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84 CHAPTER 3. POLYNOMIAL EQUATIONS
Eliminating here (using row 1 to kill the leading term in row 4, and the same
with row 2 against row 5) gives
⎛
x 2 y 2 ⎞
⎛
⎞
x
1 0 0 0 0 0 −1 0 0
2 y
x 0 1 0 0 0 0 0 −1 0
2
xy 0 0 1 0 0 0 0 0 −1
2
xy
= 0, (3.30)
⎜ 0 0 0 0 −1 0 1 0 0 ⎟
⎝
⎠
x
0 0 0 0 0 −1 0 1 0
⎜ y
0 0 0 0 1 0 0 0 −1
2
⎟
⎝ ⎠
y
1
and now row 4 can kill the leading term in row 6, to give
⎛
x 2 y 2 ⎞
⎛
⎞
x
1 0 0 0 0 0 −1 0 0
2 y
x 0 1 0 0 0 0 0 −1 0
2
xy 0 0 1 0 0 0 0 0 −1
2
xy
= 0. (3.31)
⎜ 0 0 0 0 −1 0 1 0 0 ⎟
⎝
⎠
x
0 0 0 0 0 −1 0 1 0
⎜ y
0 0 0 0 0 0 1 0 −1
2
⎟
⎝ ⎠
y
1
The last line of this corresponds to y 2 −1. To use this to deduce an equation for
x, we would need to consider x times this equation, which would mean adding
further rows and columns to the matrix.
No-one would actually suggest doing this in practice, any more than any-one
would compute a g.c.d. in practice by building the Sylvester matrix (which is
actually the univariate case of this process), but the fact that it exists can be
useful in theory, as we will find that the Sylvester matrix formulation of g.c.d.
computation is useful in chapter 4.
3.3.6 Example
Consider the three polynomials below.
g 1 = x 3 yz − xz 2 ,
g 2 = xy 2 z − xyz,
g 3 = x 2 y 2 − z.
The S-polynomials to be considered are S(g 1 , g 2 ), S(g 1 , g 3 ) and S(g 2 , g 3 ). We
use a purely lexicographical ordering with x > y > z. The leading terms of
g 2 = xy 2 z − xyz and g 3 = x 2 y 2 − z are xy 2 z and x 2 y 2 , whose l.c.m. is x 2 y 2 z.
Therefore
S(g 2 , g 3 ) = xg 2 − zg 3 = (x 2 y 2 z − x 2 yz) − (x 2 y 2 z − z 2 ) = −x 2 yz + z 2 .