Throughout this chapter, rings are commutative.

A graded ring R is the direct sum of additive subgroups Rd, as d runs from 0 to infinity. Each d extracts a slice of R. An element in Rd is called homogeneous of degree d. Thus R is the direct sum of homogeneous elements. A typical entry might be x+y+z, where x is in R3, y is in R5, and z is in R9.

Each Rd is closed under addition, and the product of two homogeneous elements adds their degree, or yields 0. The product rule tells us 1 lives in R0, along with all the integers up to the characteristic of R. In fact R0 is a subring of R. From here, each Rd is an R0 module.

At first this looks like a valuation ring, (which will be addressed in a later chapter), but there is a difference. Yes, the valuation of xy is the valuation of x plus the valuation of y, but if x and y have the same valuation, the valuation of x+y could be larger. Not so with homogeneous elements in a graded ring. x and y in Rd means x + y is in Rd.

The typical example is the ring of polynomials in one or more indeterminants over some other ring K. A polynomial lives in Rd if every term has degree d. Thus x2+xy+yz lives in R2. Verify that every polynomial is a unique direct sum of these homogeneous polynomials, and the product of two homogeneous polynomials adds their degrees. The subring R0 is equal to K, the polynomials of degree 0.

A graded ring can be given a topology. Let the nth ideal Hn be the direct sum of Rd as d runs from n to infinity. Note that H0 = R. The next ideal H1 is sometimes denoted R+. H1n lies in Hn. The converse need not hold. Add, to the polynomials in x, the polynomials in y starting with y7. One cannot get y7 from H17. However, in the standard polynomial ring, and most graded rings, H1n = Hn. Verify that each term of degree n is the product of n terms of degree 1, and you're home free.

Let c be a real number strictly between 0 and 1, and let the distance between x and y be cn, for the greatest n such that Hn contains x-y. This is determined by the homogeneous component of x-y with lowest degree.

Since 0 is in every Rd, let 0 be homogeneous with infinite degree. Hence its norm is c, or 0. The distance from x to y is 0 iff x = y.

Consider the triangular inequality on three points x y and z. Write z-x as z-y + y-x. The valuation of the sum, on the left, is at least as large as the lesser valuation on the right. The distance on the left is bounded by the larger distance on the right. The distance metric is valid, and R becomes a metric space, with the open ball topology.

Given a cauchy sequence s, the component from Rd settles down to a constant value. This holds for every d, hence s defines a canonical series of elements from Rd for all d. The partial sums of this infinite series defines another cauchy sequence t. Verify that s-t converges to 0. In other words, t is the same as s. Treat the completion of R as the set of formal power series in a dummy variable w, where the coefficient on wd comes from Rd.

To be backward compatible with R, two power series in w are added in the usual way (term by term), and multiplied in the usual way. The result is a continuous ring. In other words, addition and multiplication are continuous maps from R′ cross R′ into R′, where R′ is the completion of R. Let v be a point in R′, and place an open ball around v of radius ε. Let x+y = v, hence x,y is a point in the preimage of v. Let a and b be bounded by ½ε. This is an open ball cross an open ball, which is open in the product topology. x+a + y+b = v + a+b, and is within ε of v. Multiplication is essentially the same. Bound a and b by ¼ε, and ay and bx and ab are bounded by ¼ε. Thus (x+a) * (y+b) is within ε of v. Operations are continuous, and R′ is a continuous ring.

When R is the polynomials over K, its completion is the formal power series, in the same indeterminants, with coefficients in K.

A graded R module M is based on a graded ring R, as described above. Write M as the direct sum over Md. Once again, each Md is closed under addition. Furthermore, the degree of cx is the degree of c (in R) plus the degree of x (in M), or cx = 0.

Let Hn be the submodule that is the direct sum of Md, as d runs from n to infinity. Build a metric on M, just as we did on R. This turns M into a metric space. The completion of M becomes formal power series in w, with coefficients in Md. These series are not multiplied together, but they can be multiplied by a polynomial that represents an element of R. When a coefficient in Rd is multiplied by a coefficient in Md, invoke the action of R on M. In other words, the action is determined by the action of each homogeneous element of R on each homogeneous element of M.

M′ is a continuous R′ module, wherein addition and scaling are both continuous operators.

A graded ring R is a graded R module; simply let R act on itself.

A homomorphism between graded modules takes homogeneous elements of degree d (outside the kernel) to homogeneous elements of degree d + d0, for some offset d0 that is specific to the homomorphism. The graded R modules form a subcategory of R modules.

