The Resultant, Episode 4

This episode has one sole purpose: to show that the two formulas for the resultant are equivalent. The next episode, the finale, will tie up some loose ends.

The formulas:

(1)

(2)

where

*E*(*x*) = *a _{m}x^{m}*+···+

*a*

_{0}=

*a*(

_{m}*x–u*

_{1})···(

*x–u*) (3a)

_{m}*F*(

*x*) =

*b*+···+

_{n}x^{n}*b*

_{0}=

*b*(

_{n}*x–v*

_{1})···(

*x–v*) (3b)

_{n}Eq.(2) follows immediately from (1) once we expand either *F*(*u _{i}*) or

*E*(

*v*) using the right hand side of (3). We assume that

_{j}*m,n*>0.

We learned in Episodes 2 and 3 that the equation

*PE+QF*=det(*S*) (4)

*always* has a solution with deg *P*<*n*, deg *Q*<*m*, *P* and *Q* nonzero, coefficients in *R*. (Succinctly: *P*∈*R _{n}*[

*x*],

*Q*∈

*R*[

_{m}*x*]. We had one proof in the singular case, another for nonsingular

*S*.) Eq.(4) provides a crucial ingredient.

Here are the bones of the proof of eq.(1); flesh on the bones to follow. For some pair *u _{i}*,

*v*, set

_{j}*u*=

_{i}*v*. If

_{j}*x*=

*u*=

_{i}*v*, then

_{j}*E*=

*F*=0 from the right hand side of (3). Then (4) tells us that det(

*S*)=0. From the factor theorem we conclude that (

*u*–

_{i}*v*) is a factor of det(

_{j}*S*). Since

*u*and

_{i}*v*were arbitrarily chosen,

_{j}*all*the factors of the product divide det(

*S*), and so the product does. Comparing degrees, we can show that det(

*S*) equals the product.

Now let’s dot some i’s and cross some t’s (switching to a less visceral metaphor). We can cast the argument in a concrete 19th century style, or take a more modern structural approach. We’ll do both together. Start with a special case, where the leading coefficients *a _{m}* and

*b*are both 1. First piece of business: treat the

_{n}*u*‘s and

_{i}*v*‘s as formal symbols. Expanding out the right hand sides of eq.(3) we get expressions for the

_{j}*a*‘s and

_{i}*b*‘s as polynomials in them, the so-called elementary symmetric polynomials:

_{j}(5a)

where the hat means “omit this factor”, and likewise for the *b _{j}*‘s (eq.(5b), which I won’t bother to write out). Plugging these

*a*‘s and

_{i}*b*‘s into det(

_{j}*S*), we obtain a big, elaborate polynomial in the

*u*‘s and

_{i}*v*‘s with integer coefficients.

_{j}Structurally, we’re working in the ring *R*=*k*[*u*_{1}*,…,u _{m}*,

*v*

_{1},…,

*v*], where the

_{n}*u*‘s and

_{i}*v*‘s are variables. The

_{j}*a*‘s and

_{i}*b*‘s and det(

_{j}*S*) are all elements of

*R*.

What happens to this det(*S*) if, say, you replace *u*_{1} everywhere with *v*_{1}? We have to rewrite eq.(5a), but (5b) doesn’t change. Likewise for (3a) and (3b). Next, let’s substitute *v*_{1} for *x*. The right hand side of (3b) becomes identically 0, implying that *F*(*v*_{1}), fully expanded, is identically 0. The *modified* right hand side of (3a) (with *v*_{1} in place of *u*_{1}) also is identically 0, so *E*(*v*_{1}), is identically 0. And finally from (4) we conclude that if you replace *u*_{1} with *v*_{1} everywhere in the polynomial det(*S*), the result is identically 0.

Structurally, we pull out an old trick from our toolbox: we regard *R*=*k*[*u*_{1}*,…,u _{m}*,

*v*

_{1},…,

*v*] as

_{n}*R*=

*k*[

*u*

_{2},…,

*u*,

_{m}*v*

_{1},…,

*v*][

_{n}*u*

_{1}]. Write

*R*

_{1}for

*k*[

*u*

_{2}

*,…,u*,

_{m}*v*

_{1},…,

*v*], so

_{n}*R*=

*R*

_{1}[

*u*

_{1}]. Replacing

*u*

_{1}with

*v*

_{1}just means evaluating the polynomials of

*R*

_{1}[

*u*

_{1}] at the element

*v*

_{1}of

*R*

_{1}. Previously we’ve called this an

*evaluation homomorphism*

*R*

_{1}[

*u*

_{1}]→

*R*

_{1}. It extends canonically to a homomorphism from

*R*[

*x*] to

*R*

_{1}[

*x*]. Under this map,

*E*(

*x*) goes to the polynomial (

*x–v*

_{1})(

*x–u*

_{2})···(

*x–u*). We also have an evaluation homomorphism from

_{m}*R*

_{1}[

*x*] to

*R*

_{1}, where we set

*x*=

*v*

_{1}. That sends both

*E*and

*F*to 0, so the image of det(

*S*) is 0, by eq.(4). But det(

*S*) has no

*x*‘s in it, so this is the same as applying the homomorphism

*R*

_{1}[

*u*

_{1}]→

*R*

_{1}to det(

*S*). The upshot: the polynomial det(

*S*) of

*R*=

*R*

_{1}[

*u*

_{1}] has the root

*v*

_{1}in

*R*

_{1}.

