The Jacobian Criterion

Part IV Local Rings

13.2 The Jacobian Criterion

13.2 The Jacobian Criterion 195

Proof. (a) The proof follows Mac Lane [35]. LetKbe a perfect field, which we may assume to have positive characteristicp. We will show the following by induction on n: If L is a finitely generated extension of K with a transcendence basisT such thatLhas degreenover the separable closure ofK(T) inL, thenLis separable overK. There is nothing to show for n= 1, so assumen >1. This means that there existsα∈Lwhich is not separable overK(T), sog := irr (α, K(T))∈K(T)[x^p]. We writeT^p for the set of allp-th powers of elements ofT. If all coefficients of g lay in K(T^p), thengwould be ap-th power of a polynomial inK(T)[x], since by hypothesis every element ofKhas ap-th root inK. This contradicts the irreducibility ofg, so g /∈K(T^p)[x]. Applying Lemma 13.9below yields a new transcendence basis T⁰ such that the separable closure of K(T⁰) in Lcontains that of K(T). Since α∈T⁰, the inclusion is strict, so the result follows by induction.

(b) If char(K) = 0, every transcendence basis is separating. In the case char(K) =p > 0 we proceed by induction on the transcendence degree n:= trdeg(L/K). Ifn= 0, thenT =∅ is a separating transcendence ba- sis by hypothesis. So we may assumen >0. By hypothesis there exists a separating transcendence basisT. For every elementα∈L, the minimal polynomial irr (α, K(T)) is separable, so if all it coefficients lie in the sub- fieldK(T^p), thenαis separable over K(T^p). Assume that this happens for every element of a given generating setS ofLoverK. ThenLwould be separable overK(T^p) (see Lang [33, Ch. VII, Theorem 4.8]). But an elementtfromT has irr (t, K(T^p)) =x^p−t^p, which is inseparable. This contradiction shows that there existsα∈S such that g:= irr (α, K(T)) does not lie inK(T^p)[x]. Applying Lemma13.9below yields a new tran- scendence basisT⁰ such that L is separable overK(T⁰), and α∈T⁰. So viewed as an extension ofK(α),Lhas the separating transcendence basis T⁰\ {α}, and it is still generated byS. By induction,Lhas a separating transcendence basisT⁰⁰ ⊆S over K(α), so as an extension of K it has the separating transcendence basisT⁰⁰∪ {α} ⊆S. ut The following lemma was used in the proof.

Lemma 13.9. Let L be an extension of a field K of characteristic p > 0.

Let T be a finite transcendence basis, and write write T^p for the set of all p-th powers of elements of T. If the minimal polynomialg := irr (α, K(T)) of an α ∈ L does not lie in K(T^p)[x], then there exists t ∈ T such that T⁰ := (T\ {t})∪ {α} is a transcendence basis, and all elements fromLthat are separable overK(T)are also separable overK(T⁰).

Proof. SinceK[T] is factorial, there exists 06=h∈K[T] such thatf :=hg∈ K[T][x] is a primitive polynomial, so by the Gauss Lemma,f is irreducible (see Lang [33, Ch. V, Theorem 6.3]). Since h is the leading coefficient of f (as a polynomial in x), f does not lie in K[T^p][x], so there exists t ∈ T such that f, viewed as a polynomial in t, is separable. This shows that t

13.2 The Jacobian Criterion 197 is separable over K(T⁰). Therefore T⁰ is a new transcendence basis, and all elements ofT are separable overK(T⁰). This implies that all elements fromL that are separable overK(T) are separable overK(T⁰) (see Lang [33, Ch. VII,

Theorem 4.9]). ut

We come back to the goal of calculating the singular locus in the spec- trum of an affine algebra A. If A = K[x1, . . . , xn]/I is given as a quotient ring of a polynomial ring over a field, then an element of X := Spec(A) is given as P/I, where P ⊂ K[x1, . . . , xn] is a prime ideal with I ⊆ P. The main goal of this section is to prove the Jacobian criterion for the regu- larity of the local ring A_{P /I}. This criterion involves an irreducible compo- nent of X containingP/I. Such a component corresponds to a prime ideal Q⊂K[x1, . . . , xn] which is minimal overI, and contained inP. The criterion also involves the rank of a matrix of polynomials modulo P, defined as fol- lows: If (gi,j)∈K[x1, . . . , xn]^m×k, then rank (gi,j modP) denotes the rank of the matrix (gi,j+P)∈Quot (K[x1, . . . , xn]/P)^m×k. The matrix that ap- pears in the Jacobian criterion is made up of the (formal) partial derivatives of polynomials generatingI. This is often called theJacobian matrix.

