Algebraic Geometry Jottings 13

The Resultant, Episode 3: Inside the Episode

So we have, at long last, several expressions for the resultant:

$\det(S)=\text{res}_x(E,F) = a_m^n b_n^m\prod_{i=1}^m\prod_{j=1}^n (u_i-v_j)$    (1)
$= a_m^n \prod_{i=1}^m F(u_i) = (-1)^{mn}b_n^m \prod_{j=1}^n E(v_j)$    (2)

where

E(x) = amxm+···+a0 = am(x–u1)···(x–um)    (3a)
F(x) = bnxn+···+b0 = bn(x–v1)···(x–vn)        (3b)

and we assume m and n are greater than 0. S is the Sylvester matrix. At the moment, though, we don’t really care about the matrix or its determinant. (It will matter when we get to Kendig’s proof of Bézout’s Theorem.) Leave the proof that det(S) equals the product for the next episode. Algebraic curves occupy us; that’s the special case where R=k[y], so R[x]=k[x,y]. (Of course, you can always interchange x and y.) The key facts that make the resultant useful here:

1. While E and F belong to k[x,y], the ui‘s and vj‘s of (3) typically belong to an extension field L of K=k(y).  But the resultant belongs to k[y]—a polynomial in y alone. Let’s write resx(y) for it. (Or resy(x) when x and y trade places.)
2. The resultant is 0 whenever E(x0,y0)=F(x0,y0)=0, x0,y0k. In other words, at the intersections of E and F.
3. The roots of the resultant are places where either (a) an intersection of E and F lies on the horizontal line y=y0 through the root; or else (b) the leading coefficients of E and F are both 0. That is, resx(y0)=0, y0k, implies either (a) there is an x0k such that E(x0,y0)=F(x0,y0)=0, or else (b) am(y0)=0 and bn(y0)=0. (Remember that the coefficients belong to k[y].)

We’ll look into the proofs of these later. Two more facts, one mentioned in post 4 (with x and y switched):

1. The order of resx(y) (the degree of its lowest nonzero term) is the sum of the multiplicities of all intersections on the x-axis, provided that the leading coefficients am(y) and bn(y) are constants. (Likewise with x and y switched.)
2. The degree of resx(y) (the degree of its highest nonzero term) is the sum of the multiplicities of all intersections in the affine plane, with the same proviso about leading coefficients.

Post 10 had our lone example so far: the two ellipses x2+y2 = 1 and 2x2+y2 = 1. We regarded these as polynomials in y with coefficients in k(x), and computed resy(x)=x4 explicitly. The ui‘s are $y=\pm\sqrt{1-x^2}$  and the vj‘s are $y=\pm\sqrt{1-2x^2}$, elements of an algebraic extension field. Both intersections lie on the y-axis, so the order and degree of resy(x) are the same: 2+2=4, since they both have multiplicity 2. No intersections at infinity.

Let’s look at our favorite pair of curves from post 1, the roses:

E(x,y)=(x2+y2)2+3x2yy3
F(x,y)=(x2+y2)3–4x2y2

The resultant, using SageMath, and regarding E and F as polynomials in x: resx(y)=16y14(16y2—5)2. The leading coefficients are both 1. (E has leading term x4 and F has leading term x6.) The order of resx(y) is 14, reflecting the lone intersection on the x-axis with multiplicity 14, as we saw earlier. The degree is 18, reflecting the four additional intersections of multiplicity 1 in the affine plane.

To find all those affine intersections, we set resx(y)=0, solve for y, and plug into both E and F to find the corresponding x‘s. (As you can see from the picture, this procedure will turn up some x‘s for points on E or F but not both.) The results:

$\langle 0,0\rangle$

$\langle\pm\frac{\sqrt{5-2\sqrt{5}}}{4},\frac{\sqrt{5}}{4}\rangle\approx\langle\pm 0.18,0.56\rangle$

$\langle\pm\frac{\sqrt{5+2\sqrt{5}}}{4},-\frac{\sqrt{5}}{4}\rangle\approx\langle\pm0.77,-0.56\rangle$

Incidentally, those last four intersections occur at the angles 72°, 108°, 216°, and 324°, all multiples of 36°. In principle we have a construction of a regular decagon, though not with ruler and compass.

What about the intersections at infinity? In post 2 we pictured them by “moving the line at infinity”, and displaying the ix-axis instead of the x-axis.