Any ring could be graded, simply by dumping the entire ring into R0. This satisfies the definition, but we don't usually consider this a graded ring. We need some homogeneous c in R with degree > 0. Similarly, a graded module has some x with degree > 0.

The descending chain of ideals in R, or submodules in M, cannot stabilize at anything other than 0. Multiply by R1 again and again, and push the ideal or submodule down forever - unless it becomes 0. Thus R, or M, is artinian iff its nth ideal or module is 0.

The graded ring R has zero divisors iff it has homogeneous zero divisors. If xy = 0, the highest degree term of x times the highest degree term of y has to equal 0.

If R is an integral domain, let x have nonzero degree. The powers of x are nonzero, the chain of descending ideals cannot stabilize at 0, and R is not artinian.

The product of homogeneous elements is homogeneous by definition. The converse holds in an integral domain. If xy is homogeneous, look at the lowest and highest degree terms of x times y, and conclude x and y are homogeneous.

Let R be a noetherian graded ring. Since R0 is the quotient of R by R+, R0 is also noetherian.

Find a finite set of generators for the ideal R+, and separate these into homogeneous elements, which also live in R+. Thus R+ is generated by finitely many homogeneous elements g1 through gn.

Let S be the subring of R generated, as a ring, by R0 and g1 through gn. This includes R0. Proceed by induction on d.

For y ∈ Rd, y is in R+, and is a linear combination of g1 through gn using coefficients from R. Split coefficients into homogeneous components, and y is the sum over ci,j*gi. Ignore those terms in the sum whose degree is not equal to d. They all cancel to 0 anyways. Now each ci,j has degree < d, and is present in S by induction, hence y is included in S. Therefore S = R, and R is a finitely generated R0 algebra.

Conversely, assume R0 is noetherian, and R is a finitely generated R0 algebra. Now R is the quotient of a polynomial ring, R0 adjoin finitely many indeterminants, one indeterminant for each generator. The latter is noetherian by hilbert's basis theorem, and after taking the quotient ring, R is noetherian. A graded ring R is noetherian iff R0 is noetherian and R is a finitely generated R0 algebra.

Let R be a graded ring. A homogeneous ideal is an ideal of R that can be generated by homogeneous elements.

An ideal H is homogeneous iff every y in H is a sum of homogeneous elements that also lie in H. The backward direction is easy; H is generated by its homogeneous elements, and is homogeneous. Conversely, write an element y in H as the sum of aigi, where gi are homogeneous generators of H. Separate each ai into homogeneous components and distribute gi across these components. Now each ai,j*gi is homogeneous, and lies in H. Group these together by degree to find the homogeneous components of y.

We are really slicing H up into additive subgroups Hd, whence H becomes the direct sum over Hd.

A homogeneous submodule U is generated by homogeneous elements, and y is in U iff the homogeneous components of y are in U. The proof is the same.

Let H be homogeneous in K[x], where K is a field. Since the ring is a pid, let f generate H. Everything in H is f times a polynomial. If f has degree d, none of the lower terms of f are part of H, so there can be no lower terms. Thus H is generated by xd for some d. Any other ideal, seeded by any other generator, is not homogeneous.

Let Q be a quotient ring of R, with a homogeneous kernel H, and Q will become a graded ring. To map an element in R to an element in Q, reduce each component from Rd mod Hd. This preserves degree, from R down to Q.

As a special case of the above, let the kernel begin with Rd and go up from there. The quotient is now truncated at degree d.

Homogeneous elements in R map to homogeneous elements in Q, and each homogeneous element in Q has at least one homogeneous pullback in R. Combine the homogeneous pullbacks with the homogeneous kernel H to build a homogeneous preimage. This sets up a correspondence - homogeneous ideals in Q correspond 1 for 1 with homogeneous ideals in R containing H.

Test whether the kernel H is prime by seeing if the quotient Q is an integral domain. If Q is an integral domain it will have homogeneous zero divisors. If homogeneous elements are 0 divisors in Q, their pullbacks prove H is not prime. Therefore a homogeneous ideal is prime iff x*y in H implies x or y is in H for homogeneous x and y.

