Projective Varieties, Radical/Prime Ideals and Closed Sets

Homogeneous Radical Ideals and Closed Regions

A homogeneous ideal is just another ideal, so apply nullstellensatz, and rad(H) becomes the closure of H. Remember that rad(H) is another homogeneous ideal, so there is no trouble here.

In normal space, radical ideals and closed regions correspond one for one. We would like to make the same statement about homogeneous radical ideals and closed regions in projective space, and it's almost true. The only trouble occurs when two different homogeneous radical ideals, say H and J, lead to different closed regions in normal space (as they must), which become the same closed region in projective space. Let's see if this ever happens.

Assume R, in projective space, is nonempty, and H′ = J′ = R. Suppose p is a homogeneous polynomial in H, and not in J. Push J up to its radical ideal, which brings in every polynomial that vanishes on R, including p. Yet J is already a radical ideal, hence H contains J. By symmetry, J contains H, and H = J.

Now assume the region in projective space is empty. Suppose b is a point, other than the origin, that satisfies the homogeneous polynomials of H. The line determined by b satisfies H, and this is a contradiction. The only point that might vanish on H or J is the origin. So H and J could be different, if H′ is the origin (in normal space) and J′ is empty. The former is attained when H is the maximal ideal spanned by the indeterminants, with quotient field K. The latter is realized when J is the entire ring of polynomials.

To maintain a perfect correspondence, the radical ideal that is spanned by x₁ through x_n, denoted S⁺, is excluded from consideration. Without S⁺, homogeneous radical ideals and closed regions in projective space correspond one for one.

Remember that every proper homogeneous ideal is contained in S⁺. A chain of homogeneous prime ideals can always be augmented by 0 at the bottom, and S⁺ at the top.

Topology

Normal space, where algebraic sets are closed, forms a noetherian space, and each closed set is uniquely a finite union of irreducible components. Projective closed sets are algebraic closed sets, so we're really talking about a subspace of the aforementioned space. It too is noetherian, and the same properties apply. A closed region R in projective space becomes a union of algebraic varieties, which are defined by prime ideals, that intersect in the radical ideal, that defines R. The primes that intersect in a homogeneous ideal are homogeneous, so the algebraic components are indeed projective components. These irreducible components are called projective varieties, and they correspond to homogeneous prime ideals. If S⁺ is set aside, the correspondence is one for one.

As expected, a quasi variety is an open subset of a variety, according to the zariski topology of projective space.

A Topology for the Homogeneous Primes

Any graded ring G, such as our ring of polynomials, can be given a topology based on its homogeneous prime ideals. This is known as proj G, and it is a subspace of spec G. The projective topology omits G⁺. Thus the points in proj G correspond to prime ideals that do not contain G⁺, which correspond (in our world) to projective varieties. Furthermore, closed sets in proj G correspond to closed regions in projective space. The former is an intersection of homogeneous prime ideals, while the latter is a union of projective varieties.

Maximal Homogeneous Ideals

Since S⁺ is the largest homogeneous ideal, containing all other proper homogeneous ideals, Let M be a maximal homogeneous ideal inside S⁺. Since the ring of polynomials is noetherian, we can always raise an ideal H up to such a maximal ideal. If H′ = 0, we might wind up with S⁺, as described above. Let's avoid those ideals, and look for a maximal homogeneous ideal M such that M′ yields a nontrivial region R in projective space.

Let the indeterminants x₀ through x_n build n dimensional projective space. Let a maximal ideal M vanish on a region R. Let R contain the point b with a nonzero coordinate x₀ = b₀. Let the n polynomials b_i/b₀*x₀-x_i generate the ideal W. These generators vanish on b, just like M. If W is a radical ideal, then W includes M, W = M, and we're done.

The generators represent n linearly independent equations in n+1 unknowns, hence the solution space is a line. We have already found this line; it is the point b in R. Thus W′ = b.

If one of the generators is 7x₀-x₃, map this over to the indeterminant y₃. We don't need x₃ in the destination ring; y₃ will suffice. This is a ring isomorphism. Do this for all n generators and our ring is equivalent to K adjoin x₀ and y₁ through y_n. The image of W is generated by y₁ through y_n, and the quotient is the integral domain K[x₀]. Therefore W is prime. This makes W a radical ideal, and that completes the proof. Any additional functions, beyond W, would eliminate the point b. The maximal ideals inside S⁺, that correspond to individual points in projective space, have been characterized.

Zero Dimensional

Consider the polynomials in one indeterminant x over a field K. Let P be prime and let g be a homogeneous generator of P. Thus g = cxⁿ. If n > 1, write g as x times cx^n-1. One or the other is in P. If not x then apply the same argument to cx^n-1. Finally P contains x and P = S⁺ - the only homogeneous nonzero prime ideal. Of course S contains other prime ideals, such as that generated by x+1, but these are not homogeneous prime ideals. Well this is zero dimensional in projective space, so is not very interesting.

One Dimensional

Let S be the polynomials in x and y. Let P be prime and let g be a homogeneous generator of P having lowest degree d. If g does not contain a term y^d then g is x times something smaller. Thus g includes x^d and y^d.

Let g = x²+y² in a field where -1 is not a square, such as the reals or Z₇. Try to factor g into two linear terms, which we can make monic, x+a times x+b. You'll see that a = -b, and a² = -1, which doesn't work. Thus g generates a homogeneous prime ideal. g′ is empty (since 0 is not part of projective space), but that's because K is not closed. In C the closure of K we have two points x = ay and x = -ay. Over C, g would no longer be prime. Write it as the product of linear polynomials and each corresponds to one of the two projective points mentioned above.

In general, let C be the closure of K. Let g be homogeneous with both x^d and y^d. Use synthetic division to divide x-vy into g. The remainder is y^d times a polynomial in v whose coefficients are determined by the coefficients of g. Since C is closed there is a v that makes the remainder 0. In other words, g factors, and it is not prime after all, unless it is linear. This defines a point in projective space. If there are two such generators, such as x+ay and x+by, they combine to span x and y, which is S⁺, which is maximal and prime. So we understand all the prime ideals over C[x,y]. Their images are all the points of projective space, and the empty space and the entire space.

Remember that the ring and the topology are noetherian. each closed set is the finite union of irreducible components, and each such component is a point, hence the zariski topology is the cofinite topology.

Return to the field K, and take images in projective space L, where K is a subfield of L is a subfield of C. Let g generate a prime ideal P, but g factors into a product of linear polynomials over C. These define at most d points in projective space over C, and perhaps a subset of these over L. Each closed set is finite, or it is the whole space. If K = L then each point is a closed component and we again have the cofinite topology. We saw this in one dimensional normal space as well.