2.2. Marco teórico
2.2.2. Conozca el yo
2.2.2.1. Teoría del desarrollo personal de Eríkcson
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