Definition Let X be a scheme, and let x X be a point. The stalk OX,x. is called the local ring of X at x.

Remark By construction, the local ring OX,x has a unique maximal ideal, which will be denoted mx.

Definition If x is a point of a scheme X, we define the residue field at P to be k(x) = OX,x/mx

Remark As we have seen, if X = Spec(R) and if x X is the point corresponding to the prime ideal P R, then OX,x = RP and mx = PRP. In addition, even though the integral domain R/P need not be a field, its field of fractions is equal to OX,x = RP/PRP.

Proposition Let R be a noetherian ring and let P in R be a prime ideal. Then the local ring RP is also noetherian.

Proof: If I is an ideal in RP, write I- = I R for its pullback to R. Since R is noetherian, the ideal I- is finitely generated by some elements a1,,ar. Now consider the ideal J in RP generated by the same elements. One has

J= (a1, , ar) I.

Now suppose x I. Write x=a/b with a, b R and b P. Since a = bx I R, we have a I-, and so a/b J. In particular, the same set of generators works.

Corollary Let P be a prime ideal in a noetherian ring R, and let m=PRP be the maximal ideal in the localization. Then the natural map R RP induces an isomorphism of k(P)-vector spaces P/P2 R/P k(P) m/m2.

Proof: This follows from the fact that we can take the same generating sets for the ideals P and m.

Definition Let x be a point in a scheme X, with local ring OX,x and maximal ideal m = mx. The Zariski tangent space to X at x is defined to be the dual k(x)-vector space

ΘX,x = Homk(x)(m/m2, k(x)).

Remark We have seen previously that another description of this vector space is as the set of morphisms Mor(T, X; x), where T is the structured algebraic set with coordinate ring k[ε]/ε2.

Example Let's start with a concrete description of the tangent space to An at the origin. The corresponding maximal ideal in the polynomial ring k[x1, ,xn] is M=(x1,,xn). So, the images of x1,,xn form a basis of M/M2. Let Di be elements of the dual basis in Θ=Hom(M/M2,k). As abstract algebraic objects, the Di are characterized by the relation Di(xj) = δij, the Kronecker delta. Now a general tangent vector is a linear operator on M/M2 that has the form Θ = i=1n ti Di for an arbitrary choice of ti k. In this way, we can identify the Θ with a copy of affine n-space. Moreover, we can give Di a more natural interpretation. Let F M be any function that vanishes at the origin. It is easy to see that

Di(F) = [ F / xi](0).

Now let Θ = i=1n ti Di be an arbitrary tangent vector. If F M, then

Θ(F) = i=1n ti [ F / xi](0).

After identifying the tangent space with An, the previous equation allows us to interpret F as a linear function on Θ, and hence as an element of the dual vector space Θ* = Hom(Θ, k).

Now consider the case of an algebraic set XAn, with x=(0, ,0). Let LAn be a line through the origin, given parametrically as {ta : t k} for some fixed point a distinct from the origin. If the ideal of X is generated by {f1,,fr}, then the intersection X L is defined by

f1(ta) = = fr(ta) = 0.

This is a system of polynomials in the single variable t (picking out a structured algebraic set on the line LA1). So, we can replace the set of polynomials by its greatest common divisor

f(t) = gcd(f1(ta), , fr(ta)) = c(t-αi)ni.

Definition Define the intersection multiplicity of X and L at 0 to be the multiplicity of the factor t in the polynomial f(t). Since we started with a line that passed through the origin, the multiplicity must be at least 1. We will say that L touches X at 0 if the intersection multiplicty is at least 2.

Theorem The Zariski tangent space of X at 0 is the union of the lines that touch X at 0.

Proof: Write

dF = i=1n [ F / xi](x) ti.

It is easy to check that

d(F+G) = dF + dG


d(FG) = F(x)d(G) + G(x)dF.

Now write I(X) = (F1, , Fr). Then the tangent space to X at x is defined by the equations

dF1 = = dFr = 0.

To see this, let M=(x1,,xn) k[x1,,xn] and let m be the image of this ideal in A(X). Then there is an exact sequence

I(X) M/M2 m/m2 0.

By duality, the tangent space to X at the origin is the subspace to the tangent space to An at the origin consisting of those linear forms Θ= tiDi that vanish on functions in I(X). However, the differential dF is just Θ(F) under the natural pairing; the claim follows.

Now let L={ta : t k } be a (parametric) line through the origin. Write

Fj(x) = i=1n ai xi + h.o.t. = dFj + Gj.


