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Math 711: Lecture of September 20, 2006
We shall soon return to our treatment of the dimension formula, which was stated in the Lecture of September 18, but we first want to make some additional remarks about the behavior of analytic spread.
Theorem. Let K be a field, and T a finitely generated N-graded K-algebra with T 0 = K. Let M be the homogenous maximal ideal of T. Let F 1 ,... , Fs be homogeneous polynomials of the same positive degree d in T , and let I = (F 1 ,... , Fs)T. Then an(ITM) is the Krull dimension of the ring K[F 1 ,... , Fs] ⊆ T , and hence is the same as the maximum number of algebraically independent elements in K[F 1 ,... , Fs]. over K.
Proof. We shall show that K[F 1 ,... , Fs] ∼= K ⊗TM grI (TM). We view the latter is
K ⊗TM TM[ITMt] ∼= (K ⊗T T [It])M
and the ring on the right is the same as K ⊗T T [It], because elements of T − M map to units in K and so already are invertible in this ring. Note that K here is T /M, and so this ring is also the same as T [It]/MT [It].
Now there is a map K[F 1 ,... , Fs] → T [It] that sends Fj 7 → Fj t, 1 ≤ j ≤ s. To see that this is well-defined, note that the ideal of relations on the Fj over K is homogeneous. Thus, it suffices to see that if H ∈ K[Y 1 ,... , Ys] is a homogeneous polynomial of degree μ such that H(F 1 ,... , Fs) = 0, then H(F 1 t,... , Fst) = 0. But the left hand side is tμH(F 1 ,... , Fd) = tμ^ · 0 = 0. We then get a composite map
K[F 1 ,... , Fs] → T [It] T [It]/MT [It].
This map is clearly surjective, since the image of T in the quotient is K and It is generated by the Fj t. We need only prove that the kernel is 0. It is homogeneous: let G be an element of the kernel that is homogeneous of degree h in F 1 ,... , Fs. Then G has degree hd in x 1 ,... , xn. If G is in the kernel then Gth^ is in MIhth, and G ∈ MIh. However, all nonzero elements of this ideal have components of degree at least hd + 1 in x 1 ,... , xn, a contradiction unless G = 0.
The final statement is a general characterization of Krull dimension in finitely generated K-algebras.
Remark. This result gives another way to compute the analytic spread of the height one prime in a determinantal ring analyzed in the last Example (beginning at the bottom of p. 5) in the Lecture Notes of September 18. It is immediate that the analytic spread is n.
Example. The Example discussed in the Remark just above shows that height one primes that have arbitrarily large analytic spread. In a regular local ring a height one prime is 1
principal, and so its analytic spread is 1. But there are height two primes of arbitarily large analytic spread. Let X be an n × (n + 1) matrix of indeterminates over a field K and let P be the ideal generated by the size n minors of X in the polynomial ring K[X]. Then the analytic spread of P in K[X]M, where M is generated by the entries of X, is n + 1 by the Theorem above, for the minors of algebraically independent over K. (This is true even if we specialize the leftmost n × n submatrix to be yIn. The minors are yn and, up to sign, the products yn−^1 xi,n+1, 1 ≤ i ≤ n.) These primes have height two: the algebraic set of n × (n + 1) matrices of rank at most n − 1 has dimension n^2 + n − 2. (On the open set where the first n − 1 rows are algebraically independent, the space consisting of choices for the first n − 1 rows has dimension (n − 1)(n + 1); the choices for the final row are linear combinations of the first n − 1 rows, and are parametrized by An−^1 , giving dimension n^2 − 1 + (n − 1).)
Recall that a map of quasilocal rings h : (R, m) → (S, n) is called local if h(m) ⊆ n. (The map of a local ring onto its residue class field is local, while the inclusion of a local domain that is not a field in its fraction field is not local.)
Proposition. Let (R, m, K) be local.
(a) If h : (R, m, K) → (S, n, L) is a local homomorphism, and I ⊆ m is an ideal of R, then an(I) ≥ an(IS).
(b) If I and J are proper ideals of R, then an(I + J) ≤ an(I) + an(J).
(c) Let I and J are proper ideals of R. If either an(I) or an(J) is 0, then an(IJ) = 0. If the analytic spreads are positive, an(I J) ≤ an(I) + an(J) − 1.
Proof. We replace R → S by R(t) → S(t) if necessary, and the ideals considered by their expansions. We may therefore assume the residue class fields are infinite.
For part (a), if I is integral over an ideal I 0 with a = an(I) generators, then IS is integral over I 0 S.
For part (b) simply note that if I 0 is as above and J is integral over J 0 with b = an(J) generators, then I + J is integral over I 0 + J 0 , which has at most a + b generators.