We accomplish the first maneuver by homogenizing and then dehomogenizing by setting y=1:

E: (x2+y2)2+3x2yy3 ⇒ (X2+Y2)2+3X2YZY3Z ⇒ (x2+1)2+3x2z–z
F: (x2+y2)3–4x2y2 ⇒ (X2+Y2)3–4X2Y2Z2 ⇒ (x2+1)2–4x2z2

This moves the old line at infinity, Z=0 in homogenous coordinates, to the new x-axis, and the old x-axis to the new line at infinity. Recall how points correspond:

(x,y) = (X/Z, Y/Z, 1) ↔ (X:Y:Z) ↔ (X/Y, 1, Z/Y) = (x,z)

(Check out the “art history” pictures in post 2 if this doesn’t ring a bell, or any discussion of coordinates for the projective plane. The x on the left is the “old” x, not the same as the “new” x on the right. The capital X,Y, and Z are the homogeneous coordinates.) Our two new equations

E*(x,z) = (x2+1)2+3x2zz
F*(x,z) = (x2+1)2–4x2z2

won’t show any intersections on the line z=0, if you plug them into graphing software; for that you need to replace x by ix in the equations. I did that in post 2. This time I prefer to stick with these forms, just noting that the picture displays the ix-axis rather than the real x-axis. Of course, this makes any points with real x-coordinates invisible.

SageMath computes the resolvent for us:

resx(E*,F*) = 16z6(5z2–16)2

The order is 6; two intersections of multiplicity 3 on the (complex) x-axis, check! The total degree is 10; four additional intersections of multiplicity 1 in the xz-plane, check! Maybe you’re asking, where are they? They’re the green dots in the previous picture, other than O. They’re invisible in this picture because they have real coordinates. O is on the old x-axis, so it’s been relocated to infinity.

We could compute the new (x,z)-coordinates of these points using the same procedure as before: z=0 or 5z2–16=0, giving $z=\pm 4/\sqrt{5}$. When z=0, plugging into E* and F* yields (x2+1)2=0 so xi. For $z=\pm 4/\sqrt{5}$, let’s take a shortcut. We use the transformation equations between the old and new coordinates:

$\text{new }z = Z/Y = \frac{1}{Y/Z} = \frac{1}{\text{old }y}$
$\text{new }x = X/Y = \frac{X/Z}{Y/Z} = \frac{\text{old }x}{\text{old }y}$

The value of z checks: the old y is $\pm\sqrt{5}/4$, as we saw above. (Or y=0, giving a point at infinity in the xz-plane.) For the other four points, we get

$\langle\pm\sqrt{1-\frac{2}{\sqrt{5}}},\frac{4}{\sqrt{5}}\rangle\approx\langle\pm 0.3,1.8\rangle$

$\langle\pm\sqrt{1+\frac{2}{\sqrt{5}}},-\frac{4}{\sqrt{5}}\rangle\approx\langle\pm 1.4,-1.8\rangle$

xy=2, xy=1

Obviously no intersections in the affine plane, but it looks like two at infinity: (∞,0) and (0,∞). The resultants are resx(y)=y and resy(x)=x. This kind of makes sense, but what does Fact 3 have to say? Answer: either there’s a genuine intersection in the affine plane, or the leading coefficients are 0.

Treating the curves as polynomials in x, the leading coefficients are both y. So Fact 3 is happy. Facts 4 and 5 don’t complain, because the leading coefficient proviso is not met.

It’s easy to compute the resultant explicitly via the product formula. The roots are u1(y)=2/y and v1(y)=1/y, the difference is 1/y, and multiplying by the leading coefficients gives y2(1/y) = y. Likewise with x and y switched.

If you homogenize and dehomogenize with y=1, you’ll get a pair of parabolas, x=2z2 and x=z2. We have an intersection of multiplicity 2 at (0:1:0). Bézout says we need four intersections. If you dehomogenize with x=1, the other two turn up, at (1:0:0), with the parabolas y=2z2 and y=z2.

Finally, rotate these hyperbolas by 45 and rescale a bit, getting x2y2=2 and x2y2=1:

This time the leading coefficients are 1. The resultants are also 1. So no intersections anywhere in the affine plane. Homogenize and dehomogenize with x=1, getting two ellipses y2+2z2=1 and y2+z2=1. We’ve already seen what that looks like.