
·fi.
Consider the leading term of the polynomial being subtracted from p
LT
LT(p) LT(fi)
·fi
= LT(p) LT(fi)
·LT(fi) = LT(p).
This implies p0 has a multidegree which is strictly less than p, assuming p 6= 0.
This implies whenever we redefinep during a division step the multidegree is strictly
decreasing. By the well-ordering property, eventually we must achievep= 0, in which
the algorithm terminates.
Next, we consider some examples.
Example 4.2. To illustrate division in k[x1, . . . , xn], consider the following polyno-mial f =x2yz2 −xy2−y2z+z3 which we will divide by
F = (f1, f2, f3) = (x2−z, y−z, xy)⊂C[x, y, z].
q1:
q2:
q3:
x2−z y−z
xy
x2yz2−xy2+y2z+z3
r
First denote p = f and consider LT(p) = x2yz2. Considering each LT(f
i), we find the first fi in which LT(fi)|LT(p) is f1. Then, LT(p)/LT(f1) = yz2. As such,
this value is placed with q1. Multiplying LT(p)/LT(f1)·f1 =yz2 and subtracting
p− LT(p)
LT(f1)
·f1 =−xy2+y2z+yz3+z3.
This result is how we will redefine p. We again consider LT(p) = −xy2. We see f 2
is the first divisor listed such that LT(f2)|LT(p). This division LT(p)/LT(f2) =−xy,
so this value is placed with q2. Again, LT(p)/LT(f2)·f2 =−xy2+xyz is subtracted
fromp. The result of this is
p− LT(p)
LT(f2)
·f2 =−xyz+y2z+yz3+z3.
q1 :yz2
q2 :−xy
q3 :
x2−z
y−z
xy
x2yz2−xy2+y2z+z3
r
x2yz2−yz3
−xy2+y2z+yz3+z3
−xy2 +xyz
−xyz+y2z+yz3 +z3
We continue until we encounter LT(p) which is not divisible by any LT(fi). After
three iterations, we appear to encounter a leading term −xz2 which is not divisible by any of the leading terms of the divisors. As such, we remove this term and place in the remainder column to the right. We continue this algorithm, which is shown in its entirety below.
q1:yz2
q2:−xy−xz+yz+z3+z2
q3: 0
x2−z
y−z
xy
x2yz2−xy2+y2z+z3
r
x2yz2−yz3
−xy2+y2z+yz3+z3
−xy2+xyz
−xyz+y2z+yz3+z3
−xyz+xz2
−xz2+y2z+yz3+z3 −→ −
xz2 y2z+yz3+z3
y2z−yz2
yz3+yz2+z3
yz3+z4
yz2+z4+z3
yz2−z3
z4+ 2z3 −→ −xz2+z4
2z3 −→ −xz2+z4+ 2z3
0
The result of the division implies we can write f as
f = (yz2)(x2−z) + (−xy−xz+yz+z3+z2)(y−z) + 0·(xy) + (−xz2+z4+ 2z3).
Henceforth, we will reference the result of any division using this form.
order of the divisors. When dividing, we always use the first divisor listed whose leading term divides the leading term of the dividend. Here we have three divisors, and as such there are six different ways we can permute the divisors, each of which may result in a different remainder and quite possibly a remainder of zero. It can be shown no matter how we list the divisors the remainder in this example is always nonzero. Suppose we change the order of the polynomials in the divisors to F =
(f3, f2, f1) = (xy, y−z, x2−z). By doing this, we achieve the following result
f = (xz2−y)·(xy) + (z2+yz)·(y−z) + 0·(x2 −z) + (2z3).
Unsurprisingly, comparing this result to the previous, we see the remainders differ.
As mentioned, we can determine if a polynomial f is an element of an ideal
I = hf1, . . . , fsi by dividing f by F = (f1, . . . , fs). If we find the remainder f F
= 0 this is enough to conclude f ∈ I. Consider the next example, in which changing
the order of the divisors causes us to achieve a zero remainder. This illustrates a remainder of zero is a sufficient but not necessary condition for ideal membership.
Example 4.3. (Example 5 in [4]) Letf =xy2−xand letI be the ideal generated by
I =hxy−1, y2−1i. Assuming Lex order, letF = (f
1, f2) = (xy−1, y2−1)⊂C[x, y]
and observe the result of the division of f by F
f =y·(xy−1) + 0·(y2−1) + (−x+y).
We see fF = −x+y 6= 0. This division is not sufficient enough to claim f has
of the basis elements. If we maintain the same basis but change the order of the elements to F∗ = (f2, f1) = (y2 −1, xy −1), we achieve the following result upon
division
f =x·(y2−1) + 0·(xy−1) + 0.
Now we have a remainder of zero which is sufficient for us to conclude f ∈I.
Undoubtedly, performing division in k[x1, . . . , xn] just once can prove to be a lengthy process and is certainly one we would like to avoid performing repetitively. A computer algebra system can help make this process quicker, but what happens if we have potentially tens or hundreds of divisors? Is there a way to remedy the issue of the lack of uniqueness in remainders? It will be shown that a Gr¨obner basis corrects for such an issue and as a result division need only be performed once to check for ideal membership.
