Course Notes for Math 738 (Algebraic Geometry)

16. Varieties. Differentials and Tangents. Kähler differentials are introduced and used to get an intrinsic definition of the nonsingular locus of a variety, which happens to always be an open subset. The Zariski tangent space is introduced, as well as the tangent cone which is an intrinsic affine subvariety. Finally, we define the blow-up of a variety at a point, and show how it replaces the point with its projectivized tangent cone.

Motivation: Suppose k = R or C and f_{1}, ..., f_{r} ∈ k[x_{1}, ..., x_{n}] are given, together with a point p ∈ V (f_{1}, ..., f_{r}). Then recall that the implicit function theorem tells us V (f_{1}, ..., f_{r}) (which one should think of as the fiber Φ^{−1}(0) of the map Φ : k^{n} → k^{r} given by Φ(p) = (f_{1}(p), ..., f_{r}(p))) is a submanifold of k^{n} in a neighborhood of p if the following Jacobian matrix has full rank:

| \begin{matrix}\frac{∂f_{1}}{∂x_{1}} (p) & \frac{∂f_{1}}{∂x_{2}} (p) & ⋯ & \frac{∂f_{1}}{∂x_{n}} (p) \\ \frac{∂f_{2}}{∂x_{1}} (p) & \frac{∂f_{2}}{∂x_{2}} (p) & ⋯ & \frac{∂f_{2}}{∂x_{n}} (p) \\ ⋮ & ⋮ & ⋮ & ⋮ \\ \frac{∂f_{r}}{∂x_{1}} (p) & \frac{∂f_{r}}{∂x_{2}} (p) & ⋯ & \frac{∂f_{r}}{∂x_{n}} (p) \end{matrix} |

A variety X will be nonsingular at p if “X can defined by equations with a Jacobian matrix of the right rank at p”. To get an intrinsic definition for this, we introduce some notions from commutative algebra:

Suppose A is a k-algebra and M is a module over A.

Definition: A derivation (or k-derivation) d : A → M is a homomorphism of the abelian groups under addition, which additionally satisfies:

d(aa′) = a(da′) + a′(da)

(i.e. Leibniz’ rule) and also:

dc = 0 for all c ∈ k.

Key Observation: If d : A → M is a derivation and α : M → N is an A-module homomorphism, then δ := α ∘ d : A → N is also a derivation. In fact, precomposing with α induces an A-module homomorphism:

h_{d,N} : Hom_{A}(M, N) → {derivations δ : A → N}

Definition: A derivation d : A → Ω_{A} is universal if h_{d,M} is an isomorphism for every A-module M. In other words, every derivation δ : A → M is the composition α ∘ d for a unique homomorphism α : Ω_{A} → M of A-modules.

Remark: By the usual general nonsense, if d : A → Ω_{A} and d′ : A → Ω′_{A} are universal, then Ω_{A} ≅ Ω′_{A} by a uniquely determined isomorphism. We reserve the notation Ω_{A} for “the” A-module associated to a universal derivation. This A-module is called the module of Kähler differentials on A.

Proposition 16.1: (a) Universal derivations always exist.

(b) If A ≅ k[x_{1}, ..., x_{n}]/I, then Ω_{A} is finitely generated as an A-module. Precisely, if I = ⟨f_{1}, ..., f_{m}⟩, then Ω_{A} is isomorphic to the quotient of the free module ⊕_{i=1}^{n}A(dx_{i}) by the submodule generated by the df_{j} := ∑_{i=1}^{n} \frac{∂f_{j}}{∂x_{i}} dx_{i}.

Proof: (a) Given A, let F be the free A-module generated by the symbols {da | a ∈ A} and let N ⊆ F be the submodule generated by the relations:

(i) d(a + a′) − da − da′ for all a, a′ ∈ A,

(ii) d(aa′) − a(da′) − a′(da) for all a, a′ ∈ A and

(iii) dc for all c ∈ k.

Then Ω_{A} := F/N with natural derivation d : A → Ω_{A} sending a ↦ da is universal. If A ≅ k[x_{1}, ..., x_{n}], then F/N ≅ ⊕_{i=1}^{n}A(dx_{i}), and the universal d : k[x_{1}, ..., x_{n}] → ⊕_{i=1}^{n}k[x_{1}, ..., x_{n}]dx_{i} is given by d(g) = ∑_{i=1}^{n} \frac{∂g}{∂x_{i}} dx_{i}.

