Definition
1. Let I
be an ideal in K[x1,x2,...,xn]
other than {0}.
- We define LT(I)
as the set of the leading terms of elements of I, i.e.
LT(I)
= {LT(f)
: f
is an element of I}.
- We denote by <LT(I)>
the ideal generated by the elements of LT(I).
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| Note that if I = <f1,f2,...,fs>
than <LT(f1),LT(f2),...,LT(fs)>
and <LT(I)>
may be different ideals. It is true that <LT(f1),LT(f2),...,LT(fs)>
is included in <LT(I)>,
since LT(fi)
is an element of LT(I)
by definition, but the converse is not true in general. Example.
Let I
= <f1,f2>,
where f1
= x2 + 2xy2
and f2
= xy + 2y3 - 1.
Using lex order on K[x,y], we have
LT(f1)
= x2, LT(f2) = xy
and:
y·(x2 + 2xy2)
- x·(xy + 2y3 - 1)
= x
so that x
is in I.
Moreover LT(x) = x. But x cannot be
written as h(x,y)·LT(f1)+h(x,y)·LT(f2) for
some h
in K[x,y] so x is not in
<LT(f1),LT(f2)>.
We will give a particular name to those special bases
for which <LT(<g1,g2,...,gt>)>
= <LT(g1),LT(g2),...,LT(gt)>.
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Definition 2. Fixed a
monomial ordering, a subset G
= {g1,g2,...,gt} of
an ideal I
is said to be a Groebner
basis if:
<LT(I)>
= <LT(g1),LT(g2),...,LT(gt)>.
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| This means that a subset G = {g1,g2,...,gt} of
an ideal I
is a Groebner basis if and only if the leading term of
any element of I
is divisible by one of the LT(gi). But,
does any ideal in K[x1,x2,...,xn]
have a Groebner basis? And what is the relationship
between <f1,f2,...,fs>
and <g1,g2,...,gt>?
The following result will answer our questions [3].
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| Proposition 3. Fix a
monomial ordering. Then every ideal I = <f1,f2,...,fs>
in K[x1,x2,...,xn]
other than {0} has a Groebner basis G = {g1,g2,...,gt}.
Futhermore, any Groebner basis of I is a basis of I, i.e. I=<f1,f2,...,fs>=<g1,g2,...,gt>. |
Examples.
Let's see some examples of Groebner bases.
- First consider the ideal I
from the example above, which has the basis {f1,f2}={
x2 +
2xy2, xy +
2y3 - 1}.
Then, {f1,f2}
is not a Groebner basis of I w.r.t.
lex order on K[x,y],
since we have seen that x is
in <LT(I)>
but x is
not in <LT(f1),LT(f2)>.
We will find a
Groebner basis of I
when we discuss the Buchberger's
algorithm.
- Next, consider the ideal J = <g1,g2> =
<x+z,y-z>. We claim that {g1,g2}
is a Groebner basis using lex order
in R[x,y,z].
Thus, we must show that LT(J)
is included in <LT(g1),LT(g2)>=<x,y>,
since the converse is always true. So pick up
a nonzero polynomial f in
J.
Suppose on the contrary that LT(f) is not
expressible as A(x,y,z)x +
B(x,y,z)y,
i.e. is not divisible by neither x
nor y.
So, by definition of lex order, f
must be a polynomial in z
alone. However f
vanishes on the subspace of R3
L={(x,y,z):
x+z=0,y-z=0}={(-t,t,t)
for any t
in R}; but the only polynomial in z
alone which vanishes in all R is the
zero polynomial, which is a contradiction. It
follows that {g1,g2}
is a Groebner basis w.r.t. lex order
in R[x,y,z].
- We must point out that the relative order of
the variables (as well as the ordering on the
monomials) plays a foundamental role in
establishing if a subset {g1,g2,...,gt}
of an ideal is a Groebner basis of that ideal
(see Monomial
orderings).
If we reconsider {g1,g2}
of the example 2 w.r.t. the lex with z>y>x
we can see that {g1,g2}
is not a Grobner basis of J.
In fact, we have:
<LT(g1),LT(g2)>=<z>,
y+x=g1+g2
(so y+x is in J)
and LT(y+x)=y,
but y
is not in <z>.
Greobner bases have some other
nice properties that we will investigate further.
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