Coaxial circles, triangle centers, and the nullstellensatzvipul/studenttalks/coaxial... · Coaxial circles, triangle centers, and the nullstellensatz Vipul Naik Families with common

Coaxial circles,triangle centers,

and thenullstellensatz

Vipul Naik

Families withcommonintersection points,and Hilbert’snullstellensatz

Triangle centersand geometry overarbitrary fields

The point-idealcorrespondence

Why stay withpolynomials?

Coaxial circles, triangle centers, and thenullstellensatz

Vipul Naik

November 14, 2006



Vipul Naik





Outline

Families with common intersection points, and Hilbert’snullstellensatz

Triangle centers and geometry over arbitrary fields

The point-ideal correspondence

Why stay with polynomials?



Vipul Naik





Family of lines

let l1 and l2 be two lines in the plane. There are threepossibilities:

I l1 and l2 are the same line.

I l1 and l2 are parallel lines.

I l1 and l2 meet at exactly one point.

The third case is, in some sense, the most likely (in whatsense?)Question: Can we give the equation of a general line withthe property that its intersection set with l1 is the same asthe intersection set of l1 and l2?



Vipul Naik





Family of lines








Vipul Naik





Family of lines








Vipul Naik





Linear combination

Let L1 denote a linear expression such that l1 is the locus ofL1 = 0, and L2 denote a linear expression such that l2 is thelocus of L2 = 0.

Then the set of lines passing through the intersection of L1

and L2 correspond to equations a1L1 + a2L2. wherea1, a2 ∈ R.Why is that? The proof will have to show three things:

I a1L1 + a2L2 defines the equation of a line.

I The equation a1L1 + a2L2 is satisfied by the intersectionof l1 and l2.

I Any line which passes through the intersection satisfiesan equation of the form a1L1 + a2L2.



Vipul Naik





Linear combination

Let L1 denote a linear expression such that l1 is the locus ofL1 = 0, and L2 denote a linear expression such that l2 is thelocus of L2 = 0.Then the set of lines passing through the intersection of L1

and L2 correspond to equations a1L1 + a2L2. wherea1, a2 ∈ R.

Why is that? The proof will have to show three things:






Vipul Naik





Linear combination








Vipul Naik





Linear combination








Vipul Naik





Why each of these holds

I Both L1 and L2 are linear expressions. Hence, any linearcombination of them is also a linear expression.

I Let (x0, y0) be the intersection of l1 and l2. Then,L1(x0, y0) = 0 and L2(x0, y0) = 0. Anda1L1(x0, y0) + a2L2(x0, y0) = 0. Thus (x0, y0) lies onthe line defined by a1L1 + a2L2.

I Let p be the point of intersection of L1 and L2. Then,given any other point q, there is a unique line through pand q. But it is also true that we can find suitable a1

and a2 such that a1L1 + a2L2 is 0 on q. Hence, everyline through p is given by an equation of the forma1L1 + a2L2.



Vipul Naik












Vipul Naik












Vipul Naik





In the other cases

When l1 and l2 are intersecting lines given by equations L1

and L2, then the R-linear combinations of L1 and L2

correspond to lines passing through the intersection of L1

and L2.

What happens when l1 and l2 are parallel lines?What happens when l1 and l2 are the same line?



Vipul Naik





In the other cases




and L2.What happens when l1 and l2 are parallel lines?

What happens when l1 and l2 are the same line?



Vipul Naik





In the other cases




and L2.What happens when l1 and l2 are parallel lines?What happens when l1 and l2 are the same line?



Vipul Naik





Aside: projective plane

The collection of all R-linear combinations of L1 and L2 is atwo-dimensional vector space over R. Note that everyelement in this two-dimensional vector space does not definea different line. The expressions a1L1 + a2L2 andb1L1 + b2L2 define the same equation if a1b2 = a2b1.

The collection of all lines through the intersection of L1 andL2 is thus a quotient of the two-dimensional vector space byan equivalence relation. This quotient is often termed theprojective line(first used).



Vipul Naik





Aside: projective plane

The collection of all R-linear combinations of L1 and L2 is atwo-dimensional vector space over R. Note that everyelement in this two-dimensional vector space does not definea different line. The expressions a1L1 + a2L2 andb1L1 + b2L2 define the same equation if a1b2 = a2b1.The collection of all lines through the intersection of L1 andL2 is thus a quotient of the two-dimensional vector space byan equivalence relation. This quotient is often termed theprojective line(first used).



Vipul Naik





Other curves passing through intersection of twolines

Given two lines l1 and l2 described by equations L1 and L2,can we describe the family of all circles passing through l1and l2? How about the family of all parabolas through theintersection of l1 and l2?

The idea is to again look at linear combinations of L1 andL2, but this time, allow the coefficients to themselves belinear polynomials. More specifically, the space defined by:

p1L1 + p2L2

where p1 and p2 are linear polynomials covers all the conicspassing through L1 and L2.



Vipul Naik





Other curves passing through intersection of twolines

Given two lines l1 and l2 described by equations L1 and L2,can we describe the family of all circles passing through l1and l2? How about the family of all parabolas through theintersection of l1 and l2?The idea is to again look at linear combinations of L1 andL2, but this time, allow the coefficients to themselves belinear polynomials. More specifically, the space defined by:

p1L1 + p2L2

where p1 and p2 are linear polynomials covers all the conicspassing through L1 and L2.



Vipul Naik





Intersection of two circles

Given two circles Γ1 and Γ2, whose equations are S1 and S2,there are three qualitative possibilities:

I The circles intersect at two points.

I The circles are tangent to each other at one point.They may be internally tangent or externally tangent.

I The circles are disjoint. This may happen either withone circle completely contained in the interior of theother, or with both of them disjoint.

Question: In the first case, can we describe the entire familyof circles that pass through the same two intersection points?



Vipul Naik













Vipul Naik













Vipul Naik













Vipul Naik





Pictorial description of the three possibilities

I Intersect at two points case

I Tangent to each other at one point case

I Disjoint case



Vipul Naik





Linear combination (again)

The circles that pass through the two intersection points ofΓ1 and Γ2 are precisely those given by the equationa1S1 + a2S2 where a1, a2 ∈ R.

