Today’s Goal: Proof of Extension Theorem If a partial solution fails to extend, then Corollary. If...

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!