Algebraic Geometry Jottings 16

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The Resultant, Episode 5: Inside the Episode

The double-product form for the resultant:

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

implies Fact 3:

  1. resx(y0)=0 ⇔ [the line y=y0 passes through an intersection of E and F, or am(y0)=bn(y0)=0].

provided it makes sense to evaluate the ui‘s and vj‘s at y0. If you’ll grant me that, then here goes the argument. In one direction: if resx(y0)=0 and am(y0)≠0 and bn(y0)≠0, then some factor ui(y0)–vj(y0) must equal 0. (What if am(y0)≠0 but bn(y0)=0? See below.) Let x0=ui(y0)=vj(y0). Plugging into the factored forms for E(x,y) and F(x,y):

E(x,y) = am(y)(x–u1)···(x–um)    (3a)
F(x,y) = bn(y)(x–v1)···(x–vn)      (3b)

gives E(x0,y0)=F(x0,y0)=0, so (x0,y0) is an intersection of E and F. This argument is easily reversed to prove the other direction.

This near-proof or plausibility proof has some wrinkles worth investigating. First off, it seems from eq.(1) that am(y0)=0 or bn(y0)=0 should force resx(y0)=0—why do we need them both to be zero? The simplest example exhibiting the flaw in this ointment: x=y, xy=1, i.e., E(x,y)=x–y, F(x,y)=yx–1. When y0=0, the leading coefficient of F is 0. The double product is (y–(1/y))=(y2–1)/y. The resultant is y times that, that is y2–1; thus plugging in y0=0 gives resx(y0)=–1. If we hadn’t canceled y first. we’d have gotten 0·∞.

A more elaborate example: E(x)=ax2+bx+c, F(x)=dx+e. Using the approximation \sqrt{b^2-4ac}\approx b-2ac/b, valid for |b|≫|ac|, the roots of E(x)=0 are approximately


Keeping b and c fixed (with b≠0) while a→0, one of the roots of E(x)=0 tends to –c/b while the other wanders off to infinity. F(x)=0 has the fixed root –e/d. When a=0, E morphs into a linear equation, sharing a root with F precisely when c/b=e/d, i.e., when becd=0. We’ve seen this before: it’s the Sylvester determinant:

\begin{vmatrix}b & c\\d & e\end{vmatrix}=be-cd

But if we plug a=0 into the Sylvester determinant for the original pair E and F, we have:

\begin{vmatrix}{} & b & c\\d & e\\{} & d & e \end{vmatrix}=-d(be-cd)

Let’s look at the general case, but let’s get rid of those pesky leading coefficients by dividing them out:

E′(x) = xm + (am–1/am) xm–1 +···+ (a0/am)    (3a′)
F′(x) = xn  +   (bn–1/bn) xn    +···+  (b0/bn)    (3b′)

Say the original coefficients are polynomials in y. We’ve introduced poles at any of the roots of am(y)=0 or bn(y)=0. If, say, bn(y0)=0, then the actual degree of F(x,y0) in x is less than n, but (you might say) its formal degree is still n. (Some people do say this.) Whereas F(x,y0)=0 has fewer than n roots, F′(x,y0)=0 is blessed (or cursed) with additional roots at infinity.

Now, if am(y0) is not 0, then all the roots of E′(x,y0)=0 are finite. If we’re asking after common roots, we can ignore the infinite roots of F′(x,y0).

How does the resultant view this? The roots at infinity bedevil the double product in eq.(1), but the leading coefficient factors smooth the troubled waters. If we use the actual degrees, we’re rewarded with a smaller Sylvester determinant. As long as am(y0)≠0 and bn(y0)=0 (or vice versa), the “formal” and “actual” determinants give the same verdict for the existence of common roots: one equation has only finite roots, the other just some extra roots at infinity. If both am(y0) and bn(y0) are 0, then we have common roots at infinity, and det(S) is 0.