Theorem 13.10 (Jacobian criterion). Let I= (f1, . . . , fm)⊆K[x1, . . . , xn] be an ideal in a polynomial ring over a field, and let P ⊂K[x1, . . . , xn] be a prime ideal containing I. Furthermore, let Q ⊂K[x1, . . . , xn] be a prime ideal which is minimal overI, and which is contained inP. Then

(a)

rank ∂f_i

∂xj

modP

≤ht(Q).

(b) If equality holds in(a), then the local ring(K[x1, . . . , xn]/I)_{P /I} is regular.

(c) IfQuot (K[x₁, . . . , x_n]/P) is a (not necessarily finite) separable field ex- tension ofK, then the converse of (b)holds. The separability hypothesis is automatically satisfied ifKis a perfect field or ifP= (x₁−ξ₁, . . . , x_n− ξ_n)corresponds to a point (ξ₁, . . . , ξ_n)∈ V_Kn(I).

If the separability hypothesis of Theorem13.10(c) is not satisfied, it does happen that the converse of (b) fails (see Exercise13.7).

Before we turn to the proof of Theorem 13.10, let us note that it can be reformulated as follows: The nullity of the Jacobian matrix modulo P is greater than or equal to the dimension of every irreducible component of Spec (K[x1, . . . , xn]/I) which containsP, with equality if and only if the local ring is regular (provided the hypothesis of part (c) holds). Moreover, notice that ifP = (x1−ξ1, . . . , xn−ξn) corresponds to a point (ξ1, . . . , ξn)∈ VKⁿ(I), then the Jacobian matrix moduloP is

∂fi

∂x_j(ξ1, . . . , ξn)

∈K^m×n. The kernel of the Jacobian matrix moduloP can be interpreted as the tangent space at the pointP. So for an affine varietyX we have a tangent space attached to every point, and its dimension is greater than or equal to the dimension of an

irreducible component on which the point lies. The singular points are those where the dimension of the tangent space exceeds that lower bound.

We need two lemmas for proving Theorem13.10.

Lemma 13.11. Let P ⊂ K[x₁, . . . , x_n] be a prime ideal of height m in a polynomial ring over a field.

(a) There exist f1, . . . , fm ∈ P generating the localized ideal PP ⊆ K[x1, . . . , xn]P

(b) IfQuot (K[x1, . . . , xn]/P)is a separable field extension of K, then rank

∂fi

∂xj

modP

=m.

Proof. As a field extension ofK,L:= Quot (K[x1, . . . , xn]/P) is generated by α1, . . . , αn withαi:=xi+P. We express this by writingL=K(α1, . . . , αn).

By Corollaries 5.7 and 8.23, K[x1, . . . , xn]/P has dimension k := n−m, so by Theorem 5.9, L has transcendence degree k. Since every generating set of a field extension contains a transcendence basis, we may assume that α1, . . . , αk form a transcendence basis, so they are algebraically independent, andL is a finite extension ofL0:=K(α1, . . . , αk). IfL is separable overK, we may use Proposition13.8(b) and additionally assume thatLis separable overL₀. For everyl∈ {0, . . . , m}, consider the map

ϕ_l:K(x₁, . . . , x_k)[x_k+1, . . . , x_k+l]→L, x_i7→α_i.

We claim that im(ϕl) = K(α1, . . . , αk+l) =: Ll and ker(ϕl) = (f1, . . . , fl) with fi ∈ K[x1, . . . , xk+i] ∩ P. Additionally, if L is separable over L0, then ∂fi/∂xk+i ∈/ P. All this is true for l = 0. Using induction on l, we may assume that l > 0 and that f1, . . . , f_l−1 have already been found.

Since αk+l is algebraic over L_l−1, it follows that Ll = L_l−1[αk+l] (see Lang [33, Ch. VII, Proposition 1.4]), and so Ll = L0[αk+1, . . . , αk+l] by induction. This shows im(ϕl) = Ll. Set g := irr (αk+l, Ll−1) ∈ Ll−1[xk+l].

