Graded Rings, Homogeneous Ideal

Homogeneous Ideal

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 obvious; H is generated by its homogeneous elements, and is homogeneous. Conversely, write an element y in H as the sum of a_ig_i, where g_i are homogeneous generators of H. Separate each a_i into homogeneous components and distribute g_i across these components. Now each a_i,j*g_i 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 H_d, whence H becomes the direct sum over H_d.

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

Quotient Ring

Let Q be a quotient ring of R, with a homogeneous kernel H, so that Q becomes a graded ring. To map an element in R to an element in Q, reduce each component from R_d mod H_d. This preserves degree, from R down to Q.

As a special case of the above, let the kernel begin with R_d 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. This sets up a correspondence - homogeneous ideals in Q correspond to homogeneous ideals in R containing H.

Prime Ideal

Test whether the kernel H is prime by seeing if the quotient Q is an integral domain. Look for 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, generated by the homogeneous elements (if any) in P. By the above test, this homogeneous ideal is also prime. It is the largest homogeneous ideal contained in P.

Closure

Homogeneous ideals are closed under addition, intersection, product, and radical.

Multiply generators to generate the product.

Take the union of generators to generate the sum.

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 is the same. When H is homogeneous, rad(H) is the intersection of all the homogeneous prime ideals containing H.

At the Top

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 maximal ideal, relative to homogeneous ideals. Thus R looks a bit like a local ring.

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 our ideal is all of R. A complete chain of homogeneous ideals starts with 0, and ends with M and R.