Why is that? We again have to show three things:

I Any expression of the form a1S1 + a2S2 defines theequation of a circle.

I If p is an intersection point of Γ1 and Γ2, then p alsolies on the circle corresponding to a1S1 + a2S2.

I Any circle passing through the intersection point of Γ1

and Γ2 can be expressed as a linear combination of S1

and S2.



Vipul Naik






The circles that pass through the two intersection points ofΓ1 and Γ2 are precisely those given by the equationa1S1 + a2S2 where a1, a2 ∈ R.Why is that? We again have to show three things:





and S2.



Vipul Naik











and S2.



Vipul Naik











and S2.



Vipul Naik






I S1 is given as x2 + y2 + 2g1x + 2f1y + c1 while S2 isgiven by x2 + y2 + g2x + f2y + c2. Then a1S1 + a2S2 isa quadratic expression with equal coefficients of x2 andy2. This gives the expression of a circle.The only exception is when a1 = −a2, in which case weget the straight line passing through the two points ofintersection.

I Suppose p = (x0, y0) is a point of intersection of thecircles Γ1 and Γ2. Then S1(p) = 0 and S2(p) = 0.Hence, a1S1(p) + a2S2(p) = 0, so p lies on the circlegiven by the expression a1S1 + a2S2.

I Let p and q be the two intersection points of Γ1 and Γ2.Note that for any point s, there is a unique circlepassing through p, q and s. Note also that there is acircle of the form a1S1 + a2S2 that also passes throughs. Hence, every circle passing through p and q can beexpressed as the locus of a1S1 + a2S2.



Vipul Naik











Vipul Naik











Vipul Naik





Power of a point: abstract concept

Given a circle Γ with center O and radius r , the power of apoint P is given by |OP|2 − r2.

Thus, the power of a point on the circle is 0, the power of apoint inside the circle is negative, and the power of a pointoutside the circle is positive.In coordinate geometry terms, if the normalized equation ofΓ is S = 0, the power of (x0, y0) is simply S(x0, y0).



Vipul Naik






Given a circle Γ with center O and radius r , the power of apoint P is given by |OP|2 − r2.Thus, the power of a point on the circle is 0, the power of apoint inside the circle is negative, and the power of a pointoutside the circle is positive.

In coordinate geometry terms, if the normalized equation ofΓ is S = 0, the power of (x0, y0) is simply S(x0, y0).



Vipul Naik






Given a circle Γ with center O and radius r , the power of apoint P is given by |OP|2 − r2.Thus, the power of a point on the circle is 0, the power of apoint inside the circle is negative, and the power of a pointoutside the circle is positive.In coordinate geometry terms, if the normalized equation ofΓ is S = 0, the power of (x0, y0) is simply S(x0, y0).



Vipul Naik






Given a circle Γ with center O and radius r , the power of apoint P is given by |OP|2 − r2.Thus, the power of a point on the circle is 0, the power of apoint inside the circle is negative, and the power of a pointoutside the circle is positive.In coordinate geometry terms, if the normalized equation ofΓ is S = 0, the power of (x0, y0) is simply S(x0, y0).



Vipul Naik





Radical axis

We define the radical axis(defined) of two non-concentriccircles Γ1 and Γ2 as: the locus of a point whose power withrespect to Γ1 equals its power with respect to Γ2.

Quick checks:

I The radical axis of two circles is a straight line (can beseen geometrically as well as algebraically). If thenormalized equations are S1 = 0 and S2 = 0, theequation of the radical axis is S1 = S2.

I The radical axis is perpendicular to the line joining thetwo centers.

I If the two circles meet at two points, the radical axis isthe common chord. if the two circles are tangential, theradical axis is the common tangent.



Vipul Naik





Radical axis

We define the radical axis(defined) of two non-concentriccircles Γ1 and Γ2 as: the locus of a point whose power withrespect to Γ1 equals its power with respect to Γ2.Quick checks:






Vipul Naik





Radical axis







Vipul Naik





Radical axis







Vipul Naik





The two typical non-degenerate casesLet Γ1 and Γ2 be distinct circles with equations S1 and S2.Then, we have the following possibilities:

1. S1 and S2 intersect in two points: The collectiona1S2 + a2S2 defines loci of all circles passing throughthose two intersection points. When a2 = −a1, we alsoget the equation of the line through the two intersectionpoints (called the radical axis).Such a family of circles is termed a family of coaxialcircles(defined) with intersection points.

2. S1 and S2 do not meet, but have different centers: Thecollection a1S1 + a2S2 defines a collection of circleswhose centers all lie on the same line, such that thepairwise radical axis of any two circles is the same asthat of S1 and S2.The family a1S1 + a2S2 also includes two point circles.These are called the limit points of the family. Thus,such a family of coaxial circles is termed a family ofcoaxial circles(defined) with limit points.



Vipul Naik





The two typical non-degenerate casesLet Γ1 and Γ2 be distinct circles with equations S1 and S2.Then, we have the following possibilities:

1. S1 and S2 intersect in two points: The collectiona1S2 + a2S2 defines loci of all circles passing throughthose two intersection points. When a2 = −a1, we alsoget the equation of the line through the two intersectionpoints (called the radical axis).Such a family of circles is termed a family of coaxialcircles(defined) with intersection points.

2. S1 and S2 do not meet, but have different centers: Thecollection a1S1 + a2S2 defines a collection of circleswhose centers all lie on the same line, such that thepairwise radical axis of any two circles is the same asthat of S1 and S2.The family a1S1 + a2S2 also includes two point circles.These are called the limit points of the family. Thus,such a family of coaxial circles is termed a family ofcoaxial circles(defined) with limit points.



Vipul Naik





The borderline non-degenerate case

In one case S1 and S2 intersect in two distinct points. Thiscorresponds intuitively to the real and distinct roots case ofa quadratic equation.

In another case, S1 and S2 do not intersect at all. Thiscorresponds to the imaginary and distinct roots case of aquadratic equation.The borderline case corresponds to the real and equal rootscase of a quadratic equation. This is the case below:S1 and S2 are tangent at one point: The collectiona1S1 + a2S2 defines loci of all circles tangent to S1 at thatsame point. In the special case a2 = −a1, we get theequation of the common tangent, again called the radicalaxis.Such a family of circles is termed a family of coaxialcircles(defined) with a point of tangency.