Every prime ideal P contains a homogeneous ideal H, generated by all the homogeneous elements (if any) in P. If xy is in H, where x and y are homogeneous, then x or y is in P, and x or y is in H. That proves H is prime. And H is the largest homogeneous ideal contained in P. However, H could be 0. Let R be the pid K[x]. x+1 is a prime element, and generates a prime ideal. Multiply x+1 by any polynomial and the result is not homogeneous. The homogeneous prime ideal inside {x+1} is 0.

Homogeneous ideals are closed under addition, intersection, product, and radical. Take the union of generators to generate the sum. Multiply generators to generate the product.

If T and U are homogeneous ideals then let V = T intersect U. An element lies in V iff it is in T and U, iff its homogeneous components are in T and U, iff its homogeneous components are in V. Therefore the intersection is homogeneous. The same proof works for an arbitrary intersection of ideals, not just a finite intersection.

Let H be a homogeneous ideal. Let xn lie in H, with x′ the highest homogeneous component of x. Now x′n is the highest homogeneous part of xn, and x′n lies in H. Thus x′ lies in rad(H), and so does x-x′. Repeat until all the homogeneous components of x lie in rad(H). This makes rad(H) a homogeneous ideal.

Remember that rad(H) is the intersection of the primes containing H. Restrict each prime to the largest homogeneous ideal within that prime. This is also a prime ideal. Thus we have ratcheted each prime down to a (possibly smaller) prime, containing H, that happens to be homogeneous. The intersection of these smaller prime ideals lies in rad(H). If x is a homogeneous generator of rad(H), it lies in each prime ideal over H, and in each homogeneous prime ideal over H, and in their intersection. Thus rad(H) is the intersection of the homogeneous prime ideals containing H. The radical is the same whether you consider prime ideals or homogeneous prime ideals containing H, as long as H is homogeneous.

Let R0 be a field, and let M be the ideal generated by all homogeneous elements of positive degree. Remember that M is sometimes denoted R+. In this case M is the only maximal ideal relative to homogeneous ideals, and R looks a bit like a local ring. R has other maximal ideals, but they aren't homogeneous. If a homogeneous ideal contains x outside of M, it contains all the homogeneous components, including the component of degree 0. This is a unit in R, whence the ideal is all of R. A complete chain of homogeneous ideals starts with 0, and ends with M, and then R if you like.

If x is a homogeneous element of a graded module M, then H, the annihilator of x, is homogeneous. Let y be an element of H, hence yx = 0. Let y′ be the highest degree component of y. Now y′x = 0, whence y′ lies in H. This means y-y′ lies in H. Continue this process, and all the components of y lie in H. This holds for every y in H, hence H is a homogeneous ideal.

Not so for the annihilator of an arbitrary element. Let R be K adjoin four indeterminants w, x, y, and z, and mod out by the homogeneous ideal generated by w2y2, xz, and w2z + y2x. The quotient is still a graded ring. By construction, w2+x is in the annihilator of y2+z, but x is not, hence the annihilator is not homogeneous.

The annihilator of a homogeneous submodule U of M is also homogeneous. Take the intersection of the annihilator of x, for each homogeneous generator x in U, to build the annihilator of U. The intersection of homogeneous ideals is homogeneous.

The maximal annihilator of a submodule U is a prime ideal. Consider the set of ideals in R that act as annihilators for U. Since M is unitary, these annihilators are all proper ideals. The union of an ascending chain of annihilators is another annihilator. If there is anything at all that kills U, then it rises to a maximal annihilator H, which is a proper ideal of R.

Let H kill x in U. Suppose a*b lies in H, and let I be the annihilator of bx. If b is not in H then b*x is nonzero and I is proper. Since I contains H, I = H. I includes a, so H contains a. This makes H a prime ideal. If U is homogeneous then so is H.

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

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 are properly inside R+. Unlike spec R, proj R could be empty - this will be addressed later.

R+ is usually a homogeneous prime ideal, but sometimes there are additional primes beyond this point. Let R = Zx, polynomials with integer coefficients, having the usual gradation. R+ is generated by x, and since the quotient Z is an integral domain, this ideal is prime. Yet 2 and x generate a larger homogeneous prime ideal. Both these primes are excluded from proj R.

For any homogeneous ideal H in R, vH is the set of primes in proj R that contain H, and vH 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 those primes that contain every homogeneous ideal at the core of each closed set . These are the primes that contain the union of the underlying ideals. The union of homogeneous ideals is homogeneous, so there is no trouble here. The arbitrary 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 vH. 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 vH is the intersection over vf, where f is a homogeneous generator of H. Turn this around, and oH is the union over of. Let of, where f is any homogeneous element, be a base set, and every open set is the union of base sets.