This is easy to see from the determinant itself, using expansion by minors down the first column. Say am(y0)≠0 and bn(y0)=0. Letting m=2, n=3 as usual, to ease formatting, we have:

\begin{vmatrix}  a_2 & a_1 & a_0\\  {} & a_2 & a_1 & a_0\\  {} & {} & a_2 & a_1 & a_0\\  0 & b_2 & b_1 & b_0\\  {} & 0 & b_2 & b_1 & b_0  \end{vmatrix} =a_2\cdot\begin{vmatrix}  a_2 & a_1 & a_0\\{} & a_2 & a_1 & a_0\\  b_2 & b_1 & b_0\\  {} & b_2 & b_1 & b_0  \end{vmatrix}

with the “formal” determinant on the left, and the “actual” one on the right. (At least if b2(y0)≠0. If not, rinse, repeat.) So one determinant is 0 if and only if the other is.

OK, how about the other dodgy aspect of the plausibility proof, evaluating the ui‘s and vj‘s at a value y0 of k? The ui‘s and vj‘s belong to an algebraic extension L of k(y). We sort of have an evaluation map from k(y) to k, at least when we steer clear of poles. It seems like one should be able to extend this to a suitable map from L to k, since k is algebraically closed. When k=ℂ, the theory of Riemann surfaces handles the matter. For algebraically closed k in general, I gather Puiseaux series step into the breach. Kendig discusses this in §§3.3–3.4.

While we’re on the topic of alternate proofs, Lang’s Algebra concludes the justification of eq.(1) (det(S)=double product) this way, adjusting his notation to agree with ours:

From (2) we see that [the double product] is homogeneous and of degree n in [the ai‘s], and homogeneous and of degree m in [the bj‘s]. Since [det(S)] has exactly the same homogeneity properties, and is divisible by [the double product], it follows that [det(S)=c(double product)] for some integer c. Since both [det(S) and the double product] have a monomial [amn b0m] occurring in them with coefficient 1, it follows that c=1, and our proposition is proved.

To my mind, the logic suffers from two gaps, although I think I see how to fill them. Recall the corresponding spot in the proof from Episode 4.  We were looking at the special case with am=bn=1.  We’d set R=k[u1,…,um, v1,…,vn]. We’d shown

\det(S)=h\cdot \prod_{i=1}^m\prod_{j=1}^n (u_i-v_j), with hR.

Let’s write Rab for the subring k[am–1,…,a0, bn–1,…,b0] of R. Note that det(S) belongs to Rab. If we knew that the double product and h also belonged to Rab, we’d be home free.

Galois theory shows that the double product belongs to Rab. The equivalent expression \prod_{i=1}^m F(u_i) belongs to Rab[u1,…,um], and is symmetric in the ui‘s; that should do the trick. We know that h results from dividing one element of Rab by another, without remainder in R. This ought to force h∈Rab. Ideals, Varieties, and Algorithms (Cox, Little, and O’Shea) provides a multivariable polynomial division algorithm; I haven’t checked the details, but I think this offers what we need.

So we can validate Lang’s argument, but it takes a bit of work. The approach in Episode 4 avoids these potholes, and as a bonus, introduces the “packing with t‘s” technique we’ll need later.

One last matter. We’ve already made use of the formula PE+QF=det(S), for some PRn[x], QRm[x]. Proving this took some effort. It’s much easier to derive this result:

for some PR[x], QR[x], and DR

Here’s the proof. Let K be the fraction field of R, as usual. Then K[x] is a PID. So if E and F are coprime, then for some p,qK[x],


Since the coefficients of p and q all belong to K, we can clear denominators and get PE+QF=D. Note that D≠0. When R=k[y], D is a polynomial in y alone, and the y-coordinates of the intersections of E and F are roots of D.

Now let’s think about the ideal (E,F) = {PE+QF : P,QR[x]}. Intersecting this with R gives an ideal of R. When R=k[y], R is a PID, so (E,F)∩k[y] is principal. It’s nonzero because D belongs to it. In other words, we can eliminate x from the equations E(x,y)=0 and F(x,y)=0 and get something non-trivial.

We might as well write D for a generator of (E,F)∩k[y]. I used to believe that resx(y) was such a generator. Not true. We’ve seen an example already in Inside Episode 2 (with x and y switched): the two ellipses x2+y2=0 and 2x2+y2=0. Easy congruence calculations show that x2 ≡ 0 mod (E,F), and as we saw way back in post 6, this amounts to showing that x2 belongs to (E,F). In fact (E,F)∩k[x] is generated by x2. But resy(x) , as we saw, is x4.

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