Vipul Naik






In one case S1 and S2 intersect in two distinct points. Thiscorresponds intuitively to the real and distinct roots case ofa quadratic equation.In another case, S1 and S2 do not intersect at all. Thiscorresponds to the imaginary and distinct roots case of aquadratic equation.

The borderline case corresponds to the real and equal rootscase of a quadratic equation. This is the case below:S1 and S2 are tangent at one point: The collectiona1S1 + a2S2 defines loci of all circles tangent to S1 at thatsame point. In the special case a2 = −a1, we get theequation of the common tangent, again called the radicalaxis.Such a family of circles is termed a family of coaxialcircles(defined) with a point of tangency.



Vipul Naik






In one case S1 and S2 intersect in two distinct points. Thiscorresponds intuitively to the real and distinct roots case ofa quadratic equation.In another case, S1 and S2 do not intersect at all. Thiscorresponds to the imaginary and distinct roots case of aquadratic equation.The borderline case corresponds to the real and equal rootscase of a quadratic equation. This is the case below:S1 and S2 are tangent at one point: The collectiona1S1 + a2S2 defines loci of all circles tangent to S1 at thatsame point. In the special case a2 = −a1, we get theequation of the common tangent, again called the radicalaxis.Such a family of circles is termed a family of coaxialcircles(defined) with a point of tangency.



Vipul Naik





The three degenerate cases

There are three of them, corresponding to the threenon-degenerate cases:

1. Family of concurrent lines: We can go from any familyof coaxial circles with intersection points to such afamily, by inverting about a circle centered at one of theintersection points.

2. Family of concentric circles: We can go from any familyof coaxial circles with limit points, to such a family, byinverting about one of the limit points.

3. Family of parallel lines: We can go from any family ofcircle with a point of tangency to such a family, simplyby inverting about the point of tangency.



Vipul Naik












Vipul Naik












Vipul Naik





Orthogonality: an aside

Given two points, we can define the family of coaxial circleswith those two as intersection points; we can also define thefamily of coaxial circles with those two as limit points. Thetwo families of circles intersect orthogonally at every point.Because inversion preserves local angles, orthogonal familiesgo to orthogonal families. That’s why inverting about one ofthe two points takes the intersecting coaxial family toconcurrent lines, and the coaxial family with limit points tothe concentric circles family.



Vipul Naik





A somewhat different situation

Question: Given the equation of three lines, try finding theequation of the circumcircle of these three lines.

Clever method: Let L1, L2 and L3 be the equations of thesethree lines. Consider a general curve given bya1L2L3 + a2L3L1 + a3L1L2 = 0 where a1,a2, and a3 aresuitable real numbers. Clearly, any such curve passesthrough the three vertices of the triangle. If we can find a1,a2, a3 such that the equation above becomes the equation ofa circle, then we have found the equation of the circumcircle.



Vipul Naik





A somewhat different situation

Question: Given the equation of three lines, try finding theequation of the circumcircle of these three lines.Clever method: Let L1, L2 and L3 be the equations of thesethree lines. Consider a general curve given bya1L2L3 + a2L3L1 + a3L1L2 = 0 where a1,a2, and a3 aresuitable real numbers. Clearly, any such curve passesthrough the three vertices of the triangle. If we can find a1,a2, a3 such that the equation above becomes the equation ofa circle, then we have found the equation of the circumcircle.



Vipul Naik





The general intersection questionQuestion: Given a collection of curves given by equationsC1, C2 and so on, such that the intersection of all curves inthe family is a set F of points. Then, can we express theequations for all curves (of a certain kind) passing through Fin terms of the equations for C1, C2 etc.?

The special cases we have seen so far:

I Given a pair of lines, can we express the equation of alllines passing through their intersection in terms of theequations of the two lines? Answer: yes. In fact, everyline through the intersection is a linear combination ofthe two lines (with constant coefficients).

I Given a pair of circle, can we express the equation of allcircles passing through their intersection in terms of theequations of the two circles? Answer: yes, provided thetwo circles intersect in two points. In fact, we cansimply use linear combinations.

To formulate and explore the question in a more generalsetting, we need some more concepts.



Vipul Naik





The general intersection questionQuestion: Given a collection of curves given by equationsC1, C2 and so on, such that the intersection of all curves inthe family is a set F of points. Then, can we express theequations for all curves (of a certain kind) passing through Fin terms of the equations for C1, C2 etc.?The special cases we have seen so far:






Vipul Naik











Vipul Naik











Vipul Naik





Affine curves

Let R[x , y ] denote the set of polynomials in twoindeterminates x and y . To each (nonzero) polynomialp ∈ R[x , y ] is associated a nonnegative integer called itsdegree (guess how?).

An affine curve(defined) is a curve obtained as the solution setof such a p ∈ R[x , y ]. The degree of the curve is defined asthe degree of the polynomial.Two questions:

I Given an affine curve of degree m and an affine curve ofdegree n, how many points do they intersect in?

I Can we describe all affine curves that pass throughthose intersection points in terms of these two curves?



Vipul Naik





Affine curves

Let R[x , y ] denote the set of polynomials in twoindeterminates x and y . To each (nonzero) polynomialp ∈ R[x , y ] is associated a nonnegative integer called itsdegree (guess how?).An affine curve(defined) is a curve obtained as the solution setof such a p ∈ R[x , y ]. The degree of the curve is defined asthe degree of the polynomial.

Two questions:





Vipul Naik





Affine curves

Let R[x , y ] denote the set of polynomials in twoindeterminates x and y . To each (nonzero) polynomialp ∈ R[x , y ] is associated a nonnegative integer called itsdegree (guess how?).An affine curve(defined) is a curve obtained as the solution setof such a p ∈ R[x , y ]. The degree of the curve is defined asthe degree of the polynomial.Two questions:





Vipul Naik





Affine curves

Let R[x , y ] denote the set of polynomials in twoindeterminates x and y . To each (nonzero) polynomialp ∈ R[x , y ] is associated a nonnegative integer called itsdegree (guess how?).An affine curve(defined) is a curve obtained as the solution setof such a p ∈ R[x , y ]. The degree of the curve is defined asthe degree of the polynomial.Two questions:





Vipul Naik





The polynomial ring

The set R[x , y ] is equipped with a ring structure in thefollowing sense:

I Given two polynomials, their sum is the polynomialobtained by term-wise addition. The additive identity isthe zero polynomial and the “negative” of a polynomialis obtained by negating each coefficient.

