1 Commutative algebra 19 May 2015 Jan Draisma TU Eindhoven and VU Amsterdam
Goal: Jacobian criterion for regularity 2 Recall A Noetherian local ring R with maximal ideal m and residue field K := R/m is called regular if dim R = dim K m/m 2. (The lhs equals the minimal number of elements needed to generate an ideal whose radical equals m. The rhs equals the minimal number of elements needed to generate m.) Today A concrete criterion for testing whether the localisation A P of an affine algebra A at a prime P is regular. Analogy: a consequence of implicit function theorem ϕ : R n R m with n > m, C near p R n, with Jacobian matrix of rank m at p. Then open U p: ϕ 1 (p) U is a smooth (n m)-dimensional manifold.
Statement of the theorem 3 Theorem (Jacobian criterion) Let A = K[x 1,..., x n ]/( f 1,..., f m ), let P SpecA, and let Q SpecA be a mimal prime contained in P. Consider J := ( f i x j ) ij L m n with L := Quot(A/P). Then: 1. dim A/Q dim L ker J 2. equality implies that A P is regular (i.e., P is a nonsing pt) 3. if L is a (not necessarily finite) separable field extension of K, then the converse to 2 also holds. Example 1. A = Q[x, y]/(y 2 x 3 ) has dimension 1, and J = [3x 2, 2y] has nullity 1 mod P unless P = (x, y); Sing(Spec(A)) = {(x, y)} 2. K = F p (t), A = K[x]/(x p t) has dimension 0 but J = [0] has nullity 1 for the unique prime P = (0); and yet P is regular note that L = A is not separable over K.
Definition An extension L/K is separable if t 1,..., t l L, algebraically independent over K, s.t. L is finite over F := K(t 1,..., t l ) and the minimal polynomial p(t) = i c i T i F[T] of any element of L has a non-zero derivative p (T) := i ic i T i 1. Proof in special case, see book for general case We will assume that Q = 0, i.e., that SpecA is irreducible, and, moreover, that L = Quot(A/P) is separable over K. I. Interpretation of ker J as tangent space to SpecA at P Definition S an R-algebra, then D : R S is called a derivation if a, b R : D(ab) = ad(b) + bd(a). If R is a K-algebra, write Der K (R, S) := {D : R S D a K-linear derivation} 4
Examples/properties 1. Der K (R, S) is an S-module via (sd)(r) := sd(r). 2. If ϕ : T R is a homo and D : R S a derivation, then D ϕ : T S a derivation. In particular, if R = T/I, then Der K (R, S) {D Der K (T, S) D I = 0}. 3. R = K[x 1,..., x n ], then D Der K (R, S) is determined uniquely by s j := D(x j ), j = 1,..., n, and these can be prescribed freely. Notation: D = n j=1 s j x j. 4. F = K(t 1,..., t l ) with t 1,..., t l alg ind/k, and L any field extension of F, then Der K (F, L) is the L-vector space with basis t 1,..., t l. 5. If U R multiplicative and its image in S consists of invertible elements, then any derivation R S factors through a unique derivation U 1 R S. 5
Lemma 1 If L/K separable, then dim L Der K (L, L) = trdeg K L. 6 Proof (extending derivations to separable extensions) Take t 1,..., t l L algebraically independent s.t. L is finite and separable over F := K(t 1,..., t l ). By induction on dim F E, we prove that for each E between F and L we have dim L Der K (E, L) = l; by the above this holds if E = F. For the induction step, let D Der K (E, L) and a L \ E. We argue that D has a unique extension to an element of Der K (E(a), L). Let p = c 0 + + c d T d E[T] be the minimal polynomial of a, so that E(a) = E[T]/(p(T)). We need to define D(a) such that D(p(a)) = 0. Expanding the lhs yields p (a)d(a) + i D(c i )a i. Since a is separable over E, p (a) 0, so there is a unique solution in L for D(a). Letting D : E[T] L be the derivation extending D that maps T to that solution, we find that D (p(t)) = 0, and we have constructed the extension.