5
Gr¨
obner Bases
5.1
Gr¨
obner Basis and Hilbert Basis Theorem
Before defining a Gr¨obner basis, we discuss the Ascending Chain Condition which will prove to be a useful result.
Definition 5.1. An ascending chain of ideals is a nested increasing sequence of ideals Ii ⊆k[x1, . . . , xn] such that
I1 ⊆I2 ⊆I3 ⊆. . .
Example 5.2. The following is a finite ascending chain of ideals
hx1i ⊆ hx1, x2i ⊆ · · · ⊆ hx1, . . . , xni.
If we attempt to extend this chain further by including some f ∈ k[x1, . . . , xn]
so that hx1, . . . , xni ⊆ hx1, . . . , xn, fi, then either hx1, . . . , xni = hx1, . . . , xn, fi or
hx1, . . . , xn, fi=k[x1, . . . , xn]. Iff ∈ hx1, . . . , xnithen clearly f need not be included
in hx1, . . . , xni and we are done. If f 6∈ hx1, . . . , xni, then let f ∈ k[x1, . . . , xn]. We
will show hx1, . . . , xn, fi=k[x1, . . . , xn]. By the division algorithm we can write
f =q1x1+· · ·+qnxn+r
such that none of the monomials of r are divisible by x1, . . . , xn. Therefore, since
f ∈k[x1, . . . , xn] this would implyr must be a constant. Observe that we can write
r=f −q1x1− · · · −qnxn ∈ hx1, . . . xn, fi.
Since r is a constant in hx1, . . . , xn, fi, then clearly 1∈ hx1, . . . , xn, fi and therefore
hx1, . . . , xn, fi=k[x1, . . . , xn].
This example illustrates there is a maximal ideal for every polynomial ring.
Definition 5.3. An idealI ⊆k[x1, . . . , xn] is said to bemaximalifI (k[x1, . . . , xn], and any ideal J containing I is eitherI itself or k[x1, . . . , xn].
The following theorem was first applied to ring theory by Amelia Emmy Noether, who is often referred to as “the mother of modern algebra”. Noether, who was invited
to G¨ottingen by David Hilbert in 1919, helped develop much of the theory behind commutative algebra, especially in her famous 1921 paper “Theory of Ideals in Ring Domains” [1].
Theorem 5.4 (The Ascending Chain Condition). Let
I1 ⊆I2 ⊆I3 ⊆. . .
be an ascending chain of ideals in k[x1, . . . , xn]. Then, there exists an N ≥ 1 such
that
IN =IN+1 =IN+2 =. . . .
Proof. This condition is equivalent to saying every ideal ink[x1, . . . , xn] can be finitely generated, which is precisely the statement of the Hilbert Basis Theorem (Theorem 5.7).
Before we define a Gr¨obner basis, a few more definitions and propositions are in order. We will also discuss the Hilbert Basis Theorem, which guarantees us the property of finiteness. A Gr¨obner basis will also allow us to perform division a single time to see the result of the remainder, as these special bases afford us a uniqueness property of remainders regardless of which order we choose for our divisors. Recall division ink[x1, . . . , xn], in general, did not enjoy this property. This made answering the question of ideal membership a potentially lengthy process if our basis for the ideal consisted of many polynomials.
Definition 5.5. LetI ⊆k[x1, . . . , xn] be a nonzero ideal with some monomial order-ing on k[x1, . . . , xn].
i. We denote LT(I) as the set of leading terms of nonzero elements ofI
LT(I) = {cxα : there exists f ∈I\ {0}with LT(f) = cxα}.
ii. We denote hLT(I)i as the ideal generated by the elements of LT(I).
The following is an important and powerful proposition, as it states for an ideal I ⊆k[x1, . . . , xn],hLT(I)i, a potentially infinite generating set, can instead be gener-ated by a finite collection of leading terms of polynomials inI.
Proposition 5.6. Let I ⊆k[x1, . . . , xn] be a nonzero ideal. Then,
i. hLT(I)i is a monomial ideal;
ii. There are g1, . . . , gt ∈I such that hLT(I)i=hLT(g1), . . . ,LT(gt)i.
Proof. Clearly, since elementsg ∈LT(I) are monomials with some constant multiple
then hLT(I)i is a monomial ideal since leading terms and leading monomials will
generate the same ideal. SincehLT(I)iis a monomial ideal, then by Dickson’s lemma
(Theorem 3.20), hLT(I)i has a finite basis generated by monomials g1, . . . , gt, and
therefore hLT(I)i=hLT(g1), . . . ,LT(gt)i.
ForG={g1, . . . , gt}, we often denote
and
hLT(G)i=hLT(g1), . . . ,LT(gt)i.
Theorem 5.7 (Hilbert Basis Theorem). Every ideal I ⊆ k[x1, . . . , xn] has a finite
generating set. In other words, I =hGi for some G={g1, . . . , gt} ∈I.