If A ≅ k[x_{1}, ..., x_{n}]/⟨f_{1}, ..., f_{m}⟩, let K ⊆ ⊕_{i=1}^{n}A(dx_{i}) be the submodule generated by the images of the df_{j}. If g = ∑_{j=1}^{m} h_{j}f_{j} ∈ I, then

∑_{i=1}^{n} \frac{∂g}{∂x_{i}} dx_{i} = ∑_{j=1}^{m}(∑_{i=1}^{n} \frac{∂h_{j}}{∂x_{i}} dx_{i})f_{j} + ∑_{j=1}^{m} h_{j}(∑_{i=1}^{n} \frac{∂f_{j}}{∂x_{i}} dx_{i})

so δ : A → D/K given by δ(a) = ∑_{i=1}^{n} \frac{∂g}{∂x_{i}} dx_{i} (for any g ∈ k[x_{1}, ..., x_{n}] which maps to a ∈ k[x_{1}, ..., x_{n}]/I) is a well-defined universal derivation.

Example: If A = k[x, y]/⟨xy⟩, then Ω_{A} = Adx⊕Ady/⟨xdy+ydx⟩. Note that the nonzero element ω := xdy = −ydx has the property that xω = yω = 0. Thus Aω = kω ⊂ Ω_{A}. The quotient module Ω_{A}/kω ≅ k[x]dx ⊕ k[y]dy looks like the sum of the modules of differentials on the x and y axes respectively.

Here is a more natural construction of Ω_{A}. Given a k-algebra A, let μ : A ⊗_{k} A → A be the multiplication homomorphism μ(∑ a ⊗ a′) = ∑ aa′, and consider the ideal J := ker(μ) ⊂ A ⊗_{k} A. Since J is an A ⊗_{k} A-module, it follows that J/J^{2} is an A = A ⊗_{k} A/J-module.

Proposition 16.2: The map d : A → J/J^{2} defined by setting d(a) to be the image of 1 ⊗ a − a ⊗ 1 is a universal derivation.

Proof: I’m going to pass freely between elements of J/J^{2} and their representatives in J. For instance, if m ∈ J/J^{2} and a ∈ A, then we can compute am by taking a lift of m to J, multiplying by a ⊗ 1 or 1 ⊗ a and passing back down to J/J^{2} (why?).

From the identity: a⊗a′ +a′ ⊗a = 1⊗aa′ +aa′ ⊗1 (modulo J^{2}) it follows that d(aa′) = 1⊗aa′−aa′⊗1 = (a⊗1)(1⊗a′−a′⊗1)+(a′⊗1)(1⊗a−a⊗1) = a(da′)+a′(da). Of course, dc = 1⊗c−c⊗1 = 0 for c ∈ k, so d is a derivation.

To see that d : A → J/J^{2} is universal, suppose we are given a derivation δ : A → M. We obtain a ring R = A ⊕ M by setting:

(a, m) · (a′, m′) = (aa′, am′ + a′m).

The map f : A × A → R given by f(a, a′) = (aa′, a(δa′)) is bilinear and induces a ring homomorphism f : A ⊗_{k} A → R because δ is a derivation. Moreover, f(J) ⊂ M, by definition, and it follows that f(J^{2}) ≡ 0, so we obtain an A-module homomorphism \overline{f} : J/J^{2} → M. Since \overline{f}(1 ⊗ a − a ⊗ 1) = δa, (check this!) we see that δ = \overline{f} ∘ d. Exercise: Why is \overline{f} the unique A-module homomorphism with this property?

We need to extend our notion of localizing to modules:

Definition: If M is an A-module and S ⊂ A is a multiplicative set, then:

M_{S} := \{\frac{m}{s} | m ∈ M, s ∈ S\}/ ∼

where \frac{m}{s} ∼ \frac{m′}{s′} iff a(ms′ − m′s) = 0 for some a ∈ S. Observe that M_{S} is naturally an A_{S}-module.

Universal derivations localize to universal derivations:

Proposition 16.3: If d : A → Ω_{A} is a universal derivation and S ⊂ A is a multiplicative set, then d_{S} : A_{S} → (Ω_{A})_{S} defined by the quotient rule:

d_{S}(\frac{a}{s}) := \frac{s(da) − a(ds)}{s^{2}}

is a universal derivation. So (Ω_{A})_{S} ≅ Ω_{A_{S}} by a canonical isomorphism.