Lemma 2. Let R be a local K-algebra with maximal ideal m, and assume that L := R/m is separable over K. Then the homo π : R/m 2 R/m = L has a section σ : L R/m 2, i.e. a K-algebra homo such that π σ = id L. Proof (Hensel lifting) Take F = K(t 1,..., t d ) L as before. Choosing arbitrary lifts t 1,..., t d R/m 2 of t 1,..., t d, we get a K-algebra homo K[t 1,..., t d ] R/m 2 by setting t i t i. Since this is an injection, we get a section F R/m 2. Suppose we have constructed a section E R/m 2 for some E between F and L, and let a L \ E with minimal polynomial p E[T]. Let b R/m 2 be any lift of a. Following Newton, set ã := b p(b) p (b) R/m2, where the coefficients of p are interpreted as elements of R/m 2 via the section E R/m 2. Note that ã maps to a, and compute p(ã) = 0 R/m 2. 7
Back to the Jacobian criterion. If J (g 1,..., g n ) T = 0 where g 1,..., g n L := Quot(A/P), then D := j g j x j Der K (K[x 1,..., x n ], L) maps each f i to 0, so D Der K (A, L) and even D Der K (A P, L). Conversely, every element of the latter space pulls back to a derivation i g i x i mapping each f i to 0. This proves: Lemma 3 The nullity of J equals dim L Der K (R, L), where R = A P. Remark Let m := P P be the maximal ideal of the local ring R. The map A R has kernel P, hence we get an isomorphism L R/m. 8
II. Proof of the Jacobian criterion. Consider the L-linear map ϕ : Der K (R, L) (m/m 2 ) determined by D (a D(a)). 1. This is well-defined because D(m 2 ) = 0. 2. ker(ϕ) consists of all D that vanish on m, i.e., ker(ϕ) = Der K (R/m, L) = Der K (L, L). 3. ϕ is surjective: let ψ : R/m 2 R/m = L be a section to π : R/m 2 R/m (Lemma 3, use that L/K is separable), and let x (m/m 2 ). Then define the map D : R/m 2 L by D( f ) = x( f σ(π( f ))). This restricts to x on m/m 2 and is a K-linear derivation. (Special case: if L = K, then this reads σ( f f (P)).) 4. Thus we obtain dim L Der K (R, L)) = dim L (m/m 2 ) + dim L Der K (L, L). 5. The first term is at least the Krull dimension of R, i.e., the height of P, and the latter term equals the Krull dimension of A/P. Since A is equidimensional, these Krd. add up to dim A. Equality iff A P is regular. 9
Consequences of the Jacobian criterion 10 Theorem For any affine algebra A over a perfect field K, the set {P Spec(A) P singular } is closed. Proof If A = K[x 1,..., x n ]/I is a domain of Krull-dimension d, then the complement is the set of all primes P containing all (n d) (n d)-subdeterminants of the Jacobian matrix J. We use here that any extension of K is separable. Other facts 1. the nonsingular locus in Spec(R) is dense for any reduced, Noetherian ring R. 2. the singular locus in an affine variety is closed 3. the nonsingular locus in an affine variety over an algebraically closed field K is open and dense.
11 Examples 1. A = R[x, y]/(x 2 + y 2 ) has Jacobi matrix [2x, 2y]. So its only real point (0, 0) is in fact singular. 2. A = K[x ij i, j = 1,..., n]/(det(x)) has Jacobi matrix a row consisting of all (n 1) (n 1)-subdets of x. So the singular locus of V(det(x)) in K n n consists of all matrices of rank < n 1. 3. A = R[x, y, z]/(y 2 x 2 (t 2 x)) has J = [3x 2 2t 2 x, 2y, 2tx 2 ]. Together with the equations, these elements generate an ideal whose radical is (x, y). Exercises: 13.6 and 13.9 and the following: Let X K n n be the affine variety of matrices of rank k, where 0 < k < n is fixed. Determine the singular locus of X.