I Given two polynomials, their product is defined usingthe convolution multiplication on coefficients. Themultiplicative identity is the constant polynomial 1.



Vipul Naik





Points and polynomials

Now, we have two sets: the set R2 of points and the setR[x , y ] of polynomials. Given a polynomial and a point, thenatural thing to do is to evaluate the polynomial at thepoint.Given a point (x0, y0), consider the map:

R[x , y ] → R

given by:

p 7→ p(x0, y0)

This map is a homomorphism from the ring R[x , y ] to thering R.



Vipul Naik





The vanishing relation

Define the following relation between points andpolynomials: a point v is related to a polynomial p ifp(v) = 0. This relation is termed the vanishing relation.Then, define the following:

I The zero set for a collection S of polynomials (that is, asubset of R[x , y ]) is defined as the set of points v suchthat p(v) = 0 for every p ∈ S . This is also called thenull set of S and is denoted as Z(S).

I The vanishing ideal for a collection P of points (that is,a subset of R2) is the set of polynomials p such that forevery v ∈ P, p(v) = 0. This is denoted as I(P).



Vipul Naik





The vanishing ideal is an ideal

Given a ring R, a subset I of R is termed an ideal(defined) if:

I I is closed under addition.

I The product of any element in I and any element in Ris again in I .

Now, the following very important observation: thevanishing ideal for any collection of points is actually anideal. This observation captures the essence of the “linearcombinations” approach we have seen so far.In fact, the vanishing ideal for any set of points has an evenstronger property. If p is a polynomial such that pn belongsto the vanishing ideal, then p also belongs to the vanishingideal. That is, the vanishing ideal is a radical ideal(defined).



Vipul Naik









Now, the following very important observation: thevanishing ideal for any collection of points is actually anideal. This observation captures the essence of the “linearcombinations” approach we have seen so far.

In fact, the vanishing ideal for any set of points has an evenstronger property. If p is a polynomial such that pn belongsto the vanishing ideal, then p also belongs to the vanishingideal. That is, the vanishing ideal is a radical ideal(defined).



Vipul Naik









Now, the following very important observation: thevanishing ideal for any collection of points is actually anideal. This observation captures the essence of the “linearcombinations” approach we have seen so far.In fact, the vanishing ideal for any set of points has an evenstronger property. If p is a polynomial such that pn belongsto the vanishing ideal, then p also belongs to the vanishingideal. That is, the vanishing ideal is a radical ideal(defined).



Vipul Naik





Zero sets are closed sets

Every polynomial defines a continuous function on R2.Hence, the inverse image of 0 under any polynomial is aclosed set. The intersection of inverse images of 0 under acollection of polynomial maps is an intersection of closedsets, and hence a closed set.

Another way of looking at it is: if a polynomial vanishes on asequence of points, it also vanishes on the limit of thatsequence. Hence, the set of points on which the polynomialvanishes is a closed set.



Vipul Naik





Zero sets are closed sets

Every polynomial defines a continuous function on R2.Hence, the inverse image of 0 under any polynomial is aclosed set. The intersection of inverse images of 0 under acollection of polynomial maps is an intersection of closedsets, and hence a closed set.Another way of looking at it is: if a polynomial vanishes on asequence of points, it also vanishes on the limit of thatsequence. Hence, the set of points on which the polynomialvanishes is a closed set.



Vipul Naik





Going back and forthThere’s a general notion of Galois correspondence:

I two sets A and B and a relation R ⊆ A× B.

I Let f : 2A → 2B be a map that sends S ⊆ A to the setof those elements in B that are related to every elementof S . Similarly, define g : 2B → 2A.

I We have the following:

I S ⊆ g(f (S)) for every S ⊆ A and T ⊆ f (g(T )) forevery T ⊆ B

I S ⊆ S ′ =⇒ f (S ′) ⊆ f (S)I f (S) = f (g(f (S))) for every S ⊆ A and

g(T ) = g(f (g(T ))) for every T ⊂ B

Thus sets that arise as the image of f are invariantunder the operation f .g and sets that arise as the imageof g are invariant under the operation f .g . These arerespectively termed the closed sets in B and in A.

In our case, A = R2 is the set of points and B = R[x , y ] isthe set of polynomials. R is the vanishing relation.



Vipul Naik
















Vipul Naik
















Vipul Naik
















Vipul Naik





What are the closed sets for the Galoiscorrespondence?

In this Galois correspondence, what are the closed sets inR[x , y ]? In other words, what are the subsets of R[x , y ] thatoccur as the vanishing ideal for a certain point set?

We have shown that any such ideal must be a radical ideal.Is the converse true? That is, is every radical ideal preciselythe vanishing ideal of a certain point set?



Vipul Naik





What are the closed sets for the Galoiscorrespondence?

In this Galois correspondence, what are the closed sets inR[x , y ]? In other words, what are the subsets of R[x , y ] thatoccur as the vanishing ideal for a certain point set?We have shown that any such ideal must be a radical ideal.Is the converse true? That is, is every radical ideal preciselythe vanishing ideal of a certain point set?



Vipul Naik





If only it were true...

If it were true that every radical ideal were precisely thevanishing ideal of a certain point set, then we couldapproach our original problem as follows.

Given two affine curves C1 and C2, let F be the set of pointsin both C1 and C2. Consider the smallest radical idealgenerated by the polynomials for C1 and C2. This vanisheson F and nowhere else. Hence, it should be precisely thevanishing ideal of F . Thus, we have described the collectionof all affine curves passing through the intersection of C1

and C2.



Vipul Naik





If only it were true...