Proof: Exercise.

We will need the following standard results from field theory:

Existence of a primitive element: If K ⊂ L is a separable field extension, then there is an element α ∈ L such that L = K[α].

Existence of a separating basis: If k ⊂ K is a field extension of finite type (see §4) and k is algebraically closed (or more generally “perfect”), then there is a trancendence basis α_{1}, ..., α_{n} for K such that the field extension k(α_{1}, ..., α_{n}) ⊂ K is separable. Such a basis is referred to as separating.

We assume k is algebraically closed for the rest of this section.

Let X be a variety over k. To each point p ∈ X, we have the module of Kähler differentials Ω_{O_{X},p} over the local ring O_{X,p} (with maximal ideal m_{p}). This is finitely generated by Propositions 16.1(b), 16.3 and HW #1(a). Thus

Ω_{X}(p) := Ω_{O_{X,p}}/m_{p}Ω_{O_{X,p}}

is a finite dimensional vector space over k = O_{X,p}/m_{p}O_{X,p}.

Theorem 16.4: (a) For each p ∈ X, there is a canonical isomorphism:

Hom_{k}(Ω(p), k) ≅ Hom_{k}(m_{p}/m_{p}^{2}, k)

(this k-vector space is called the Zariski tangent space to X at p.)

(b) The function e : X → R defined by:

e(p) := dim_{k}(m_{p}/m_{p}^{2})

is upper-semi continuous. Moreover, e(p) ≥ dim(X) for all p ∈ X and e(p) = dim(X) on a nonempty open subset of X.

Proof: Let A = O_{X,p}, and k = A/m_{p} as an A-module. We start with the canonical isomorphism of A-modules:

Hom_{A}(Ω_{A}, k) ≅ {derivations δ : A → k}

On the left side, Hom_{A}(Ω_{A}, k) = Hom_{k}(Ω(p), k) (HW # 1(e)).

On the right side, a derivation δ : A → k has the property that δ(m_{p}^{2}) ≡ 0, so it certainly determines a linear map \overline{δ} : m_{p}/m_{p}^{2} → k. On the other hand, given l ∈ Hom_{k}(m_{p}/m_{p}^{2}, k), define δ_{l}(a) := l(a − a(p)). Then

δ_{l}(ab) = l(ab − a(p)b(p)) = a(p)l(b − b(p)) + b(p)l(a − a(p))

as you can easily check, so δ_{l} is a derivation, and (a) follows.

Next, we use (a) to prove (b) since now e(p) = dim_{k}(Ω_{X}(p)). Choose an affine neighborhood V of p and suppose k[V] ≅ k[x_{1}, ..., x_{n}]/⟨f_{1}, ..., f_{m}⟩. There are two important observations to make. First, by Proposition 16.3, we have (Ω_{k[V]})_{m_{p}} ≅ Ω_{O_{X},p}, and it follows from HW #1(b) that:

Ω_{X}(p) ≅ (Ω_{k[V]})_{m_{p}}/m_{p}(Ω_{k[V]})_{m_{p}} ≅ Ω_{k[V]}/m_{p}Ω_{k[V]}

(we sloppily let m_{p} stand for both the maximal ideal in k[V] and in O_{X,p}.) Second, using 16.1(b) and HW # 1(c), we have:

Ω_{k[V]}/m_{p}Ω_{k[V]} ≅ ⊕_{i=1}^{n}k(dx_{i})/⟨∑_{i=1}^{n} \frac{∂f_{1}}{∂x_{i}} (p)dx_{i}, ..., ∑_{i=1}^{n} \frac{∂f_{m}}{∂x_{i}} (p)dx_{i}⟩

From this we conclude that dim_{k}(Ω_{X}(p)) = n − rank(|\frac{∂f_{j}}{∂x_{i}} (p)|) where |\frac{∂f_{j}}{∂x_{i}} (p)| is the Jacobian matrix associated to the f_{j}. By the same reasoning, this is true for all q ∈ V. The set of q ∈ V such that rank(|\frac{∂f_{j}}{∂x_{i}} (q)|) ≤ n − α is closed in V since it is given by the vanishing of the appropriate minors of a matrix of polynomials, hence e : X → R is upper-semi continuous on each affine open V and it follows that e is upper-semi continuous.

