Topics in Algebra: Cryptography - The theorem of Cayley–Bacharach

Topics in Algebra: Cryptography - The theorem of Cayley–Bacharach

http://www.mat.univie.ac.at/~gagt/crypto2019

Martin Finn-Sell

martin.finn-sell@univie.ac.at

Here we present a proof of the Cayley–Bacharach theorem. We begin by recalling some useful facts, and then proving the result.¹

Theorem 1. (Bézout) LetC₁,C₂ be two plane curves over a fieldk whose defining polynomials F1, F2are relatively prime and have degreesd1andd2. Then there intersectionC1∩ C2inP²(k⁰), wherek’ is an algebraically closed field containingk, counted with their multiplicities, consists of d₁d₂ points.

This version differs from the notes of Tao. We remark that the version he is working with follows from this result, since the number of points (up to multiplicities) actually included in kwill be smaller thand₁d₂, and the notion of “common component” agrees with the notion of

“relatively prime” we use above.

The second thing we need to know is that given any pair of points, there is a unique line between them - similarly, for any five points there is a quadric and 9 points a cubic. For a justification of these points, think about the linear algebraic formulation of these statements - for instance given two points A = (a₀, a₁)and B = (b₀, b₁), then the line between them L={(x, y)|ax+by =c}is overdefined - we have four values and three unknowns. The same is the case for higher degree curves as mentioned above.

Lemma 2. Letk[x, y, z]dbe the space of homogeneous polynomials of degreed². Thendimkk[x, y, z]_d =

(d+1)(d+2)

2 .

Proof. Count the polynomials by figuring out a basis.

Let R_d = P(k[x, y, z]d), and P(R_d(P₁, ..., P_n)) = {M ∈ R_d | M(P_i) = 0for alli = {1, ..., n}}. Whenkis algebraically closed, andn≤ ^(d+1)(d+2)₂ −1, the subspaceR_d(P₁, ..., P_n) ofR_dhas dimension at most ^(d+1)(d+2)₂ −n−1.

1https://terrytao.wordpress.com/2011/07/15/pappuss-theorem-and-elliptic-curves/

would provide an alternative source - as an exercise translate that version into this one.

2recall that elements ofk[x, y, z]ddefineplanecurvesofdegreedoverk

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IfΩ={P₁, ..., P_n}⊂P²(k)is a set of points, thenΩdefineslconditionson polynomials of degreedif the codimension ofRd(Ω)isl. We say thatΩdefines independent conditions if the codimension ofR_d(Ω)inR_dis exactly|Ω|=n.

Proposition 3. Let Ω = {P1, ..., Pn} ⊂ P²(k) be any collection of n ≤ 2d+2 points. Then the points ofΩfail to determine independent conditions on curves of degreedif and only if either d+2of the points are colinear, orn=2d+2andΩis contained in a conic.

Proof. The “if” direction: Ifd+2points of Ωlie on a lineL, then by Bézout’s theorem any curve of degreedmust containL. The subset of curves of degreedcontaingLhas dimension

d+2 2

− ^d+1₂

= d+1. The remaing n− (d+2) points can impose at mostn − (d +2) conditions, so we see that Ω imposes at mostn−1conditions. A similar argument for the second case completes that direction.

The “only if” direction. We must do induction on both the degree d and the number of pointsn. The induction hypothesis fornwill allow us to assume thatno proper subset ofΩ does not impose independent conditions on curves of degreed. If we were supposing thatΩ does not impose independent conditions, then this hypothesis states that any curve of degree dthat contains all but one point ofΩin fact contains all ofΩ.

We note that ford=1, the result is satisfied (The reader should check this).

For arbitrarydsatisfyingn≤d+1the result is also easy - to exhibit a curve of degreed containing all but one pointPn ∈ Ω- and we do this by taking the union of general linesLi

throughP_ifori ∈{1, ..., n−1}and any curve of degreed−n+2not passing throughP_n. Now we take arbitrarydsatisfyingn > d+1.

Suppose first thatΩcontainsd+1points on a lineL. Suppose that no further points belong toL, and letΩ⁰be the complementary set ofn− (d+1)points ofΩ. We claim thatΩ⁰ must fail to impose independent conditions on curves of degree d−1- otherwise we could find a curveMof degreed−1containing all but one point ofΩ⁰ and thenL∪Mwould be a curve of degreedcontaining all but one point ofΩ.

By induction,Ω⁰must consist of exactlyd+1points on a lineL⁰, and thus eitherLcontains d+2points, orn=2d+2andΩlies on the conicL∪L⁰.

Next, suppose that only some lineLcontainsl≥3points ofΩ. By the previous argument, the remainingn−lpoints ofΩmust fail to impose independent conditions on curves of degree d−1, and so must include at leastd+1colinear points - which is precisely what we ended up considering in the previous paragraph.

We are done now, except for the case thatΩcontains no three colinear points. Choose any three pointsP1, ..., P3inΩand letΩ⁰ be the complement of these three points inΩ. If for any ithe points ofΩ⁰∪{P_iimpose independent conditions on curves of degreed−1, we are done - for then we can find a curveCof degreed−1containingΩ⁰ but notP_i, and then the union ofCwith the line joining the remainingPj and Pk is a curve of degreedcontaining all of Ω exceptP_i.

Thus, we can suppose thatΩ⁰∪{P_i}fails to impose independent conditions on curves of degreed−1. Since it cannot containd+1colinear points, we have by induction thatn=2d+2 and for eachithe setΩ⁰∪{P_i}is contained in a conicC_i. In the cased=2, we are done since six points fail to impose indepenent conditions on conics if and only if they lie on a conic.

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Ifd ≥ 3then Ω⁰ contains at least5points, no three colinear and so there can be at most one conic coontainingΩ, thus all the conicsCi must be equal to a single conic curveCwhich then contains all ofΩ.

Now we can prove the Cayley–Bacharach theorem.

Theorem 4. (Cayley–Bacharach) LetP₁, ..., P₈ be points inP²(k), no 4 on a line and no 7 on a conic then there is a 9th pointQsuch that any cubic throughP1, ..., P8also passes throughQ.

Proof. We apply the above proposition whend = 3,Ω ={P₁, ..., P₈}(son = 8). In this case, Ωmust determine independent conditions on cubics - as if we suppose not then we will find our other hypotheses in contradiction to the equivalence in Proposition 1.

Unpacking what this definition means -R₃(Ω)has codimension8, but the dimension ofR₃ is9. So there is a cubic containingP₁, ..., P₈in general position, and it must also pass through one more pointQsatisfying the conditions by Bézout’s theorem.

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