If it were true that every radical ideal were precisely thevanishing ideal of a certain point set, then we couldapproach our original problem as follows.Given two affine curves C1 and C2, let F be the set of pointsin both C1 and C2. Consider the smallest radical idealgenerated by the polynomials for C1 and C2. This vanisheson F and nowhere else. Hence, it should be precisely thevanishing ideal of F . Thus, we have described the collectionof all affine curves passing through the intersection of C1

and C2.



Vipul Naik





Alas...

It isn’t true.There is one major problem: there may not be anyintersection points at all!

Consider two circles which do not intersect. Then the radicalideal generated by the equations of the circle is definitely notthe whole of R[x , y ]. However, the collection of polynomialswhose corresponding curves pass through the intersection ofthe two circles is the whole of R[x , y ].How do we fix this problem?What is the problem?The problem is that when we try solving the equation of thetwo circles, we get a quadratic equation which has complexroots but does not have real roots.So can we fix the problem by going to complex numbers?



Vipul Naik





Alas...

It isn’t true.There is one major problem: there may not be anyintersection points at all!Consider two circles which do not intersect. Then the radicalideal generated by the equations of the circle is definitely notthe whole of R[x , y ]. However, the collection of polynomialswhose corresponding curves pass through the intersection ofthe two circles is the whole of R[x , y ].

How do we fix this problem?What is the problem?The problem is that when we try solving the equation of thetwo circles, we get a quadratic equation which has complexroots but does not have real roots.So can we fix the problem by going to complex numbers?



Vipul Naik





Alas...

It isn’t true.There is one major problem: there may not be anyintersection points at all!Consider two circles which do not intersect. Then the radicalideal generated by the equations of the circle is definitely notthe whole of R[x , y ]. However, the collection of polynomialswhose corresponding curves pass through the intersection ofthe two circles is the whole of R[x , y ].How do we fix this problem?

What is the problem?The problem is that when we try solving the equation of thetwo circles, we get a quadratic equation which has complexroots but does not have real roots.So can we fix the problem by going to complex numbers?



Vipul Naik





Alas...

It isn’t true.There is one major problem: there may not be anyintersection points at all!Consider two circles which do not intersect. Then the radicalideal generated by the equations of the circle is definitely notthe whole of R[x , y ]. However, the collection of polynomialswhose corresponding curves pass through the intersection ofthe two circles is the whole of R[x , y ].How do we fix this problem?What is the problem?

The problem is that when we try solving the equation of thetwo circles, we get a quadratic equation which has complexroots but does not have real roots.So can we fix the problem by going to complex numbers?



Vipul Naik





Alas...

It isn’t true.There is one major problem: there may not be anyintersection points at all!Consider two circles which do not intersect. Then the radicalideal generated by the equations of the circle is definitely notthe whole of R[x , y ]. However, the collection of polynomialswhose corresponding curves pass through the intersection ofthe two circles is the whole of R[x , y ].How do we fix this problem?What is the problem?The problem is that when we try solving the equation of thetwo circles, we get a quadratic equation which has complexroots but does not have real roots.

So can we fix the problem by going to complex numbers?



Vipul Naik





Alas...

It isn’t true.There is one major problem: there may not be anyintersection points at all!Consider two circles which do not intersect. Then the radicalideal generated by the equations of the circle is definitely notthe whole of R[x , y ]. However, the collection of polynomialswhose corresponding curves pass through the intersection ofthe two circles is the whole of R[x , y ].How do we fix this problem?What is the problem?The problem is that when we try solving the equation of thetwo circles, we get a quadratic equation which has complexroots but does not have real roots.So can we fix the problem by going to complex numbers?



Vipul Naik





Yes!!

Hilbert’s nullstellensatz for complex numbers: Every radicalideal in C[x , y ] is precisely the vanishing set of a collectionof points in C2.

More generally (higher dimensional version): Every radicalideal in C[x1, x2, . . . , xn] is precisely the vanishing set of acollection of points in Cn.



Vipul Naik





Yes!!

Hilbert’s nullstellensatz for complex numbers: Every radicalideal in C[x , y ] is precisely the vanishing set of a collectionof points in C2.More generally (higher dimensional version): Every radicalideal in C[x1, x2, . . . , xn] is precisely the vanishing set of acollection of points in Cn.



Vipul Naik





Outline







Vipul Naik





Triangle centers

My favourite triangle centers:

1. The centroid(defined) of a triangle is defined as theintersection of the medians of the triangle.

2. The orthocenter(defined) of a triangle is defined as theintersection of the altitudes of the triangle.

3. The circumcenter(defined) of a triangle is defined as theintersection of the perpendicular bisectors of thetriangle’s sides.

4. The incenter(defined) of a triangle is defined as theintersection of the internal angle bisectors of thetriangle.

Question: Which of these definitions makes sense when weare working over an arbitrary field?



Vipul Naik





Triangle centers

My favourite triangle centers:

1. The centroid(defined) of a triangle is defined as theintersection of the medians of the triangle.

2. The orthocenter(defined) of a triangle is defined as theintersection of the altitudes of the triangle.

3. The circumcenter(defined) of a triangle is defined as theintersection of the perpendicular bisectors of thetriangle’s sides.

4. The incenter(defined) of a triangle is defined as theintersection of the internal angle bisectors of thetriangle.

Question: Which of these definitions makes sense when weare working over an arbitrary field?



Vipul Naik





What does arbitrary field mean?

Let k be a field. Consider the plane k2 and use the termpoint for an element of k2. Then, a triangle in k2 is a set ofthree non-collinear (?) points in k2.

Then, can we make sense of centroid, orthocenter,circumcenter and so on within k?

1. Centroid?: Midpoint? Yes. Line joining two points?Yes. Intersection of two lines? Yes. Concurrence?

2. Orthocenter: Perpendicularity? Yes. Intersection of twolines? Yes. Concurrence?

3. Circumcenter?: Midpoint? Yes. Perpendicular bisector?Yes. Intersection of two lines? Yes. Concurrence?

4. Incenter? No. How do we define angle bisector?



Vipul Naik






Let k be a field. Consider the plane k2 and use the termpoint for an element of k2. Then, a triangle in k2 is a set ofthree non-collinear (?) points in k2.Then, can we make sense of centroid, orthocenter,circumcenter and so on within k?