To prove the last assertions, it suffices by upper-semi continuity to prove that e(p) = n := dim(X) for all p in an open (hence dense) subset of X. Let V ⊂ X be an open affine subset. The algebra results listed above imply that k(V) ≅ k(x_{1}, ..., x_{n})[x_{n+1}]/f(x_{n+1}), and we may as well assume, clearing denominators, that f ∈ k[x_{1}, ..., x_{n+1}] is irreducible. It follows that k(V (f)) ≅ k(V) for the hypersurface V (f) ⊂ A_{k}^{n+1}.

Thus there is a birational isomorphism Φ : V − − > V (f) ⊂ A_{k}^{n+1}, hence a birational regular map Φ : U → V (f) from the domain of Φ. By Proposition 15.13, such a map is an isomorphism from some open subset of U to an open subset of V (f), so it suffices to prove that dim_{k}(Ω_{V (f)})(q) = n for all q in an open subset of V (f). Here the Jacobian matrix is the “gradient” ∇f := (\frac{∂f}{∂x_{1}} , ...\frac{∂f}{∂x_{n+1}} ) so e(q) = n iff some \frac{∂f}{∂x_{i}} (q) ≠ 0 (e(q) = n + 1 otherwise). Suppose f has degree N. Then each \frac{∂f}{∂x_{i}}, if nonzero, is a polynomial of degree N − 1, and by the irreducibility of f, it is impossible that V (f) ⊆ V (\frac{∂f}{∂x_{i}}). In characteristic zero, the gradient is never identically zero. But in characteristic p > 0, this happens exactly when every variable appears in every term of f as a pth power(!) But this would imply f = g^{p} for some g ∈ k[x_{1}, ..., x_{n+1}], violating the irreducibility of f. Thus the open subset of V (f) where the gradient of f fails to vanish is not empty and the proof is complete.

Remarks: Some important results are embedded in this proof(!)

(i) Every variety over k has a nonempty open subset which is isomorphic to an open subset of an affine hypersurface.

(ii) If p ∈ V ⊂ X is an affine neighborhood of p, then we can compute dim(m_{p}/m_{p}^{2}) as the corank in k^{n} of the Jacobian matrix |\frac{∂f_{j}}{∂x_{i}} (p)| for any set of generators f_{1}, ..., f_{m} for the ideal of V ⊂ A_{k}^{n}.

Definition: A point p ∈ X is nonsingular if dim_{k}(m_{p}/m_{p}^{2}) = dim(X). The variety X is nonsingular if every point is nonsingular.

By the theorem, the set of nonsingular points of a variety is always open.

Examples: (a) A_{k}^{n} and P_{k}^{n} are nonsingular.

(b) If f ∈ k[x_{1}, ..., x_{n}] is irreducible then V (f, ∇f) := V (f, \frac{∂f}{∂x_{1}} , ...) = ∅ if and only if V (f) ⊂ A_{k}^{n} is nonsingular.

(c) (Exercise) If F ∈ k[x_{0}, ..., x_{n}]_{d} is homogeneous and irreducible, then V (∇F) = ∅ if and only if V (F) ⊂ P_{k}^{n} is nonsingular. For example, the “Fermat” hypersurfaces V (x_{0}^{m} + ... + x_{n}^{m}) ⊂ P_{k}^{n} are nonsingular. If char(k) divides m, then what happens?

(d) If V = V (f_{1}, ..., f_{m}) ⊂ A_{k}^{n} is a veriety of dimension n − m, then p ∈ V is a nonsingular point if and only if the Jacobian matrix |\frac{∂f_{j}}{∂x_{i}} | has rank m at p (hence also in an open neighborhood of p). Compare this with the motivation at the beginning of this section.

Corollary 16.5: The local ring O_{X,p} is a DVR if and only if dim(X) = 1 and p is a nonsingular point of X.

Proof: If dim(X) > 1 or p is a singular point, then dim_{k}(m_{p}/m_{p}^{2}) > 1 by the theorem, hence it certainly requires more than one element to generate m_{p} (why?). On the other hand, if p is a nonsingular point and dim(X) = 1, then dim_{k}(m_{p}/m_{p}^{2}) = 1, and if a ∈ m_{p} is an element with a nonzero image in m_{p}/m_{p}^{2}, then the map O_{X,p} → m_{p} given by b ↦ ba must be surjective by Nakayama’s lemma, since it is surjective after taking the quotient by m_{p}(!) So m_{p} is principal, and O_{X,p} is a DVR.

