Download Quadratic Transforms in Local Rings: Properties and Applications - Prof. Melvin Hochster and more Study notes Algebra in PDF only on Docsity!
Math 711: Lecture of November 6, 2006
To finish our comparison of symbolic powers in a regular ring, we shall make use of quadratic transforms (also called quadratic transformations or quadratic dilatations) in a more general context than in the proof of the Lipman-Sathaye Jacobian Theorem.
Let (R, m, K) be a local domain with R ⊆ (V, n) a local map, where V is a not necessarily Noetherian valuation domain. The first quadratic transform of R along V is the localization (R 1 , m 1 ) of R[m/x] at the contraction of n, where x is any element of m such that xV = mV. This ring is again a local ring with a local map R 1 → V.
The quadratic transform is independent of the choice of the element x. To see this, suppose that xV = yV , where y, x ∈ m. Then y/x ∈ R[m/x] is a unit in V , so its inverse x/y ∈ R 1. Since m/y = (m/x)(x/y), it follows that R[m, y] ⊆ R 1. Moreover, each element of R[m/y] that is invertible in V has an inverse in R 1 , so that if Q is the contraction of n to R[m/y] we have an induced inclusion map R[m/y]Q → R 1. An exactly symmetric argument gives the opposite inclusion.
As in our earlier situation, we may take iterated quadratic transforms
R ⊆ R 1 ⊆ · · · ⊆ Rk ⊆ · · · ⊆ V.
Note that if m = x 1 ,... , xh, then mV = (x 1 ,... , xh)V , so that x may be chosen from among the xi. Putting this together with the Lemma on p. 2 of the Lecture Notes of September 29, we have:
Proposition. Let (R, m, K) be regular local with x 1 ,... , xd a regular system of param- eters and suppose that R ⊆ (V, n) is local where V is a valuation domain. If the xi are numbered so that xj V ⊆ x 1 V for all j > 1 , then the quadratic transform R 1 is a localiza- tion of the ring S = R[x 2 /x 1 ,... , xd/x 1 ], which is regular of dimension d. In particular, R 1 has dimension at most d. Moreover, S/x 1 S ∼= K[X 2 ,... , Xd], where Xi is the image of xi/x 1 , 2 ≤ i ≤ d.
Here is another important example:
Theorem. Let R be a one dimensional local domain whose integral closure (V, n) is local and module-finite over R. (This is always the case if R is a complete one-dimensional local domain.) Let R ⊆ R 1 ⊆ · · · ⊆ Rk ⊆ · · · ⊆ V
be the sequence of iterated quadratic transforms. Then for all sufficiently large k, Rk = V.
Proof. Since V is module-finite over R, it cannot have an infinite ascending chain of R- submodules. It follows that the chain Ri is eventually stable. But if the maximal ideal of 1
Ri is not principal and has minimal generators y 1 ,... , yh with y 1 of least order in V , then for some j > 1, yj /y 1 ∈ V − Ri, and yj /y 1 ∈ Ri+1. Therefore, for sufficiently large i, the maximal ideal of Ri is prinicpal. But then Ri is a DVR, and is a normal ring inside the fraction field of R and containing R. It follows that Ri = V.
We also note:
Theorem. Let (R, m, K) be a local domain with R ⊆ (V, n) a local inclusion, where V is a valuation domain of the fraction field of R. Let q be a prime ideal of V lying over P 6 = m in R. Let R ⊆ R 1 ⊆ R 2 ⊆ · · · ⊆ Rk ⊆ · · · ⊆ V
be the sequence of quadratic trransforms of R along V. Let Pi be the contraction of P to Ri. Then R/P ⊆ R 1 /P 1 ⊆ R 2 /P 2 ⊆ · · · ⊆ Rk/Pk ⊆ · · · ⊆ V /q
is the sequence of quadratic transforms of R/P ⊆ V /q along V /q.
Proof. By induction on k, it suffices to see this when k = 1. Let x 1 ,... , xh generate the maximal ideal m of R with x 1 V = mV. Some x ∈ m is not in n, and since xV ⊆ mV = x 1 V , x 1 ∈/ q and so x 1 ∈/ P. Moreover, x 1 (V /q) = (m/P )(V /q). It follows that the quadratic transform of R/P along V /q is the localization at the contraction of n of (R/P )[ m/x˜ 1 ], where m˜ is m/P and x 1 is the image of x 1 in R/P. The stated result follows at once. Note that we again have P 1 6 = m 1 , the maximal ideal of R 1 , since x 1 ∈ m 1 −P 1.
We next observe:
Lemma. Let (R, m, K) be a regular local ring with algebraically closed residue class field, and suppose R ⊆ (V, n) is local, where V is a valuation domain and R/m → V n is an isomorphism. Then there is a regular system of parameters x 1 ,... , xd for R such that the first quadratic transform is the localization of R[x 2 /x 1 ,... , xd/x 1 ] at the height d maximal ideal generated by x 1 , x 2 /x 1 ,... , xd/x 1 , so that these elements are a regular system of parameters in the first quadratic transform.
Proof. Let x 1 , y 2 ,... , yd be one regular system of parameters for R such that x 1 V = mV. n contains x 1 , and so n lies over a prime ideal of R containing x 1. Hence, the quotient of R[m/x 1 ] by the contraction of n is also a quotient of K[Y 2 ,... , Yd], where Yi is the image of yi/x 1 , 2 ≤ i ≤ d. The resulting quotient domain imbeds embeds K-isomorphically in K = V /n, and so is equal to K. It follows that the contraction of n corresponds to a maximal ideal of K[Y 2 ,... , Yd], which must have the form (Y 2 − c 2 ,... , Yd − cd) for elements c 2 ,... , cd ∈ K. Therefore we may let xi = yi − cix 1 for each i, 2 ≤ i ≤ d.
Proof of the theorem on comparison of symbolic powers. We want to show that if R is regular and P ⊆ Q are prime, then P (n)^ ⊆ Q(n)^ for all n. By considering a saturated chain of primes joining P to Q we immediately reduce to the case where the height if Q/P in
Therefore, for some t, we have
f /xn 1 ∈ (1/xn 1 +1)(x^21 , x 2 ,... , xd)n+1mt/xt 1
and so xt 1 +1 f ∈ (x^21 , x 2 ,... , xd)n+1mt.
Each of the obvious generators obtained by expanding the product on the right that in- volves x^21 has degree at least n + t + 2. Hence, in the degree n + t + 1 part of grm(R) = K[X 1 ,... , Xd], we have that
X 1 t+1 F ∈ (X 2 ,... , Xd)n+1(X 1 ,... , Xd)t,
where F is the image of f in mn/mn+1, and is supposedly not 0. By taking homogeneous components in degree n + t + 1 we see that xt 1 +1 F must be in the K-vector space span of the obvious monomial generators of
(X 2 ,... , Xd)n+1(X 1 ,... , Xd)t.
But this is clearly impossible with F 6 = 0, since none of these monomials is divisible by Xt 1 +.
This completes the proof, once we have shown that for primes generated by a regular sequence, symbolic powers are the same as ordinary powers.