Fj(at) = (i=1n aia)t + t2*stuff.

So, t2 divides Fj(ta) if and only if dFj(a) = 0. In other words, the line L touches X at 0 if and only if the point a defining the line lies in the tangent space to X at 0.

Definition Let X be a structured algebraic set in An defined by the ideal I(X)=(F1,,Fr). Let x X be any point. The Jacobian matrix of X at x is defined to be JX(x) = (( Fi / xj)(x)).

Lemma Let XAn and let x X. If mx is the maximal ideal in A(X) of a point x X, then

dim(mx/mx2) = n - rank(JX(x)).

Proof: We have just seen that the tangent space to X at x is the null space of the Jacobian matrix. Since the tangent space is dual to mx/mx2, the result follows.

Example Let X=Z(F)An be a hypersurface. Then the tangent space to X at a point x=(x1,,xn) is defined by the single linear equation

i=1n [ F / xi](x) ti = 0.

Thus, the tangent space is either An-1 (if at least one coefficient is nonzero) or An (if all the coefficients xs vanish). One can show that if X is a variety, then there exists at least one point (and hence an entire open set) where the tangent space equals An-1.

Definition Let XAn be an affine variety of dimension d. A point x X is called nonsingular if rank(JX(x)) = n-d.

Definition Let A be a local ring with maximal ideal M. We say that A is regular if dim(M/M2) = dim(A).

Proposition Let XAn be an affine variety and let x X. Then x is a nonsingular point of X if and only if the local ring OX,x is regular.

Proof: Let M A(X) be the maximal ideal of x, and let mOX,x be the maximal ideal in the local ring. We already know that

m/m2 M/M2 Θ*X,x

and ΘX,x is isomorphic to the null space of the Jacobian. So,

dim(m/m2) + rank(JX(x)) = n.

One also has A(X)OX,x k(X). Thus, all three rings have the same transcendence degree d=dim(A(X))=dim(OX,x). So, OX,x is regular if and only if d=dim(m/m2) if and only if rank(JX(x)) = n-d if and only if x is a nonsingular point on X.

Theorem Let X be a variety. Then the set of singular points in X is a proper closed subset.

Proof: We first show that the set of singular points is closed. This is a local property. Since the affine open subsets form a basis for the topology, we can assume XAn is an affine variety of dimension d. The proof of the previous result shows that rank(JX(x)) n-d for all points x X. Thus, the set of singular points is the set where the rank is strictly less than n-d. So, the set of singular points is the algebraic set defined by the ideal generated by I(X) together with the determinants of all (n-d)×(n-d) submatrices of the Jacobian.

Next, we show that the set of singular points is proper. Since birationally isomorphic varieties have isomorphic open subsets, this property is birational. So, we can assume that X=Z(F) is an irreducible hypersurface in An. If the singular set is all of X, then one has an equality of ideals (F) = I(X) = (F, F/ x1, , F/ xn). But the partial derivatives have degree smaller than the degree of F. The equality of ideals can only hold if all the partials vanish. In characteristic 0, this can only happen if F is a constant. In characteristic p, it can only happen if F is a p-th power. Both cases are impossible, since X is an irreducible hypersurface.

  1. y2=x3+x2 has a singularity at (0,0), and no other singular points.
  2. y2=x3 has a singularity at (0,0), and no other singular points. Even though the tangent space is the same as that of the previous case, we want to think of these as different kinds of singularities; we need better invariants to do so.
  3. (x2+y2)2 + 3x2y-y3 is a three-leaved rose. It has a singularity at (0,0), and no other singular points. This singularity is a triple point.

Definition Let A be a (local) ring with (maximal) ideal M. The completion of A along M is defined to be the universal object that maps compatibly onto the natural sequence

A/Mi+1 A/Mi A/Mi-1 A/M.

The completion will be denoted by I^. The usual category theoretic notation is to write

I^ = lim A/Mi;

this is an example of an inverse limit.

Remark We can put a topology on A by taking the Mi to be the open neighborhood of 0, and making the topology translation invariant. Use Cauchy sequences to complete A in the topological sense; the result is naturally a ring, and is canonically isomorphic to the completion of A along M. It is clear either from this construction or from the universal property that the completion comes equipped with a natural map AI^.