Here is another characterization of a nonsingular point:

Proposition 16.6: p ∈ X is nonsingular iff Ω_{O_{X,p}} is a free O_{X,p}-module.

Proof: Suppose dim_{k}(m_{p}/m_{p}^{2}) = dim_{k}(Ω_{X}(p)) = r. Then as in the corollary above, if we choose elements a_{1}, ..., a_{r} ∈ O_{X,p} such that the images of da_{1}, ..., da_{r} are a basis for Ω_{X}(p), then Nakayama’s lemma applied to:

⊕_{i=1}^{r}O_{X,p} → Ω_{O_{X,p}} given by (b_{1}, ..., b_{r}) ↦ b_{1}da_{1} + ... + b_{r}da_{r}

shows that the da_{i} generate Ω_{O_{X,p}}. If V is an affine neighborhood of p and Ω_{k[V]} has generators α_{1}, ..., α_{n}, then we can solve α_{j} = ∑_{i=1}^{r} b_{i,j}da_{i} for b_{i,j} ∈ O_{X,p} and taking f to be the product of denominators of the b_{i,j} and da_{i} (thought of as elements of (Ω_{k[V]})_{m_{p}}), it follows that the da_{i} generate Ω_{k[U_{f}]}.

Consider the kernel K of the surjective homomorphism of k[U_{f}]-modules:

⊕_{i=1}^{r}k[U_{f}] → Ω_{k[U_{f}]} given by (b_{1}, ..., b_{r}) ↦ b_{1}da_{1} + ... + b_{r}da_{r}.

If r = dim(X), then since the da_{i} generate Ω_{O_{X,q}} for every q ∈ U_{f}, it follows in particular that U_{f} is contained in the nonsingular locus of X. Moreover, if κ = (g_{1}, ..., g_{r}) ∈ K, then the functions g_{i} ∈ k[U_{f}] must evaluate to zero on every q ∈ U_{f}, otherwise we would have dim(Ω_{X}(q)) < dim(X). But this implies κ = 0, hence Ω_{k[U_{f}]} is free, and so is every Ω_{O_{X,q}} for q ∈ U_{f}.

If r > dim(X), then Ω_{O_{X,p}}, if free, would require r generators. As above, we would then be able to find an affine neighborhood p ∈ U_{f} such that Ω_{k[U_{f}]} is generated by the same elements. Then Ω_{k[U_{f}]} would be free of rank r since any relation among the generators would persist after localizing at m_{p}. Thus dim_{k}(Ω_{X}(q)) = r for all q ∈ U_{f} which contradicts the fact that U_{f} intersects the nonsingular locus of X (which is dense in X).

Next, I want to take a more concrete point of view of the Zariski tangent space as a “linear” approximation to a variety X at a point p ∈ X. We will also introduce the tangent cone to X at p which turns out to be a better nonlinear approximation when p is a singular point. We initially work locally.

Suppose p = (0, ..., 0) ∈ V ⊂ A_{k}^{n} and let I = I(V) ⊂ k[x_{1}, ..., x_{n}]. Notice that if f ∈ I and f = f_{0} + f_{1} + ... + f_{d} is its decomposition as a sum of homogeneous polynomials, then f_{0} = 0. If f = f_{e} + ... + f_{d} and f_{e} ≠ 0, then we will call f_{e} the first homogeneous term of f.

Definitions: (a) Let I_{lin} be the ideal generated by the linear terms of the elements of I. It follows that V (I_{lin}) is an sub-vector space of k^{n} = A_{k}^{n}. Let

T_{p}V := V (I_{lin}) ⊂ A_{k}^{n}.

be this “explicit tangent space” to V at p.

(b) Let I_{init} be the ideal generated by the first homogeneous terms of the elements of I. Then V (I_{init}) is a cone over a closed subset of P_{k}^{n−1}. Let

C_{p}V := V (I_{init}) ⊂ A_{k}^{n}

be this “explicit tangent cone” to V at p.

Observations: (a) Since I_{lin} ⊆ I_{init}, we have C_{p}V ⊆ T_{p}V.

(b) The linear terms in a collection of generators for I will also generate I_{lin}. However, the first nonvanishing homogeneous terms in a collection of generators for I need not generate I_{init}. If I is principal, though, then I_{init} is also principal, as you can check.

