Graphics Reference
In-Depth Information
1
2
1
6
2
3
Ê
ˆ
100
---
Á
Á
Á
Á
Á
˜
˜
˜
˜
˜
11 10 0 7
210 101
03 0 0 1 5
-
-
Ê
ˆ
1
3
5
3
Á
Á
˜
˜
-
-
Æ
010 0
Ë
¯
--
1
2
1
2
001
--
5
Ë
¯
The last row leads to the equation
1
2
1
2
x
-
x
-
x
+
50
=
,
1
2
3
which is an implicit representation of the plane defined parametrically by equations
(10.59).
Let us see how we can interpret what we did in these examples in polynomial
terms. For example, in Example 10.10.1 the row reductions in the matrices corre-
spond to showing that
()
=
(
)
=
(
)
Vfg
,
Vfg
,
Vf g
,
,
1
1
1
where g
1
= g - 2f and f
1
= f + 2g
1
. Replacing g by g
1
will be called a
reduction
. It corre-
sponds to eliminating the X term in g. How this is done is via a long division of g by f:
2
)
XYZXYZ
XYZ
YZ
-+
2
2 3 4
242
2
-+
-+
+
If there are more equations, then more divisions may be needed and by more than
one polynomial. In analyzing what is going on here, there are two points to observe
as we prepare to generalize this process:
(1) We
ordered
the terms of the linear polynomials - we decided to eliminate X
first in Example 10.10.1 and we listed the x
i
after the t
i
in Example 10.10.2.
(2) We subtract multiples of a polynomial to eliminate the current “leading” terms
from remaining polynomials.
In the nonlinear case for arbitrary polynomials in k[X
1
,X
2
,...,X
n
] we shall try to
use similar steps, but things get more complicated. We address question (1) first. If an
ideal I is generated by polynomials g
1
, g
2
,..., g
m
, then to determine whether or not the
polynomial f belongs to I reduces to finding polynomials a
1
, a
2
,..., a
m
, so that
fag ag
=
+
2 2
...
+
+
a
mm
.
(10.61)
11
Finding such polynomials a
i
or showing that they do not exist is not as easy as in the
linear case, especially, since we cannot assume anything about their degree even if we
