Graded Rings, Proj R

proj R

Like spec R, proj R converts a graded ring into a topological space. You should be familiar with spec R before you proceed.

Remember that spec R is the prime "spectrum" of a ring. The word "proj" is used here, because this spectrum is used to define certain "projective" spaces in algebraic geometry.

Let R be a graded ring, and let R⁺ be the ideal generated by all elements of degree > 0. The points of proj R are the homogeneous prime ideals of R that do not contain R⁺. Unlike spec R, proj R could be empty - this will be addressed later.

For any homogeneous ideal H in R, v_H is the set of primes in proj R that contain H, and v_H is closed by definition. Since 0 and R⁺ are homogeneous ideals, proj R is open and closed, as it should be.

The arbitrary intersection of closed sets consists of primes that contain every homogeneous ideal H at the core of each closed set v_H. The ideal spanned by an arbitrary union of homogeneous ideals is also homogeneous, hence the intersection of closed sets is closed.

Let H₁ and H₂ define two closed sets. If a prime P contains their product it contains one or the other, and if P contains either H₁ or H₂ it contains their product. The product H₁*H₂ is also homogeneous, hence the finite union of closed sets is closed. We have a valid topology for proj R.

Recall that rad(H) is homogeneous, hence one can raise H up to its radical without changing the closed set v_H. The same thing occurred in spec R.

If H is homogeneous, it can be generated by homogeneous elements. A prime P contains H iff it contains all the homogeneous generators. Thus v_H is the intersection over v_f, where f is a homogeneous generator of H. Turn this around, and o_H is the union over o_f. Let o_f, where f is any homogeneous element, be a base set, and every open set is the union of base sets.

The union of v_e and v_f is v_ef, as described above. Thus the intersection of o_e and o_f is o_ef, another base set, and we have a valid base for the topology.

Subspace

Proj R is a subspace of spec R. Let a collection of homogeneous primes be closed in spec R, hence each prime contains a common ideal H. Raise H up to the intersection of the aforementioned primes. Now H becomes homogeneous, and our set becomes closed in proj R, which is what we would expect from the subspace topology.

Nil Radical

The nil radical is a radical ideal, and is homogeneous. Mod out by the nil radical and find another graded ring R′. Prime ideals correspond, and homogeneous ideals correspond, hence homogeneous primes correspond. Closed sets also correspond. Therefore proj R = proj R′. The same thing occurred in spec R.

As a sanity check, make sure a prime doesn't contain R⁺ after the ring has been reduced. Given P, select x outside of P, with positive degree. Mod out by the nil radical, which lives inside P, and x persists, as an element outside of P, having positive degree in the reduced ring. Thus no primes are lost when R is reduced.

When proj R is Empty

If R⁺ is a nil ideal, then it lies in the nil radical. Every prime contains R⁺, and there are no primes in proj R.

Conversely, assume R⁺ contains x, where x is not nilpotent. The powers of x form a multiplicatively closed set, and some prime P misses the powers of x. Ratchet P down to the ideal generated by the homogeneous elements inside P. This is homogeneous, and prime, and it does not contain R⁺. Therefore proj R is nonempty.

In summary, proj R is empty iff R⁺ is a nil ideal.

Compact

Review the proof that spec R is compact. If R is a reduced graded ring, this proof can be tweaked to show that proj R is compact when R⁺ is finitely generated. The ideal referred to as E, in that proof, must contain every x in R⁺, else E could be pushed up to a prime ideal missing x. Select finitely many elements of E sufficient to generate each of finitely many generators of R⁺. This establishes finitely many base sets, and a finite subcover, whence proj R is compact.

If R is noetherian than the reduced ring is noetherian, R⁺ is finitely generated, and proj R is compact.

In contrast, let R be the direct product of infinitely many copies of K[x], graded in the usual way. Build open sets about the instances of x. This is a cover with no finite subcover.

Localization

Unless otherwise stated, the fractions of a graded ring or module have homogeneous elements in the denominator. These are closed under multiplication, so there is no trouble. Thus, R_P consists of fractions whose denominators are homogeneous elements outside of P. Similarly, R/f, for a homogeneous element f, has powers of f in the denominator, assuming f is not nilpotent.

Then, there is an additional restriction. Consider fractions where the numerator and denominator are homogeneous of the same degree. Verify that this is a subring of R/S, closed under addition and multiplication, and containing 1/1.

Let R′ be the subring of R/S, as described above. We would like to compare spec R′ with proj R.

If P intersects S then P/S includes some w/w, and becomes all of R/S. This brings in all of R′. Therefore, primes that touch S are left on the cuttingroom floor.

Let P be a homogeneous prime that misses S. It corresponds to a unique prime in R/S. Contract this into R′ and find a prime ideal, a point in spec R.

Remember that P is generated by homogeneous elements, and these turn into fractions having homogeneous numerators and denominators in S. Restrict these generators to R′.

At this point I'm going to make another assumption, which is often true. For every degree d, S includes an element whose degree is a multiple of d. If S contains f, with nonzero degree, then some power of f satisfies this requirement for every d. If x is a homogeneous generator of P, place an appropriate power of x in the numerator, and an element w ∈ S in the denominator, so that both have the same degree. This fraction survives in R′, and it allows us to reverse the process. Push this fraction back into R/S, and extend this into an ideal. Multiply by w and get a power of x. Any prime ideal lying over this ideal must include x. All the generators of P come back, and P reappears. The map from proj R into spec R′ is injective.

Is it onto? Start with a prime P′ in spec R′. Extend this into homogeneous generators over arbitrary denominators in R/S. Take the radical, and suppose the ideal spanned by the numerators is not prime in R. Remember that a homogeneous ideal fails to be prime if ab lies in P, while a and b do not, for a and b homogeneous. So ab to a certain power becomes one of the homogeneous elements in P. Pull this down into R/S, and then, with suitable denominators, into R′. A power of ab, with its attendant denominator, lies in P′, so a over its denominator, or b over its denominator, lies in P′. This forces a or b to lie in P, and P is prime after all. The map is surjective.

A closed base set v_f in proj R (where f is homogeneous) corresponds to a closed base set in spec R′, once f has been raised to a certain power and assigned an appropriate denominator. Base sets correspond, and the spaces have the same topology. Therefore proj R, restricted to the primes that miss S, is homeomorphic to spec R′, where denominators are drawn from S.

As a corollary, set S to the powers of f, and o_f in proj R is the same as spec R′.