  1. Let A=k[x] and M=(x). Then I^= k[[x]] is the ring of formal power series. The natural map k[x] k[[x]] from the universal property is the usual map identifying a polynomial with a (finite) power series.
  2. In the same way, the completion of k[x1, , xn] at the maximal ideal of the origin can be identified with the ring k[[x1,,xn]] of formal power series in n variables.
  3. Let A=Z and let M=(p) be the ideal generated by a prime p. Then I^=Zp is usually called the ring of p-adic integers. Elements of Zp can be thought of as "power series" in the variable p. More precisely, any element of Zp can be written uniquely in the form

    i=0@[infty]@ ai pi,

    where ai { 0,1,,p-1}. Actual integers are identified with their expansions in base p. You add and multiply using the usual formulas; the only essential difference is that the sum or product of two "monomials" of the same degree may not be another monomial. (This occurs because ai+bj or aibj can be bigger than p.)

We would like to try to generalize these examples, in order to think of elements in completions as power series, but in as coordinate-free a way as possible. The basic tool form commutative algebra is the following.

Theorem (Nakayama's Lemma) Let N be a module of finite type over a local ring A with maximal ideal M. If u1,,un N generate N/MN, then they also generate N.

Proof: Let S denote the submodule of N generated by the ui. Thus, the hypotheses force (N/S)/M(N/S) = 0. Since it suffices to show that (N/S)=0, we can reduce to the case that N/MN=0, where we need to show that N=0.

Let w1,,ws be generators for N. If s=1, then

N = MN = M* Aws = M* ws.

In particular, we can write ws = mws for some element m M. Hence, (1-m)ws=0. Since 1-m M, it must be a unit in the local ring A. Therefore, ws=0, and N=0.

Now suppose s>1, and use induction. Write N'=N/Aws. Then N' is generated by s-1 elements, so N'=0. Thus, N=Aws is generated by one element, and we're done.

Definition We say that a set of elements {u1,,un}OX,x forms a system of local parameters at the point x if their images form a basis of Mx/Mx2.

Corollary Let X be an algebraic set, and let {u1,,un} be a system of local parameters at a point x X. Then the homomorphism φ:k[[x1,,xn]] O^X,x that sends xi|→ ui is a surjection.

Proof: By Nakayama's Lemma, the ui generate the maximal ideal, and the monomials in the ui generate the powers of the maximal ideal.

Theorem O^X,x k[[u1,,un]] if and only if x is a nonsingular point of X.

Proof: By the previous result, φ is surjective. Now the point x is nonsingular if and only if dim(M/M2) = dim(OX,x) = dim(O^X,x). However, if φ had a kernel, the dimension would decrease. Thus, φ is also injective.

Definition Let P X and Q Y be points on algebraic sets. We say that P and Q are analytically isomrphic if there is an isomorphism O^X,P O^Y,Q.

  1. If P and Q are analytically isomorphic points on algebraic varieties X and Y, then dim(X)=dim(Y).
  2. If P and Q are nonsingular points on two varieties of the same dimension, then P and Q are analytically isomorphic.
  3. Let P be the origin on the nodal cubic curve y2=x3+x2, and let Q be the origin on the reducible curve xy=0 consisting of the two coordinate axes. Then P and Q are analytically isomorphic. To prove this, we need to exhibit an isomorphism between the two rings

    O^P = k[[x,y]]/(y2-x2-x3) and O^Q = k[[x,y]]/(xy)

    So, we have to find power series g, h k[[x,y]] such that
    g = (y+x) + g2 + g3 +
    h = (y-x) + h2 + h3 +
    gh = y2 - x2 - x3.

    We start by writing

    gh = y2-x2 +g2(y-x)+h2(y+x) + (degree3).


    g2 (y-x) + h2(y+x) = x3.

    We can take g2 = xy+[1 / 2]y2 and h2=x2-[1 / 2]y2. Next, we need to solve

    (y-x)g3 + (y+x)h3 = -g2h2.

    We can always find such a solution (because y-x and (y+x) generate the maximal ideal). Continue until you build the entire power series. In this example, it is interesting to note that the ring OP is an integral domain, but its completion O^P is not.
  4. Let XA2 be an irreducible plane curve and let P=(0,0) be a point of X. The irreducible function cutting out X has the form

    f = fr + fr+1 + ,

    where each fi is a homogeneous polynomial of degree i. In this case, we say that P is an r-fold point of X. If fr is a product of r distinct linear factors, then we say that P is ordinary. When r=1, the point P is nonsingular. Any two ordinary double points are analytically isomorphic. Any two ordinary triple points are analytically isomorphic. However, there is a one-parameter family of different ordinary fourfold points!

Comments on this web site should be addressed to the author:

Kevin R. Coombes
Department of Biomedical Informatics
The Ohio State University
Columbus, Ohio 43210