Proof. If I is the trivial ideal then I = h0i = {0}, which is clearly finite. If I is
nonempty then it must contain some nonzero polynomial in I. Then fix a monomial
ordering . Note hLT(I)i exists and by proposition 5.6 we can find a finite number
of G={g1, . . . , gt} ∈I such that
hLT(G)i=hLT(I)i
We will show I =hGi. To do this, it suffices to show hGi ⊆I andI ⊆ hGi. The first
inclusion is trivial as g1, . . . , gt∈I implies hGi=hg1, . . . , gti ⊆I .
To see thatI ⊆ hGi, letf be any polynomial inI. Then by the division algorithm
ink[x1, . . . , xn] we can divide f by Gand write
f =q1g1+· · ·+qtgt+r
such that no terms of r are divisible by any LT(gi). Then we can write
r=f−q1g1− · · · −qtgt∈I.
Note if r is nonzero, then LT(r) ∈ hLT(I)i = hLT(g1), . . . ,LT(gt)i. By lemma 3.15
this implies LT(r) must be divisible by some LT(gi). This contradicts the fact that
f =q1g1+· · ·+qtgt+ 0∈ hg1, . . . , gti.
This implies all f ∈I can be generated by hg1, . . . , gti, and thus I ⊆ hGi. Therefore
I =hGi.
Definition 5.8. Fix a monomial order on the polynomial ring k[x1, . . . , xn]. A finite subset G = {g1, . . . , gt} of a nonzero ideal I ⊆ k[x1, . . . , xn] is said to be a
Gr¨obner basis if
hLT(G)i=hLT(I)i.
An equivalent, and maybe more intuitive definition bridged by lemma 3.15, is
{g1, . . . , gt} is a Gr¨obner basis if and only if the leading term of any element in I is
divisible by some LT(gi). Gr¨obner bases have many useful properties and applications,
and fortunately, every nonzero ideal ink[x1, . . . , xn] has such a basis.
Corollary 5.9. For a fixed monomial ordering , every ideal I ⊆ k[x1, . . . , xn] has
a Gr¨obner basis and any Gr¨obner basis for an ideal I is a basis of I.
The proof follows directly from the Hilbert Basis Theorem (Theorem 5.7).
Example 5.10. Let I ⊆ C[x, y] be the ideal generated by I = hf1, f2i where f1 =
x2y−1 andf
2 =xy2−x, and use Lex ordering. Our goal is to determine ifF ={f1, f2}
is a Gr¨obner basis for the ideal I. If F is in fact a Gr¨obner basis, this implies the
leading term of any nonzero element of I must be divisible by the leading term of at
least one of the polynomials in our basis. Equivalently, by the definition of a Gr¨obner basis,
hLT(f1),LT(f2)i=hLT(I)i.
Recall any element of I can be written in the form
hf1, f2i={h1·f1+h2·f2 :hi ∈C[x, y]}.
Consider the polynomial x2−y which is an element of I, which can be shown by
yf1+ (−x)f2 =y(x2y−1) + (−x)(xy2−x)
= (x2y2 −y) + (−x2y2+x2)
=x2−y.
This result indicates F in not a Gr¨obner basis. Notice the leading term of this
polynomial LT(x2 −y) = x2 is not divisible by x2y nor xy2, which are the leading
terms LT(f1) and LT(f2) respectively.
Proposition 5.11. Let I ⊆ k[x1, . . . , xn] be an ideal and let G = {g1, . . . , gt} be a
Gr¨obner basis for I. Then given f ∈k[x1, . . . , xn], there is a unique r ∈k[x1, . . . , xn]
with the following two properties:
i. No term of r is divisible by any of the LT(G).
ii. There is a g ∈I such that f =g+r.
In particular, r is the remainder on division of f by G no matter how the elements
Proof. The properties follow by the division algorithm. Since uniqueness of remain-ders is an important part of Gr¨obner bases, we consider the proof. Assume we can writef =g+rfor someg ∈I. Suppose the contrary that we can also writef =g0+r0
for some g0 ∈I where r6=r0. Then we have
g+r=g0+r0 =⇒ g−g0 =r−r0.
Since g, g0 ∈I, then g−g0 ∈I and therefore r−r0 ∈I. This implies that
LT(r−r0)∈ hLT(I)i=hLT(G)i.
That is, LT(gi)|(r−r0) for some gi ∈ G, which is a contradiction. Therefore r =r0
which implies r is unique.
As a result of the second condition, we have the following corollary.
Corollary 5.12. Let G be a Gr¨obner basis for an ideal I ⊆ k[x1, . . . , xn] and let
f ∈k[x1, . . . , xn]. Then f ∈I if and only if the remainder on the division of f by G
is zero.
Example 5.13. Recall the previous example whereF ={x2−1, xy2−x}={f1, f2},
which we have already determined is not a Gr¨obner basis for I =hx2−1, xy2 −xi.
Because this is not a Gr¨obner basis, a unique remainder is not guaranteed. To see this, let f =x5y2−x4 −x2y+y3 and divide f by F = (x2−1, xy2 −x) = (f1, f2).