Now we can appeal to the factor theorem, and conclude that det(*S*) is divisible by (*u*_{1}*–v*_{1}) in *R*. I should mention that the factor theorem still holds even for polynomials over a ring, and not a field. The proof amounts to using long division by a linear polynomial of the form *x*–*a* (or *u*_{1}*–v*_{1} in our present circumstances); since the leading coefficient is 1, we never need inverses in the ring. This explicit long division serves as the concrete argument.

Because *u*_{1} and *v*_{1} were arbitrary, the same rigamarole shows that det(S) is divisible by (*u _{i}*–

*v*) for any

_{j}*i*and

*j*. Indeed, det(S) is divisible by the

*product*of all these factors; you can show this by induction, but perhaps a slicker approach is to note that

*R*is a UFD, and that all the (

*u*–

_{i}*v*)’s are irreducible and no two are associates.

_{j}OK, now we have , with *h*∈*R*. Next we compare degrees. The product consists of *mn* factors, each of degree 1 in the variables (the *u _{i}*‘s and

*v*‘s), so when expanded, it’s homogeneous of degree

_{j}*mn*. If we can show the same for det(

*S*), then it will follow that

*h*is a constant.

As it happens, Kendig provides the argument we need on p.66. Let’s say we replace *u _{i}* and

*v*with

_{j}*tu*and

_{i}*tv*. If we can show that det(

_{j}*S*) turns into

*t*det(

^{mn}*S*), it will follow that det(S) is homogeneous of degree

*mn*. Looking at the elementary symmetric polynomials (5a), we see that

*a*

_{0}has degree

*m*,

*a*

_{1}is homogeneous of degree

*m*–1, and in general

*a*is homogeneous of degree

_{i}*m–i*. Likewise

*b*is homogeneous of degree

_{j}*n–j*. The Sylvester matrix (for

*m*=2 and

*n*=3, with leading coefficients

*a*

_{2}=

*b*

_{3}=1) looks like:

So after we replace *u _{i}* and

*v*with

_{j}*tu*and

_{i}*tv*, the entries are multiplied elementwise by this matrix:

_{j}It’s hard to guess what happens to the determinant as a whole from this. We improve matters by (as Kendig puts it) packing the matrix with additional powers of *t*. Multiply each *row *by a power of *t* to make the *columns *uniform, thus:

Since multiplying a column multiplies a determinant by the same factor, this matrix will multiply det(*S*) by *t*^{1+2+3+4} = *t*^{10}, in our special case. We packed with additional factors *t*^{1+2+1} = *t*^{4}, again for this case. Net effect, multiplying *u _{i}* and

*v*by

_{j}*t*multiplies det(

*S*) by

*t*

^{6}, as desired. Kendig does the (straightforward) algebra for general

*m*and

*n*. So the determinant is homogeneous of degree

*mn*in the

*u*‘s and

_{i}*v*‘s.

_{j}Our goal is in sight. We’ve shown *h* is constant, now to show it’s 1. For this, we expand the product, picking a term to focus on; then we focus on the same term in the determinant.

Say we choose *v _{j}* in each factor (

*u*–

_{i}*v*) of the product; that gives us one term of the expansion. Each

_{j}*v*

_{j}*is paired with*

*m*

*u*‘s, so our result is (–1)

_{i}

^{mn}*v*

_{1}

*···*

^{m}*v*

_{n}*= ((–1)*

^{m}

^{n}*v*

_{1 }···

*v*)

_{n}*=*

^{m}*b*

_{0}

*. In det(*

^{m}*S*), if we go down the main diagonal we get

*b*

_{0}

*. All the other terms in the “sum of products” formula for the determinant involve replacing at least one of the*

^{m}*b*

_{0}‘s with a

*b*,

_{j}*j*>0. That means at least one

*v*is replaced with a

_{j}*u*. So the main diagonal is the

_{i}*only*term of the determinant formula contributing a term that is all

*v*‘s. Therefore the constant

*h*must be 1.

What about the general case, with arbitrary leading coefficients *a _{m}* and

*b*? Throw them in as additional variables; let’s just call them

_{n}*a*and

*b*. The

*a*‘s and

_{i}*b*‘s acquire an additional factor; that is, the “new”

_{j}*a*is

_{i}*aa*and the “new”

_{i}*b*is

_{j}*bb*. So the “new” det(

_{j}*S*) is

*a*times the “old” det(

^{n}b^{m}*S*). That’s exactly the factor in front of the double product in (1). Since (1) held before (without the

*a*on the right), it remains true after multiplying both sides by

^{n}b^{m}*a*.

^{n}b^{m}Closing remarks: we’ve actually shown that (1) is a *polynomial identity* in the ring *k*[*a*, *b*, *u*_{1}*,…,u _{m}*,

*v*

_{1},…,

*v*]. (We didn’t even use the fact that

_{n}*k*is a field.)