Vipul Naik













Vipul Naik













Vipul Naik













Vipul Naik





Another look at proofs of concurrence

Given a triangle 4ABC in R2, we have three curves ΓA, ΓB

and ΓC defined cyclically with respect to the vertices. Howdo we prove that ΓA, ΓB and ΓC concur?Typical proof: We find functions fA, fB and fC from R2 to Rsuch that ΓA is the locus of fB = fC , ΓB is the locus offC = fA, and ΓC is the locus of fA = fB .If we can find these functions fA, fB and fC in the generalityof a field (that is, as polynomials), we have shownconcurrence.



Vipul Naik





In each case

1. For the median, we can set fA(P) as the area of thetriangle formed by P, B and C . Note that this area is alinear polynomial in the coordinates of P, and itsvanishing locus is precisely the median. We cyclicallydefine fB and fC .

2. For the altitudes, we can set fA(P) as |PA|2 + |BC |2after first equipping with the Euclidean metric.Although this is a quadratic polynomial in thecoordinates of P, the polynomial fA − fB is linearbecause the top terms cancel. The loci are precisely thealtitudes. Note that we may not be able to define |PA|within the field but we can define |PA|2.

3. For the perpendicular bisectors, we can set fA as |PA|2.This is a quadratic polynomial, but again fA − fB is alinear polynomial, and the locus is the perpendicularbisector.



Vipul Naik





In each case






Vipul Naik





In each case






Vipul Naik





For the angle bisector

For the angle bisectors, fA comes out to be, not apolynomial, but an expression involving roots anddenominators. Since the expression involves a squareroot, wecan only do angle bisection when the field k is quadraticallyclosed, that is, when we can take squareroots of positivenumbers.

More generally, when ΓA, ΓB , and ΓC are circles, we requirea condition like quadratic closure to ensure intersection.



Vipul Naik





For the angle bisector

For the angle bisectors, fA comes out to be, not apolynomial, but an expression involving roots anddenominators. Since the expression involves a squareroot, wecan only do angle bisection when the field k is quadraticallyclosed, that is, when we can take squareroots of positivenumbers.More generally, when ΓA, ΓB , and ΓC are circles, we requirea condition like quadratic closure to ensure intersection.



Vipul Naik





Fields are linearly closed

Here’s an easy statement, that we already know (but maybenever noticed): if we have a system of linear equations withcoefficients in a field k , and that system of equations has asolution in any extension of k, the system has a solutionwithin k itself.

That is because if the system has a solution in any extensionof k, we can actually find that solution using a formula interms of addition, subtraction, multiplication and divisionfrom the coefficients. Hence, that solution must lie in k.What’s the deep thing behind this?We defined a field in such a way that any consistent linearequation in one variable that had a solution in an extension,automatically had a solution in the field.And, practically for free, we obtained that any consistentsystem of linear equations that had a solution in anextension, had a solution in the field itself.



Vipul Naik






Here’s an easy statement, that we already know (but maybenever noticed): if we have a system of linear equations withcoefficients in a field k , and that system of equations has asolution in any extension of k, the system has a solutionwithin k itself.That is because if the system has a solution in any extensionof k, we can actually find that solution using a formula interms of addition, subtraction, multiplication and divisionfrom the coefficients. Hence, that solution must lie in k.

What’s the deep thing behind this?We defined a field in such a way that any consistent linearequation in one variable that had a solution in an extension,automatically had a solution in the field.And, practically for free, we obtained that any consistentsystem of linear equations that had a solution in anextension, had a solution in the field itself.



Vipul Naik






Here’s an easy statement, that we already know (but maybenever noticed): if we have a system of linear equations withcoefficients in a field k , and that system of equations has asolution in any extension of k, the system has a solutionwithin k itself.That is because if the system has a solution in any extensionof k, we can actually find that solution using a formula interms of addition, subtraction, multiplication and divisionfrom the coefficients. Hence, that solution must lie in k.What’s the deep thing behind this?

We defined a field in such a way that any consistent linearequation in one variable that had a solution in an extension,automatically had a solution in the field.And, practically for free, we obtained that any consistentsystem of linear equations that had a solution in anextension, had a solution in the field itself.



Vipul Naik






Here’s an easy statement, that we already know (but maybenever noticed): if we have a system of linear equations withcoefficients in a field k , and that system of equations has asolution in any extension of k, the system has a solutionwithin k itself.That is because if the system has a solution in any extensionof k, we can actually find that solution using a formula interms of addition, subtraction, multiplication and divisionfrom the coefficients. Hence, that solution must lie in k.What’s the deep thing behind this?We defined a field in such a way that any consistent linearequation in one variable that had a solution in an extension,automatically had a solution in the field.

And, practically for free, we obtained that any consistentsystem of linear equations that had a solution in anextension, had a solution in the field itself.



Vipul Naik






Here’s an easy statement, that we already know (but maybenever noticed): if we have a system of linear equations withcoefficients in a field k , and that system of equations has asolution in any extension of k, the system has a solutionwithin k itself.That is because if the system has a solution in any extensionof k, we can actually find that solution using a formula interms of addition, subtraction, multiplication and divisionfrom the coefficients. Hence, that solution must lie in k.What’s the deep thing behind this?We defined a field in such a way that any consistent linearequation in one variable that had a solution in an extension,automatically had a solution in the field.And, practically for free, we obtained that any consistentsystem of linear equations that had a solution in anextension, had a solution in the field itself.



Vipul Naik





Replace linear by algebraic

We just proved the linear Nullstellensatz: a field is closedunder linear extensions in one variable, hence it is closedunder systems of linear equations.What Hilbert proved was the algebraic Nullstellensatz. Theweak form of the Nullstellensatz states: if a field is closedunder all algebraic extensions in one variable, then it isclosed under consistent systems of algebraic equations.



Vipul Naik





Is this a general philosophy?

It is!

There are results in model theory that state that if astructure is closed under “extensions” for one variable, it isclosed under extensions for finitely many variables.So we have an analogous “differential nullstellensatz”: givena differentially closed field, any system of differentialequations that is consistent has a solution within thedifferentially closed field itself.



