Nullstellensatz

Algebraic Geometry, Nullstellensatz

German

The word nullstellensatz is German for a "zero sentence", where "sentence", in a mathematical context, means theorem.  Thus a theorem about zeros.  Which polynomials become zero across various regions of n space?  A precise English translation is dificult, so I just leave it as nullstellensatz.

Maximal Ideals in the Polynomial Ring

Let C be an algebraically closed field.  Adjoin finitely many indeterminants x1 through xn to C to build the ring R.  Let M be a maximal ideal in R.  We know that R/M becomes algebraic over C.  Thus a ring homomorphism f maps R into C, taking M to 0.  Actually f maps R onto C, since it maps C onto C - and the indeterminants all land somewhere in C.

Let z1 through zn be the images of x1 through xn in C.  Note that xi-zi is a perfectly good polynomial, and an element of R.  Apply f, and get f(xi)-f(zi), or 0.  Thus the ideal H generated by the n elements xi-zi is contained in M.  Mod out by H and each generator turns into an element of C.  The quotient ring is equal to C, H is maximal, and H = M.  All maximal ideals have been characterized.

Weak Nullstellensatz

Let C be the algebraic closure of a field K.  Let H be any proper ideal in K[x1…xn].  Thus H is an ideal of polynomials taken from a polynomial ring, and we are interested in the algebraic set H′ in Cn.

Since C is integral over K, the ring C[x1…xn] is integral over K[x1…xn].  Push H up to a maximal ideal in the smaller ring, then raise this up to a maximal ideal in the larger ring.  (There are always primes over primes in an integral extension.)  Thus we can embed H in a maximal ideal M in C[x1…xn].  This maximal ideal is generated by xi-zi, so that M vanishes on the point z1,z2,z3…zn in Cn.  Since H vanishes on the same point, the algebraic set H′ is nonempty.

Note that the image space has to be Cn; the point z1,z2,…zn may not lie in Kn.

The Entire Space

As shown above, every proper ideal vanishes somewhere, which means you need 1, or the complete ring of polynomials, to produce the empty space.  But which polynomials vanish on the entire space?  If p is such a polynomial, then p+1 vanishes nowhere, which means p+1 generates 1.  Write (p+1)*q = 1, and both factors must be units.  Thus p is a unit, and p = 0.  In other words, 0′ = ∅, and ∅′ = 0.

Again, C must be closed, as shown by x7-x mod 7.

Strong Nullstellensatz

As before, let C be the algebraic closure of K, and consider polynomials with coefficients in K, evaluated on Cn.  We need this, because we're going to invoke weak nullstellensatz.

Let H be a proper ideal in the polynomial ring, hence H′ is a nonempty region in Cn.  What about H′′?  It certainly includes H.  In fact, H′′ = rad(H).

Suppose pn evaluates to 0 for every z in H′.  Rewrite pn(x) as p(x)n.  Now p(z) raised to the nth is 0 in a field, hence p(z) = 0, and p is in H′′.

Conversely, let p lie in H′′, and assume p is nonzero.  Adjoin y to produce the larger ring K[x1,x2,…xn,y].  Let H and y*p-1 generate the ideal J.  What does J′ look like?

Suppose z1,z2,…zn,y is in J′.  Since z1 through zn satisfy p, the polynomial y*p-1 can never equal 0.  Therefore J′ is the empty set.  Apply weak nullstellensatz, and J is the entire ring.  In particular, J includes 1.  If each coefficient ci is a polynomial in the extended ring, write the following.

1 = c0*(yp-1) + c1h1 + c2h2 + c3h3 + … cnhn

The polynomials h1 h2 etc come from the set H, which is not necessarily an ideal in the extended ring, but it contributes to J.

Let a ring homomorphism map the extended ring into the fraction field K(x1,x2,…xn).  Map xi to xi/1, and map y to 1/p.  (This is why p has to be nonzero.)  Apply this to our representation of 1.  The first term on the right hand side drops to 0, since y*p-1 becomes 0.

Look at the image of a coefficient ci.  After obtaining a common denominator, ci becomes a polynomial in x1 through xn, divided by a power of p.  Multiply through by a sufficiently high power of p, and pn is generated by polynomials drawn from H.  Thus p lies in rad(H), and H′′ = rad(H).

Zariski Topology

Under the zariski topology, a closed region in Cn is an algebraic set.  Now we know the ideal that vanishes on this region is a radical ideal.  Conversely, each radical ideal implies a closed region in Cn.  If H and J are different radical ideals, H′ and J′ must be different, else H′′ and J′′ would be the same, and these are H and J.  Therefore radical ideals and closed regions correspond one for one.

In any ring, the zariski topology equates every radical ideal with a closed set.  In this sense, the two zariski topologies are the same.

Depends on K

If K is replaced with some other field, whose algebraic closure is still C, the ideals in the polynomial ring could change, along with the closed regions in Cn.  Let's illustrate with n = 2, K = Q adjoin the square root of 2, and C is the closure of the rationals.  Let the polynomial p be y = sqrt(2)x, giving a line in the plane.  Then try to realize this line with rational polynomials.  To vanish on the origin, there can be no constant terms.  If p(x,y) vanishes on this line, turn x into t and y into sqrt(2)t and substitute, giving a polynomial in t.  This must be 0 for all t.  Since C is infinite, and p has finitely many roots, p must be 0.  The linear terms cancel, and the quadratic terms, and so on.  Concentrate on a subpolynomial - the terms of degree d.  Let r represent the square root of 2, divide through by td, and find a polynomial in r that must equal 0.  Now p(r) must have a root of sqrt(2).  Yet this root admits its conjugate -sqrt(2).  Both lines must be present in the plane, intersecting in the origin.  This is illustrated by the polynomial y2 = 2x2.  We can't have one line without the other.  So the closed sets depend on K.  Additional closed sets may arise, i.e. a stronger topology, when K includes more algebraic elements.

Union and Intersection

Although one may add ideals to produce the intersection of two algebraic sets, other polynomials may vanish on the resulting closed set.  We need to take the radical.

for example, let x2-yz generate a quadratic surface in 3 space.  This polynomial is irreducible in the ufd K[x,y,z], hence it is prime.  It generates a prime ideal, which is a radical ideal.  In the same way, the polynomial y generates a radical ideal that corresponds to a plane in 3 space.  The intersection is a line, with x and y set to 0.  This is demonstrated by an ideal H, generated by y and x2-yz.  This ideal contains x2, but not x, hence it is not a radical ideal.

Union is a simpler operation, at least in theory.  The ideal that vanishes on the union of two algebraic sets is the radical of the product of the two ideals.  This is equal to the radical of the intersection.  This in turn is equal to the intersection of the radicals.  Each ideal is already a radical, hence the union of two algebraic sets is realized by taking the intersection of the corresponding radical ideals.