The Resultant, Episode 1: Inside the Episode

In Episode 1 of our miniseries, “The Resultant”, the characters were introduced: integral domain *R* with fraction field *K* and extension field *L*, and polynomials *E*(*x*) and *F*(*x*) in *R*[*x*], factoring completely in *L* as *a*(*x–u*_{1})···(*x–u _{m}*) and

*b*(

*x–v*

_{1})···(

*x–v*). (Repeated roots allowed.) We had our first formulas for the resultant:

_{n}

(1)

(2)

And we learned two facts about res* _{x}*(

*E*,

*F*): it’s zero if and only if

*E*and

*F*have a common root in

*L*(obvious from formula (1)), and it belongs to

*R*(made plausible, but not actually proven).

Now let’s chew over a few things with the director.

Formula (1) doesn’t provide an obvious way to *calculate* the resultant. As hinted already, it can be expressed as a determinant. I believe though that other algorithms are used in practice. Anyway, I’ll just use SageMath as a blackbox. Let’s start with those two circles from the last chewing post:

Eliminating* x*, the resultant is res* _{x}*(

*E*,

*F*)=(

*y*

^{2}–1)

^{2}. So right off the bat, we know the curves can intersect only when

*y*

^{2}–1=0, i.e., when

*y*=±1. Eliminating

*y*, we get res

*(*

_{y}*E*,

*F*)=

*x*

^{4}, which agrees with

*x*being 0 at the two intersections.

Straightforward, but let’s roll the example over our tongues a little. It’s easy to eliminate *y* from the two equations: *x*^{2}+*y*^{2} = 2*x*^{2}+*y*^{2}, so *x*^{2} = 0. Not much harder to eliminate *x*: 2(*x*^{2}+*y*^{2})–(2*x*^{2}+*y*^{2}) = *y*^{2} and also = 1, so *y*^{2}–1 = 0. But the resultants are *x*^{4} and (*y*^{2}–1)^{2}, respectively. Let’s look at our formula to see what gives.

The polynomials factor in an extension field of *k*(*x*):

The product for res* _{y}*(

*E*,

*F*) takes the form

which high school algebra equates to

We have an illustration of how the factors of the product (1) make sense only in the extension field *L*, but the product itself lies in *K*. The computation for res* _{x}*(

*E*,

*F*) is similar.

In post 4 I mentioned Kendig’s Theorem 3.3: when the origin *O* is the *only* intersection on the y-axis, the order of res* _{y}*(

*E*,

*F*) is the multiplicity at

*O*. (There is also a proviso about leading coefficients, not a problem here.) As I said, a slight tweak to Kendig’s proof gives a stronger result: the order of res

*(*

_{y}*E*,

*F*) is the sum of the multiplicities for all intersections lying on the y-axis. Looks good here: multiplicity 2 at both (0,1) and (o,–1). What about res

*(*

_{x}*E*,

*F*)? The order at

*O*of (

*y*

^{2}–1)

^{2}is 0, and indeed no intersections lie on the x-axis. So let’s shift the x-axis up to the top intersection with the substitution

*y*↦

*y*–1. This changes res

*(*

_{x}*E*,

*F*) into (

*y*(

*y*–2))

^{2}, which has order 2 at

*O*. All good!

This bone about extension fields—it’s worth gnawing on a bit. Often I like to let the ground field *k* be ℂ (or at any rate algebraically closed). When that’s true, *E* and *F* belong to ℂ[*x*,*y*]. To compute res* _{y}*(

*E*,

*F*), we have to treat them as elements of

*R*[

*y*] with

*R*=ℂ[

*x*]. That’s not even a field! But no problem, just use the fraction field

*K*=ℂ(

*x*), rational functions of

*x*. But in the two-ellipses example, we needed square roots of rational functions. So even though we

*started*with the algebraically closed field ℂ, we still ended up working in an algebraic extension of ℂ(

*x*). In general, we could take

*L*to be the mutual splitting field of

*E*and

*F*.

While we’re at it, res* _{y}*(

*E*,

*F*) is

*not*the zero element of

*R*=ℂ[

*x*]. It’s only when we evaluate it, by plugging in a complex number for

*x*, that we might get 0. If

*c*∈ℂ then we have a canonical homomorphism from ℂ[

*x*] to ℂ, namely

*p*(

*x*)↦

*p*(

*c*). So in modern language, we’re look for “elements of ℂ such that the resultant lies in the kernel of the associated canonical evaluation homorphism”. A mouthful! Still, sometimes the distinction matters. In old-fashioned language, we’d say the resultant

*has the value*0, vs. being

*identically*0.

This whole discussion applies with any ground field *k*. I just used ℂ for concreteness.

Incidentally, people talk about *specialization*: if you plug in values for some variables in an expression, that’s specializing it. We have *specialization homomorphisms* from various rings to other rings. Simple example: *k*[*x*,*y*] to *k*[*x*], plugging in some fixed value *a* ∈ *k* for *y*. Or replace some subset of the variables of *k*[*x*_{1},…,*x _{n}*] with fixed values; notationally messier to describe, but essentially the same idea.

Does it ever happen that resultant is *identically* 0? Sure, when *E* and *F* have a common component. For example:

*E*=(*x*^{2}+*y*^{2}*–*1)(*x–**y*) and *F*=(*x*^{2}–*y*^{2})

*E* is the unit circle plus the diagonal line *x*=*y*; *F* is the two diagonals *x*=±*y*. In this case *u*_{1}(*x*) ≡ *v*_{1}(*x*) ≡ *x*. The factor *u*_{1}*–**v*_{1} is identically zero; in other words, is the zero element of *L*.

In fact, the resultant is identically 0 when and only when *E* and *F* have a common component. (Tacitly assuming “non-triviality”: neither *E* nor *F* is constant, and by “common component” we mean a nonconstant common factor.)

Ah, autoplay is already queuing up Episode 2!