Vipul Naik






It is!There are results in model theory that state that if astructure is closed under “extensions” for one variable, it isclosed under extensions for finitely many variables.

So we have an analogous “differential nullstellensatz”: givena differentially closed field, any system of differentialequations that is consistent has a solution within thedifferentially closed field itself.



Vipul Naik






It is!There are results in model theory that state that if astructure is closed under “extensions” for one variable, it isclosed under extensions for finitely many variables.So we have an analogous “differential nullstellensatz”: givena differentially closed field, any system of differentialequations that is consistent has a solution within thedifferentially closed field itself.



Vipul Naik





Outline







Vipul Naik





Knitting things together

We gave two formulations of the Hilbert nullstellensatz: onein terms of a Galois correspondence between sets in kn andideals in k[x1, x2, . . . , xn], and the other in terms ofconsistency of systems of algebraic equations.

Why are these formulations the same?



Vipul Naik





Knitting things together

We gave two formulations of the Hilbert nullstellensatz: onein terms of a Galois correspondence between sets in kn andideals in k[x1, x2, . . . , xn], and the other in terms ofconsistency of systems of algebraic equations.Why are these formulations the same?



Vipul Naik





The evaluation homomorphism

Given a point (a1, a2, . . . , an), there is an evaluation map:

eval(a1,a2,...,an)k[x1, x2, . . . , xn] → k

given by:

p 7→ p(a1, a2, . . . , an)

This map is a ring homomorphism. It is also surjective.The kernel(first used) of this map is precisely the vanishing idealfor the single point (a1, a2, . . . , an).Elementary ring theory tells us that the kernel is a so-calledmaximal ideal: there is no ideal strictly between it and thewhole ring.Upshot: In the Galois correspondence, points give rise tomaximal ideals.Does every maximal ideal arise this way?



Vipul Naik







eval(a1,a2,...,an)k[x1, x2, . . . , xn] → k

given by:

p 7→ p(a1, a2, . . . , an)

This map is a ring homomorphism. It is also surjective.

The kernel(first used) of this map is precisely the vanishing idealfor the single point (a1, a2, . . . , an).Elementary ring theory tells us that the kernel is a so-calledmaximal ideal: there is no ideal strictly between it and thewhole ring.Upshot: In the Galois correspondence, points give rise tomaximal ideals.Does every maximal ideal arise this way?



Vipul Naik







eval(a1,a2,...,an)k[x1, x2, . . . , xn] → k

given by:

p 7→ p(a1, a2, . . . , an)

This map is a ring homomorphism. It is also surjective.The kernel(first used) of this map is precisely the vanishing idealfor the single point (a1, a2, . . . , an).Elementary ring theory tells us that the kernel is a so-calledmaximal ideal: there is no ideal strictly between it and thewhole ring.

Upshot: In the Galois correspondence, points give rise tomaximal ideals.Does every maximal ideal arise this way?



Vipul Naik







eval(a1,a2,...,an)k[x1, x2, . . . , xn] → k

given by:

p 7→ p(a1, a2, . . . , an)

This map is a ring homomorphism. It is also surjective.The kernel(first used) of this map is precisely the vanishing idealfor the single point (a1, a2, . . . , an).Elementary ring theory tells us that the kernel is a so-calledmaximal ideal: there is no ideal strictly between it and thewhole ring.Upshot: In the Galois correspondence, points give rise tomaximal ideals.

Does every maximal ideal arise this way?



Vipul Naik







eval(a1,a2,...,an)k[x1, x2, . . . , xn] → k

given by:

p 7→ p(a1, a2, . . . , an)

This map is a ring homomorphism. It is also surjective.The kernel(first used) of this map is precisely the vanishing idealfor the single point (a1, a2, . . . , an).Elementary ring theory tells us that the kernel is a so-calledmaximal ideal: there is no ideal strictly between it and thewhole ring.Upshot: In the Galois correspondence, points give rise tomaximal ideals.Does every maximal ideal arise this way?



Vipul Naik





Intersections of maximal ideals

In the Galois correspondence, we had associated, to everysubset S of kn, the ideal of polynomials I(S) vanishing on S .

In hindsight:

I(S) =⋂a∈S

ker evala

Thus every vanishing ideal is an intersection of maximalideals corresponding to points.So, the Hilbert’s nullstellensatz (first version) translates tosaying: every radical ideal is an intersection of maximalideals corresponding to points.



Vipul Naik






In the Galois correspondence, we had associated, to everysubset S of kn, the ideal of polynomials I(S) vanishing on S .In hindsight:

I(S) =⋂a∈S

ker evala

Thus every vanishing ideal is an intersection of maximalideals corresponding to points.

So, the Hilbert’s nullstellensatz (first version) translates tosaying: every radical ideal is an intersection of maximalideals corresponding to points.



Vipul Naik






In the Galois correspondence, we had associated, to everysubset S of kn, the ideal of polynomials I(S) vanishing on S .In hindsight:

I(S) =⋂a∈S

ker evala

Thus every vanishing ideal is an intersection of maximalideals corresponding to points.So, the Hilbert’s nullstellensatz (first version) translates tosaying: every radical ideal is an intersection of maximalideals corresponding to points.



Vipul Naik





More on ideals

Some east observations:

I An ideal is said to be proper(defined) if it is not the wholering. In the case of the polynomial ring, an ideal isproper if and only if it does not contain any nonzeroconstant polynomial.

I Given an ideal, we can, for each polynomial p in theideal, we write the equation p(x1, x2, . . . , xn) = 0. Wehave a (possibly infinite) system of equations. The idealis proper if and only the equation 1 = 0 cannot bederived by adding, subtracting, and multiplying theseequations by polynomials. In other words, an ideal isproper if and only if the corresponding system ofequations is consistent.

I Thus, demanding that “every proper ideal has anonempty zero set” is the same as saying that “everyconsistent system of algebraic equations has a solution”.



Vipul Naik





More on ideals







Vipul Naik





More on ideals







Vipul Naik





Bringing maximal ideals in the picture

A theorem of ring theory: in a ring with multiplicativeidentity, every ideal is contained in a maximal ideal.

