advanced29 min studyLesson 551 in Mathematics

Affine varieties and vanishing ideals

Translate polynomial solution sets into ideals while distinguishing equations, point sets, and their intrinsic polynomial functions.

On this page

What you’ll understand

  • Construct a common zero locus and its full vanishing ideal.
  • Explain why equations reverse inclusions and why vanishing ideals are radical.
  • Compute a parabola's ideal by division and compare presentations with the same points.

An affine algebraic set records simultaneous polynomial constraints, but the written equations are only a presentation. Adding a redundant equation, changing generators, or squaring an equation may leave the points unchanged. Geometry therefore needs a way to recover every polynomial consequence that is visible on the solution set. Vanishing ideals provide this bridge and explain why the algebra attached to a point set naturally forgets nilpotent information.

Fix the field and the meaning of variety

Throughout this module k is algebraically closed unless an exception is stated; concrete worked examples use characteristic zero. Affine space Akn\mathbb A_k^n is the set of n-tuples with its polynomial functions. Given an ideal J in the polynomial ring, its zero locus consists of the tuples annihilating every element of J. It suffices to test generators because evaluation respects addition and multiplication. Hilbert's basis theorem makes a finite list sufficient even if the initial family of equations is infinite.

We call any such zero locus an affine algebraic set and reserve affine variety for a nonempty irreducible algebraic set. Some references use variety without irreducibility, so the convention matters when fraction fields appear later. Closed subsets in the Zariski topology are precisely algebraic sets. They are closed under arbitrary intersections and finite unions, but usually not arbitrary unions. A real drawing below displays only a real slice of the algebraically closed geometry, rather than replacing the ground field silently.

V(J)={a∈kn:f(a)=0 for all f∈J},I(S)={f∈k[x1,…,xn]:f∣S=0}V(J)=\{a\in k^n:f(a)=0\ \text{for all }f\in J\},\qquad I(S)=\{f\in k[x_1,\ldots,x_n]:f|_S=0\}
V turns equations into points; I collects every polynomial vanishing on the specified set of points.↔ Scroll wide equations horizontally.

Inclusions reverse, and powers disappear

The set I(S) is an ideal: sums still vanish, and multiplying a vanishing polynomial by an arbitrary polynomial preserves vanishing. A larger point set imposes more tests and therefore gives a smaller vanishing ideal. Conversely, a larger equation ideal cuts out fewer possible points. These two reversals are the basic logic behind the dictionary. They prevent an easy mistake: adding equations cannot enlarge a solution set, although adding generators can make its presentation look more complicated.

Every vanishing ideal is radical. If a power of f vanishes at every point, its scalar value has zero power in a field, so that value itself is zero. Consequently the equations f and f squared have the same points even though their generated ideals need not agree. The quotient by a point set's full ideal is reduced, meaning it has no nonzero nilpotents. An arbitrary equation quotient can retain a nonzero class whose power is zero; that is additional scheme information, not an extra ordinary point.

J⊆I(V(J)),S⊆V(I(S)),fr∈I(S)⟹f∈I(S)J\subseteq I(V(J)),\qquad S\subseteq V(I(S)),\qquad f^r\in I(S)\Longrightarrow f\in I(S)
The inclusions hold over any field; equality with the radical of J will require the Nullstellensatz hypotheses.↔ Scroll wide equations horizontally.

Recovering geometry rather than a preferred presentation

For a closed algebraic set X, applying I and then V recovers X: its original defining equations already belong to I(X), and all of I(X) vanish on X. For an arbitrary subset S, the same operation produces its Zariski closure. An infinite subset of the affine line is dense because a nonzero univariate polynomial has only finitely many roots. Thus closure is determined by polynomial tests and can be much larger than an intuitive picture based on isolated sampled points.

Useful set identities become ideal operations. A common zero of the sum of two ideals lies in both zero loci; a zero of their product lies in their union. To prove the latter, a point outside both loci admits one nonvanishing polynomial from each ideal, whose product also does not vanish. These observations give unions of components without introducing inequalities. They also show why deleting a point generally produces an open or locally closed set rather than another closed algebraic set in the same ambient space.

Every equation of a parabola

Over an algebraically closed field of characteristic zero, let X=V(y−x2)⊂A2X=V(y-x^2)\subset\mathbb A^2. Prove that its full vanishing ideal is exactly (y−x²), rather than merely an ideal containing this generator.

  1. Divide any polynomial f(x,y) by the monic polynomial y−x², treating y as the polynomial variable over k[x]. The remainder is a polynomial r(x) independent of y, so f=q(y−x²)+r(x).

  2. Every point of X has the form (t,t²). If f vanishes there, substitution makes r(t)=0 for every t in k. The field is infinite, so a univariate polynomial with all these roots is the zero polynomial.

  3. Therefore every vanishing polynomial is a multiple of y−x². The reverse inclusion follows by evaluation. Replacing the generator by its square preserves the points but changes the equation ideal and introduces a nilpotent class in the quotient.

The result

I(X)=(y−x²). The equation presentation using (y−x²)² has the same ordinary points but a different, nonreduced quotient ring.

Explore the model

Interactive lab

Interactive lab · function

Points constrained by one polynomial

Pause and Reset, then set coefficient a = 1 and position x = 1. Predict the displayed point (1,1) satisfies y−x²=0 and its tangent slope is two. Reset and set both a = 1 and x = −1: the point is (−1,1), with slope minus two. The quadratic instrument draws a real slice of this one zero locus; it cannot distinguish a defining ideal from its square or enumerate complex points.

Points constrained by one polynomialThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA graph can check pointwise equations, while the ideal records algebraic information that a picture alone cannot recover.

Check your understanding

Which statement holds for the vanishing ideal I(S) of any subset of affine space over a field?

Try it yourself

Over characteristic zero, find the vanishing ideal of the two points (1,0) and (−1,0). Prove the claim using a remainder, rather than only checking proposed generators.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

First discard multiples of y, then divide the remaining polynomial in x by x²−1. Evaluate its linear remainder at both points.

Compare with the solution

The proposed ideal is (y,x2−1)(y,x^2-1). Reduce a vanishing f modulo y and then modulo x²−1, leaving ax+b. Its values at x=1 and x=−1 give a+b=0 and −a+b=0. Since two is invertible, a=b=0. Hence every vanishing f belongs to the proposed ideal. Conversely both generators vanish at both points, so all their polynomial combinations do too. The ideal is also (x−1,y)∩(x+1,y)(x-1,y)\cap(x+1,y); its quotient is the product of two copies of k, which records two reduced points rather than a doubled point.

Take it with you

  • The ground field and the convention for irreducible varieties must be explicit.
  • Vanishing ideals are radical and reverse inclusions of point sets.
  • Different equation ideals can have the same points while retaining different nilpotent information.

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