Algebraic Geometry Jottings 10

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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–u1)···(x–um) and b(x–v1)···(x–vn). (Repeated roots allowed.) We had our first formulas for the resultant:

\text{res}_x(E,F) = a^n b^m\prod_{i=1}^m\prod_{j=1}^n(u_i-v_j)    (1)

                            = a^n \prod_{i=1}^m F(u_i) = (-1)^{mn}b^m \prod_{j=1}^n E(v_j)    (2)

And we learned two facts about resx(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:

x2+y2 = 1
2x2+y2 = 1

Eliminating x, the resultant is resx(E,F)=(y2–1)2. So right off the bat, we know the curves can intersect only when y2–1=0, i.e., when y=±1. Eliminating y, we get resy(E,F)=x4, 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: x2+y2 = 2x2+y2, so x2 = 0. Not much harder to eliminate x: 2(x2+y2)–(2x2+y2) = y2 and also = 1, so y2–1 = 0. But the resultants are x4 and (y2–1)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 resy(E,F) takes the form


which high school algebra equates to

[(1-x^2)-(1-2x^2)]\cdot[(1-x^2)-(1-2x^2)]=x^2\cdot x^2=x^4

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 resx(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 resy(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 resy(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 resx(E,F)? The order at O of (y2–1)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 yy–1. This changes resx(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 resy(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, resy(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 ak for y. Or replace some subset of the variables of k[x1,…,xn] 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=(x2+y21)(x–y)  and  F=(x2y2)

E is the unit circle plus the diagonal line x=y; F is the two diagonals xy. In this case u1(x) ≡ v1(x) ≡ x. The factor u1v1 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!

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