Thus, demanding that every proper ideal has a nonemptyzero set is equivalent to demanding that every maximal idealhas a nonempty zero set.However, the zero set of a maximal ideal cannot have morethan one point.Thus, demanding that every proper ideal has a nonemptyzero set is equivalent to demanding that every maximal idealis the vanishing ideal of a point.



Vipul Naik






A theorem of ring theory: in a ring with multiplicativeidentity, every ideal is contained in a maximal ideal.Thus, demanding that every proper ideal has a nonemptyzero set is equivalent to demanding that every maximal idealhas a nonempty zero set.

However, the zero set of a maximal ideal cannot have morethan one point.Thus, demanding that every proper ideal has a nonemptyzero set is equivalent to demanding that every maximal idealis the vanishing ideal of a point.



Vipul Naik






A theorem of ring theory: in a ring with multiplicativeidentity, every ideal is contained in a maximal ideal.Thus, demanding that every proper ideal has a nonemptyzero set is equivalent to demanding that every maximal idealhas a nonempty zero set.However, the zero set of a maximal ideal cannot have morethan one point.

Thus, demanding that every proper ideal has a nonemptyzero set is equivalent to demanding that every maximal idealis the vanishing ideal of a point.



Vipul Naik






A theorem of ring theory: in a ring with multiplicativeidentity, every ideal is contained in a maximal ideal.Thus, demanding that every proper ideal has a nonemptyzero set is equivalent to demanding that every maximal idealhas a nonempty zero set.However, the zero set of a maximal ideal cannot have morethan one point.Thus, demanding that every proper ideal has a nonemptyzero set is equivalent to demanding that every maximal idealis the vanishing ideal of a point.



Vipul Naik





So far...

We have shown that the following are equivalent:

I Every consistent system of algebraic equations over kn

has a solution.

I Every proper ideal in the polynomial ring of n variableshas a nonempty zero set.

I The maximal ideals in the polynomial ring of n variablesare precisely the ideals corresponding to points.

The weak form of the Nullstellensatz states that these threeequivalent conditions hold over an algebraically closed field.



Vipul Naik





So far...

We have shown that the following are equivalent:

I Every consistent system of algebraic equations over kn

has a solution.

I Every proper ideal in the polynomial ring of n variableshas a nonempty zero set.

I The maximal ideals in the polynomial ring of n variablesare precisely the ideals corresponding to points.

The weak form of the Nullstellensatz states that these threeequivalent conditions hold over an algebraically closed field.



Vipul Naik





Strong form

Original statement of the strong form: Every radical ideal isthe vanishing ideal of some set. Equivalently, every radicalideal is the intersection of maximal ideals corresponding topoints.

Statement of strong form assuming weak form: Every radicalideal is an intersection of maximal ideals.The strong form is also true over an algebraically closed field.



Vipul Naik





Strong form

Original statement of the strong form: Every radical ideal isthe vanishing ideal of some set. Equivalently, every radicalideal is the intersection of maximal ideals corresponding topoints.Statement of strong form assuming weak form: Every radicalideal is an intersection of maximal ideals.

The strong form is also true over an algebraically closed field.



Vipul Naik





Strong form

Original statement of the strong form: Every radical ideal isthe vanishing ideal of some set. Equivalently, every radicalideal is the intersection of maximal ideals corresponding topoints.Statement of strong form assuming weak form: Every radicalideal is an intersection of maximal ideals.The strong form is also true over an algebraically closed field.



Vipul Naik





Outline







Vipul Naik





Original motivation: families of curves

Recall what we started out seeking: given two curves, whatare the other curves passing through their intersectionpoints?

We began by looking at affine curves of the same degree:lines through the intersection of lines, conics through theintersection of conics, and so on. We then generalizedsomewhat to looking at all affine curves (irrespective ofdegree) passing through the intersection of the curves.Hilbert’s nullstellensatz came to our rescue there.What if we want to look at more general curves?



Vipul Naik






Recall what we started out seeking: given two curves, whatare the other curves passing through their intersectionpoints?We began by looking at affine curves of the same degree:lines through the intersection of lines, conics through theintersection of conics, and so on. We then generalizedsomewhat to looking at all affine curves (irrespective ofdegree) passing through the intersection of the curves.

Hilbert’s nullstellensatz came to our rescue there.What if we want to look at more general curves?



Vipul Naik






Recall what we started out seeking: given two curves, whatare the other curves passing through their intersectionpoints?We began by looking at affine curves of the same degree:lines through the intersection of lines, conics through theintersection of conics, and so on. We then generalizedsomewhat to looking at all affine curves (irrespective ofdegree) passing through the intersection of the curves.Hilbert’s nullstellensatz came to our rescue there.

What if we want to look at more general curves?



Vipul Naik






Recall what we started out seeking: given two curves, whatare the other curves passing through their intersectionpoints?We began by looking at affine curves of the same degree:lines through the intersection of lines, conics through theintersection of conics, and so on. We then generalizedsomewhat to looking at all affine curves (irrespective ofdegree) passing through the intersection of the curves.Hilbert’s nullstellensatz came to our rescue there.What if we want to look at more general curves?



Vipul Naik





A general ring of functions

The general situation:

I A subring R of the ring of functions R2 → R.

I Two functions f and g in R.

I The vanishing loci of f and g are respectively Γf andΓg .

We seek: the collection of all functions h in R such that Γh

passes through the intersection of Γf and Γg .R could be: the ring of continuous functions, the ring ofinfinitely differentiable functions, th ring of analyticfunctions, and so on.



Vipul Naik














Vipul Naik











passes through the intersection of Γf and Γg .

R could be: the ring of continuous functions, the ring ofinfinitely differentiable functions, th ring of analyticfunctions, and so on.



Vipul Naik














Vipul Naik





Think about these

I Can we set up the Galois correspondence?

I Does every maximal ideal arise from a point?

I Is every radical ideal an intersection of maximal ideals?

I Can we change from R to a sufficiently “closed” field inwhich the above things hold?

Think about these!

Documents

Coaxial circles, triangle centers, and the nullstellensatzvipul/studenttalks/coaxial... · Coaxial circles, triangle centers, and the nullstellensatz Vipul Naik Families with common