Since Ll−1 = L0[αk+1, . . . , αk+l−1], there exist fl ∈ K[x1, . . . , xk+l] and h ∈ K[x₁, . . . , x_k]\ P such that g = ^fl(α1_h(α^{,...,αk+l−1,xk+l)}

1,...,α_k) . It follows that fl ∈ K[x1, . . . , xk+l]∩P. Moreover, if L is separable over L0, then Ll is also separable overLl−1(see Lang [33, Ch. VII, Theorem 4.9]), and it follows thatghas no multiple roots, so_∂x^∂g

k+l(αk+l)6= 0. This implies∂fl/∂xk+l∈/ P.

Clearly f_l ∈ ker(ϕ_l). For proving ker(ϕ_l) = (f₁, . . . , f_l), take f ∈ ker(ϕ_l).

Then g divides f(α₁, . . . , α_k+l−1, x_k+l), so there exist r ∈ K[x₁, . . . , x_k+l] ands∈K[x1, . . . , xk]\P with

f(α₁, . . . , α_k+l−1, x_k+l)·h(α₁, . . . , α_k)

fl(α1, . . . , α_k+l−1, xk+l) =r(α₁, . . . , α_k+l−1, x_k+l) s(α1, . . . , αk) .

13.2 The Jacobian Criterion 199 Therefore all coefficients ofhsf−rfl(as a polynomial inx_k+l) lie in ker(ϕ_l−1).

So by inductionhsf−rf_l∈(f₁, . . . , f_l−1)_{K(x1,...,xk)[xk+1,...,xk+l]}. Sinceh, s6=

0, this implies f ∈(f₁, . . . , f_l), and the claim is proved.

For l =m we get ker(ϕ_m) = (f₁, . . . , f_m)_{K(x1,...,xk)[xk+1,...,xn]}. The alge- braic independence of α₁, . . . , α_k implies that K(x₁, . . . , x_k)[x_k+1, . . . , x_n]⊆ K[x1, . . . , xn]P, so (ker(ϕm))_K[x

1,...,xn]P = (f1, . . . , fm)_{K[x1,...,xn]P} follows.

We obtain

PP ⊆(ker(ϕm))_K[x

1,...,xn]P = (f1, . . . , fm)_{K[x1,...,xn]P} ⊆PP, showing (a).

For showing (b), consider the last m columns of the Jacobian matrix, which form the square matrixA:= (∂fi/∂xk+j)∈K[x1, . . . , xn]^m×m. Since f_i ∈ K[x₁, . . . , x_k+i], A is a lower triangular matrix with diagonal entries

∂f_i/∂x_k+i. Since these entries do not lie in P under the hypothesis of (b),

det(A)∈/P follows, showing (b). ut

Notice that in Lemma13.11,K[x₁, . . . , x_n]_P is a local ring of dimensionm with maximal idealPP. Therefore part (a) says thatK[x1, . . . , xn]Pis regular, soK[x1, . . . , xn] is a regular ring. This is generalized in Exercise13.1.

The following lemma gives an interpretation of the rank of the Jacobian determinant off1, . . . , fmmodulo a prime ideal in terms of the ideal generated by the fi.

Lemma 13.12. Let I = (f1, . . . , fm) ⊆ K[x1, . . . , xn] be an ideal in a polynomial ring over a field, and let P ⊂ K[x1, . . . , xn] be a prime ideal containing I. With L := K[x1, . . . , xn]P/PP (which is isomorphic to Quot (K[x1, . . . , xn]/P)) we have

(a)

rank ∂f_i

∂xj

modP

≤dim_L

I_P+P_P² P_P²

. (b) IfL is a separable field extension ofK, then equality holds in (a).

Proof. We will construct linear mapsϕ:L^m→P_P/P_P² andψ:P_P/P_P² →Lⁿ. First

ϕ:L^m→PP/P_P², (g1+PP, . . . , gm+PP)7→

m

X

j=1

gjfj+P_P²

gives a well-defined, L-linear map with image im(ϕ) = IP+P_P²

/P_P². To define ψ, consider f ∈ P_P². This means that hf = Pr

j=1gjhj with h ∈ K[x1, . . . , xn]\P andgj, hj∈P. For 1≤i≤nwe get

h∂f

∂x_i +f ∂h

∂x_i =

r

X

j=1

gj

∂hj

∂x_i +hj

∂gj

∂x_i

∈P,