The union of ve and vf is vef, as described above. Thus the intersection of oe and of is oef, another base set, and we have a valid base for the topology.

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.

The nil radical is the radical of 0, and is homogeneous. Mod out by the nil radical and find another graded ring. Prime ideals correspond, and homogeneous ideals correspond, hence homogeneous primes correspond. Closed sets also correspond. Therefore proj R = proj R/nil(R). The same thing occurred in spec R.

As a sanity check, make sure a prime doesn't become 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.

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. An example is the integers mod 49, where the multiples of 7 have degree 1.

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+, since it does not contain x. Therefore proj R is nonempty. In summary, proj R is empty iff R+ is a nil ideal.

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. (R is reduced, so x is not nilpotent.) 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/nil(R) is noetherian, then R+ is finitely generated, and proj R is compact.

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, RP 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.

Let S be a multiplicatively closed set containing 1, and something with degree > 0. 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 U be this subring, and build a map from proj R, or a subspace thereof, into spec U.

There is already a map from the primes of R missing S onto the primes of R/S. Let P be a homogeneous prime that misses S. Extend P into R/S, then contract this back to U, and find a prime ideal Q, a point in spec R. How is this accomplished? P is generated by homogeneous elements, and these turn into fractions having homogeneous numerators and homogeneous denominators in S. For every degree d, S includes an element whose degree is a multiple of d. Remember that S contains some f with nonzero degree, so some power of f satisfies this requirement for any 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 U, all the generators of P map into U, and P maps into U. Furthermore, this process can be reversed. Pull these fractions back to R/S, and extend this into an ideal. Multiply by w and get a power of x. Any prime ideal containing this ideal must include x. All the generators of P come back, and P reappears. If a larger prime maps to Q then an additional homogeneous generator y has yn as a numerator of Q, and pulls back to y, whence y is in P after all. The map from proj R into spec U is injective.

Is it onto? Start with a prime Q in spec U, and try to find a prime P in proj R lying over Q. Q consists entirely of fractions of homogeneous elements, where the degree of the numerator equals the degree of the denominator. Let the numerators generate a homogeneous ideal H in R. H is the least ideal that maps to Q. Suppose a linear combination of generators of H equals w, for some w in S. Separate the coefficients into homogeneous pieces. Now the sum of ci,jgi = w. We've seen this before - the terms of the sum that have a degree other than the degree of w must sum to 0. You can leave them out. So we are left with various terms ci,jgi whose degree is the same as that of w. Embed this in R/S and divide through by w, so that the sum is equal to 1. Each gi has some vi in s such that gi and vi have the same degree. After all, each gi is a numerator taken from Q. Multiply each term by vi/vi, which doesn't change anything. Each term has two factors, gi/vi, and vici,j/w. The first belongs to Q, and the second is a homogeneous fraction in U. Q is an ideal, so the terms, and the sum, belong to Q. Since Q is prime it cannot contain 1. This is a contradiction, hence the extension of Q into R, i.e. the ideal H, misses S, and is a proper ideal.

A similar proof characterizes all the homogeneous elements of H. Assume a linear combination of generators ci,jgi = e, where e is homogeneous in R. Throw away any terms whose degree is not the same as e, embed this in R/S, and raise this to the k, for a suitable value of k, so that some w in S has the degree of ek. Expand the sum raised to the k using the multinomial theorem. Each term is still something times gi, where gi is one of our original generators. Multiply each term by vi/vi, to balance the degree of gi. Divide through by w, and ek is one of the numerators of Q. Any new homogeneous element of H, when raised to some power, becomes a preexisting generator of H, a numerator of Q. If S includes anything of degree 1, this power is 1, and H has no new homogeneous elements beyond the numerators of Q. In general, k is the least power, such that the degree of e, times k, is represented by something in S.

Since H misses S, push H up to a prime ideal M missing S, then pull this down to the largest homogeneous ideal P in M, which still contains H, and is prime, and lies over Q. P does the trick, right? Not quite, because P might map to more than Q. But it's good to know H lives in at least one homogeneous prime ideal in R.

Let P be the radical of H, and suppose P is not prime. Since H is homogeneous, P = rad(H) is homogeneous. 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 a homogeneous element in H, and ab to a higher power becomes a numerator from Q. Raise ab to a higher power still, if need be, so that an and bn both have balancing denominators from S. Either an over its denominator, or bn over its denominator, lies in Q. Either an, or bn, is a numerator that was used to generate H, and certainly lies in H. This places a or b in P, and P is prime.

