Def: the Hilbert dimension of a homogeneous ideal is 1 + degree of the Hilbert polynomial, or 0 if the Hilbert polynomial is 0.
Theorem. HilbDim(I) = HilbDim(Rad(I)).
Proof. If I is not radical, we can add some homogeneous x whose square is in I.
Then do nullity plus rank on the map (R/I)_n -> (R/I)_{n+k} that multiplies by x.
Theorem. Let I be homogeneous.
1) If I is a prime ideal, and J (homogeneous) properly contains I, then HilbDim(J) < HilbDim(I).
2a) If I is not prime, then there exists J (homogeneous) properly containing it, with HilbDim(J) = HilbDim(I).
2b) That J can be taken to be prime.
Proof.
1) Even adding one element lowers the HilbDim by 1, by a calculation last week.
2a) If I is not radical, use the previous theorem.
Otherwise let a,b not in I, ab in I, and check that the map R/I -> R/I+< a > \oplus R/I+< b > is 1:1. Do nullity plus rank on that.
2b) Repeat 2a and use ACC.
We didn't get to
Theorem. HilbDim(I) = homogeneous KrullDim(I) (which is at most KrullDim(I)).
Proof. In a chain of homogeneous prime ideals, the HilbDim must drop at each step by (1) of the last theorem. That gives "HilbDim is at most homogeneous KrullDim".
Conversely, we can alternate the following two steps: increase I to a prime ideal without changing the HilbDim (using (2b)), and then increase it using one new homogeneous generator, lowering the HilbDim by 1, by the calculation last week. We get stuck only when we get to whole ring. This constructs a long sequence of prime ideals, giving "HilbDim is at least homogeneous KrullDim".