To prove part (c), first note that the analytic spread of I is 0 if and only if I consists of nilpotents. Thus, if either a or b is 0, then IJ consists of nilpotents and an(I J) = 0 as well. Now suppose that both analytic spreads are positive and that I 0 and J 0 are as above. Map the polynomial ring T = Z[X 1 ,... , Xa, Y 1 ,... , Yb] → R so that (X 1 ,... , Xa)T maps onto I 0 and (Y 1 ,... , Yb)T maps onto J 0. Since IJ is integral over I 0 J 0 , it suffices to show that an(I 0 J 0 ) ≤ a + b − 1. Let M be the inverse image of m in T. Then M is a prime ideal of T that is either (X 1 ,... , Xa, Y 1 ,... , Yb)T or pT + (X 1 ,... , Xa, Y 1 ,... , Yb)T. Let A = TM. Let I = (X 1 ,... , Xa)A and J = (Y 1 ,... , Yb)A. Then we have an induced local map TM → R such that IR = I 0 and J R = J 0. By part (a), it will suffice to show that an(I J ) ≤ a + b − 1.
Let A denote the ideal (X 1 ,... , Xa, Y 1 ,... , Yb) ⊆ T , and let B denote the ideal (X 1 ,... , Xa)(Y 1 ,... , Yb)T. There are two cases. First suppose that M = A. Then
that has length h + k. If, moreover, D is catenary then all saturated chains from m to (0) have the same length, and this is dim (D), so that h + k = dim (D).
Proof of the dimension formula. By adjoining generators of S to R one at a time, we can construct a chain of rings R = S 0 ⊆ S 1 ⊆ · · · ⊆ Sn
such that for each i, 0 ≤ i ≤ n, we have that Si+1 is generated over Si by one element. Let Qi = Q ∩ Si for each i. Note that when R is universally catenary, every Si is universally catenary. It will suffice to prove the dimension formula (whether the inequality or the equality) for each inclusion Si ⊆ Si+1. When we add the results, each term associated with Si for i different from 0 and n occurs twice with opposite signs. The intermediate terms all cancel, and we get the required result.
We henceforth assume that S = R[x], where x need not be an indeterminate over R. By replacing R and S by RP and RP ⊗R S, we may assume that (R, P, K) is local. We consider two cases, according as whether x is transcendental or algebraic over R.
Case 1. x is transcendental over R. Then the primes of S = R[x] lying over P correspond to the primes of R[x]/P R[x] ∼= K[x], a polynomial ring in one variable. There are two subcases.
Subcase 1a. Q corresponds to the prime ideal (0) in K[x], i.e., Q = P R[x]. In this case SQ ∼= R(x) has the same dimension as R, so that height Q = height P. We have that tr. deg.(G/F) = 1, and L ∼= K(x), so that tr. deg.(L/K) = 1 as well. Since 0 = 1 − 1, we have the required equality whether R is universally catenary or not.
Subcase 1b. Q is generated by P R[x] and a monic polynomial g of positive degree whose image g mod P is irreducible in K[x]. The height of Q is evidently has height height P + 1: a system of parameters for P together with g will give a system of parameters for R[x]Q. The left hand side of the inequality is therefore 1, while the right hand side is 1−0, because L ∼= K[x]/
g
. Again, we have the required equality whether R is universally catenary or not.
Case 2. x is algebraic over R. Let X be an indeterminate and map R[X] R[x] = S as R-algebras by sending X 7 → x. We have a commutative diagram:
F[X] −−−−→ G x
x
R[X] −−−−→ S
where the horizontal arrows are surjective and the vertical arrows are inclusions. By hy- pothesis, the top horizontal arrow has a kernel, which will be the principal ideal generated by a monic polynomial h of positive degree: the minimal polynomial of x over F. The kernel P 0 of R[X] S may therefore be described as hF[X] ∩ R[X]. We claim that P 0 is a height one prime of R[X]. To see this, we calculate R[X]P 0. Since R ⊆ S, P 0 does not meet R, and R − { 0 } becomes invertible in RP 0. Thus, RP 0 is the localization of F[X] at
the expansion of P 0 , which is hF[x], and is a one-dimensionsial ring. Let Q denote the inverse image of Q in R[X]. Then Q contains P and, in fact, lies over P. It also contains P 0. There are again two subcases, depending on what Q is.
Subcase 2a. Q = P R[X]. In this subcase the right hand side of the dimension formula is 0 − 1. The height of Q is the same as height P , and killing P 0 decreases it at least by 1 as required. If R is universally catenary it decreases by exactly 1.
Subcase 2b. Q has the form P R[X] + f R[X], where f ∈ R[X] is monic of positive degree and irreducible mod P. The right hand side of the dimension formula is 0 − 0. The height of Q is height P + 1. Killing P 0 decreases it by least 1, and by exactly 1 in the universally catenary case.
Remark. If S is a polynomial ring over R, we can choose the chain so that Si+1 is always a polynomial ring in one variable over Si. We are always in Case 1 of the proof, and so equality holds in the dimension formula without assuming that R is universally catenary.