Suppose P maps forward to some ideal that is larger than Q. Some homogeneous element f in P is not a numerator from Q, yet it survives in U. fl = e, for some e in H, and ek is a numerator from Q. Remember that k is the least exponent where ek has a balancing denominator from S. Lesser powers of e cannot exist in U. If f is in U then e is in U, and that is a contradiction. Therefore fl = e, where e is itself a numerator of Q. Let f have a balancing denominator v, so that f/v lives in U. Raise f/v to the l and find something in Q. Since Q is prime, f is a numerator of Q. This too is a contradiction, hence P cannot add any new numerators to Q, and the image of P is Q.

Finally, P misses S, and S has some elements with positive degree, hence P is not R+. P is a valid prime in proj R, and the map is surjective.

A closed base set vf in proj R (where f is homogeneous) corresponds to a closed base set in spec U, once f has been raised to a certain power and assigned an appropriate denominator. Conversely, let a fraction in U have numerator f, and vf in spec U pulls back to vf in proj R. Base sets correspond, and the spaces have the same topology. Therefore proj R, restricted to the primes that miss S, is homeomorphic to spec U.

As a corollary, set S to the powers of e, for some homogeneous e in R, and oe in proj R is the same as spec U.

Spec R is a contravariant functor, and so is proj R.

Let H be a graded ring homomorphism from R into S, preserving degree. Since h preserves degree, it maps homogeneous elements in R to homogeneous elements in S. Let y be in the image of h, so that h(x) = y. Separate x into homogeneous components, and each of these becomes a homogeneous piece of y. Thus the components of y are also in h(r). The image of R is a homogeneous subring of S.

If y is in the image of R, and y is homogeneous, let y = h(x), and the other components of x must map to 0. Thus y has a homogeneous pullback in R.

Like any homomorphism, h induces a reverse map g from the prime ideals of S into the prime ideals of R. Restrict this map to the subspace proj S. Restrict a homogeneous ideal of S to the image h(R) in S. This is the intersection of two homogeneous sets, and it is homogeneous. It is also an ideal in h(R). Thus the contraction of a homogeneous prime ideal in S produces a homogeneous prime ideal in h(R). This pulls back to a homogeneous prime ideal in R. The map from proj S into proj R is well defined, with one caveat. The prime in S could pull back to a prime in R that contains all of R+, which is not in proj R. This is seen when K[x] embeds in K[x,y], and x generates P.

Let proj+ include all homogeneous prime ideals, to get past this little problem. Now proj+ S maps into proj+ R. Since this map carries a subspace of spec S into a subspace of spec R, it is continuous. It also respects composition - the composition of graded ring homomorphisms yields the composition of the associated continuous functions. We have a valid functor from graded rings and homomorphisms into topological spaces and continuous maps.

Two primes in proj R that have the same homogeneous elements beyond a certain degree d0 are in fact the same prime. Let H be an ideal in proj R. If H contains an element x of degree i then it contains x*R. Conversely, if H contains all the elements in x*R of degree > d0, then, by primality, H contains x, or H contains all the homogeneous elements in R of degree > d0-i. If the latter holds, take the radical to find the containing prime. Every homogeneous element of degree > 0 is present. Thus H contains R+, which is a contradiction. Thus H contains x. This holds for each x in H, thus H is uniquely determined.

Beyond some degree d0, let a ring homomorphism f become an isomorphism, mapping Rd onto Sd. The induced function g on proj S is immediately injective, since Q in R has a tail that equals the tail of its preimage in S, and that belongs to at most one prime P in S. How about surjective? As mentioned earlier, Q does not contain all the homogeneous elements beyond d0. Let W be the multiplicatively closed set of homogeneous elements in R, and in S, beyond d0, that are not in Q. Drive the image of Q up to a prime ideal P in S missing W. Then pull P down to a homogeneous prime ideal if need be. g carries P onto Q, and is a bijection.

Since g is a bijection, it is automatically bicontinuous. Intersect the primes in a closed set in one ring, and intersect those same primes in the other. The intersection is a homogeneous ideal, and it becomes a homogeneous ideal when extended from R into S, or when contracted from S back to R. Closed sets correspond, and g is a homeomorphism. Proj R and Proj S are the same space.