Today’s Goal: Proof of Extension Theorem
If a partial solution fails to extend, then),...,,( 32 naaa
),...,(),...,,( 132 sn ggaaa V
Corollary. If is constant for some i, then all partial solutions extend.
ig
The proof makes use of the resultant...
Definition: The resultant of f and g with respect to x, denoted by Res(f, g, x), is the determinant of the Sylvester matrix, Res(f, g, x) = det(Syl(f, g, x)).
Recall from last lesson...
121
10)( ll
ll axaxaxaxf
Question What if one of the polynomials is a constant?
Let f and g be two polynomials in R[x] of positive degree:
121
10)( mm
mm bxbxbxbxg
Syl(f, g, x)) is the matrix:
Recall: Res(f, g, x) = 0 iff f and g share a common factor.m columns n columns
Resultants involving constant polys
See Exercise 14 on page 161 of your textbook for more on this.
Time for a Game...
The game is called
“Make a Conjecture” I will present some examples; then you make a conjecture.
Ready?
Example
)},...,,(thatsuch,...,,existsthere|),...,,{( 212121 mknkk fffxxxxxxx V
]][[13),(
]][[)4(6),(2
22
xykxxyyxg
xykyxyxf
1040
3104
360
006
det),,(
2
2
y
y
y
y
xgfh =Res
The Sylvester matrix is a
4 x 4 square matrix.
Res(f, g, x) is a polynomial in y; indeed, Res(f, g, x) is in the first elimination ideal ‹ f, g › k[y]
Compute Res(f, g, x).
Specialization
)},...,,(thatsuch,...,,existsthere|),...,,{( 212121 mknkk fffxxxxxxx V
1040
3104
360
006
det),,(
2
2
y
y
y
y
xgfh(y) = Res
1040
3104
0360
0006
deth(0)
104
310
036
det6
Consider the specialization of the resultant obtained by setting y = 0.
))12(36(6
180
More generally...if then Res(f, g, x1) .],...,,[, 21 nxxxkgf ],...,[ 2 nxxk
Definition If c = (c2, c3,..., cn) kn-1 and R(x2, x3, ..., xn) := Res(f, g, x1), then a specialization of the resultant Res(f, g, x1) is R(c2, c3,..., cn).
Specialization Example
)},...,,(thatsuch,...,,existsthere|),...,,{( 212121 mknkk fffxxxxxxx V
What if we plug 0 into the polynomials at the start?
1040
3104
0360
0006
deth(0)
104
310
036
det6 180
Compare this with the specialization h(0)...
We’re off by a factor of 6.
]][[13),(
]][[)4(6),(2
22
xykxxyyxg
xykyxyxf
13)0,(
46)0,( 2
xxg
xxf
Res( f(x, 0), g(x, 0), x) = Res(6x2 – 4, 3x – 1)
104
310
036
det 30
× 6
WHY 6?
The 6 was “zeroed” out in Res( f(x, 0), g(x, 0), x) because plugging 0 into g(x, y) caused the degree of g to drop by 1.
Specialization Example
)},...,,(thatsuch,...,,existsthere|),...,,{( 212121 mknkk fffxxxxxxx V
Let’s modify the previous example slightly...
1040
0104
0060
0006
deth(0)
10
01det62 36
More generally, if f, g k[x, y] and deg g = m with deg(g(x, 0)) = p then
1)0,(
46)0,( 2
xg
xxf
Res( f(x, 0), g(x, 0), x) = Res(6x2 – 4, – 1)
10
01det 1
× 62
]][[13),(
]][[)4(6),(2
22
xykxxyyxg
xykyxyxf
y
Setting y = 0 causes deg g to drop by 2.
1040
3104
360
006
det),,(
2
2
y
y
y
y
xgfh(y) = Resy
y
h(0) = LC(f)m-p Res(f(x, 0), g(x, 0), x)
Proposition 3 (Section 3.6)
)},...,,(thatsuch,...,,existsthere|),...,,{( 212121 mknkk fffxxxxxxx VThis result, which describes an “interplay” between partial solutions and resultants, will be used in the proof of the Extension Theorem...
Proof of Extension Theorem
)},...,,(thatsuch,...,,existsthere|),...,,{( 212121 mknkk fffxxxxxxx V
Let , where
and let , a partial solution.
],...,[,..., 11 ns xxffI
)(,..., 12 Iccc n V
iN
nii Nxxxxgf i degree has in which terms,..., 112
Consider the evaluation homomorphism
defined by:
][],...,[: 11 xxx n
),(),...,,()( 121 cxfccxff n
The image of under is an ideal in ,
and is a PID.sffI ,...,1 ][ 1x
][ 1x Hence .][)()( 11 xxuI
Want to show that if , then there exists such that .
),...,,(,..., 212 sn gggcc V1c )(,...,, 21 Iccc n V
Proof of Extension Theorem
)},...,,(thatsuch,...,,existsthere|),...,,{( 212121 mknkk fffxxxxxxx V where ).(},:),({)( 111 IIfxfxu V cc
][
],...,[
1
1
x
xx n
)()( 1xuI
I
We get a nice commutative diagram:
We consider two cases:
Case 1: is nonconstant. )( 1xu
Case 2: , a nonzero constant. 01)( uxu
We’ll use resultants to show that the second case
can’t actually happen.
Proof of Extension TheoremCase 1: is nonconstant. ][)( 11 xxu
In this case, the FUNdamental Theorem of Algebra
ensures there exists such that .0)( 1 cu1c
Since ),,...,(},:),({)( 211 nccIfxfxu cc
for all .0),...,,( 21 ncccf If
In particular, all generators, f1, f2 , ..., fs, of I vanish at the point . ),...,,( 21 nccc
Hence , so the partialsolution extends, as claimed.
)(),...,,( 21 Iccc n V)(),...,( 12 Icc n V
Proof of Extension TheoremCase 2: is a nonzero constant. 01)( uxu
In this case, we show leads to a contradiction.
),...,,(,..., 212 sn gggcc V
Since ),,...,(},:),({)( 211 nccIfxfxu cc
there exists such that If 011 )(),( uxuxf c
Since , there exists such that .
),...,,(,..., 212 sn gggcc Vig
0),...,( 2 ni ccg
Now consider the resultant:
h = Res(fi, f, x1) ],...,[ 2 nxx
Proof of Extension TheoremCase 2: is a nonzero constant. 01)( uxu
Applying Proposition 3 to h = Res(fi, f, x1) yields:
)),,(),,((Res)()( 111),(deg-deg 1 xxfxfgh i
xffi cccc c
),),,((Res)( 101deg xuxfg i
fi cc
= 0
iffi ug deg
0deg)(c iNf
i ug 0deg)(c 0
],...,[, 2 ni xxff But recall: h = Res(fi, f, x1)
0)( ch0)()( cc iffand
Hence .0)( ch
A contradiction!