Examples: (a) (Nodal plane cubic) If V = V (y^{2} − x^{2}(x + 1)) ⊂ A_{k}^{2}. Then T_{p}V = A_{k}^{2}, and C_{p}V = V (x^{2} − y^{2}) = V (x − y) ∪ V (x + y) is a pair of lines intersecting at the origin.

(b) (Cuspidal plane cubic) Suppose V = V (y^{2} − x^{3}) ⊂ A_{k}^{2}. Then again T_{p}V = A_{k}^{2}, but C_{p}Y = V (y^{2}). Here the definition of C_{p}V really is not right. Although the tangent cone looks linear, we will want to remember that the equation is a square. (Schemes would help us here!)

(c) Suppose I = ⟨F_{1}, ..., F_{m}⟩ where the F_{i} are homogeneous, that is, suppose V is the cone over a projective variety. Then it is easy to see that I_{init} = I in this case, so C_{p}V = V.

Suppose now that p ∈ V ⊂ X is an affine neighborhood in a variety over k, and V is identified with a closed, irreducible subset of A_{k}^{n} in such a way that p is identified with the origin. If f ∈ I(V), then f_{1} is transformed into df(p) when we replace the x_{i}’s by dx_{i}’s. Remark (ii) following Theorem 16.4 then implies that T_{p}V is isomorphic to the Zariski tangent space. In particular, the dimension of T_{p}V does not depend upon the choice of an affine neighborhood of p. In the next proposition, we show that the tangent cone, too, is intrinsic. In other words we find an affine “cone” in the Zariski tangent space at p (thought of as affine space) which is identified with C_{p}V for every choice of an affine neighborhood of p.

Proposition 16.7: If p ∈ V ⊂ X as above, then there are isomorphisms:

τ : Sym^{∗}(m_{p}/m_{p}^{2}) →^{∼} k[x_{1}, ..., x_{n}]/I_{lin} and

γ : ⊕_{d=0}^{∞}m_{p}^{d}/m_{p}^{d+1} →^{∼} k[x_{1}, ..., x_{n}]/I_{init}

as graded k-algebras such that the following diagram commutes:

Sym^{∗}(m_{p}/m_{p}^{2}) →^{ρ} ⊕_{d=0}^{∞}m_{p}^{d}/m_{p}^{d+1}
τ ↓  γ ↓
k[x_{1}, ..., x_{n}]/I_{lin} →^{r} k[x_{1}, ..., x_{n}]/I_{init}

where m_{p} ⊂ O_{X,p} is the maximal ideal, and ρ and r are the natural quotients.

In particular, the ideal J := I_{init}/I_{lin} ⊆ k[x_{1}, ..., x_{n}]/I_{lin} which has the property that V (J) = C_{p}V ⊆ T_{p}V, is mapped isomorphically by τ^{−1} to the ideal ker(ρ) ⊂ Sym^{∗}(m_{p}/m_{p}^{2}). Thus V (ker(ρ)) defines an intrinsic tangent cone sitting in the Zariski tangent space to X at p.

Proof: Let’s start with some notation:

m = ⟨x_{1}, ..., x_{n}⟩ ⊂ k[x_{1}, ..., x_{n}],

m̃_{p} = m/I ⊂ k[V] is the maximal ideal of p ∈ V, and

m_{p} is the corresponding ideal in k[V]_{m_{p}} ≅ O_{X,p}

For each d ≥ 1, we have natural isomorphisms:

m_{p}^{d}/m_{p}^{d+1} ≅ m̃_{p}^{d}/m̃_{p}^{d+1} ≅ m^{d}/(m^{d+1} + I ∩ m^{d}) = (k[x_{1}, ..., x_{n}]/I_{init})_{d}.

The first isomorphism is the isomorphism of HW # 1(b), the second is a fundamental isomorphism for ideals once we observe that m̃_{p}^{d} = m^{d}/I, and the third equality follows from the definition of I_{init}. Notice also that (k[x_{1}, ..., x_{n}]/I_{init})_{1} = (k[x_{1}, ..., x_{n}]/I_{lin})_{1}. One can check that these isomorphism carry products in ⊕_{d=0}^{∞}m_{p}^{d}/m_{p}^{d+1} to products in k[x_{1}, ..., x_{n}]/I_{init}, yielding the isomorphism γ. The isomorphism τ of graded k-algebras is uniquely determined by its degree one isomorphism m_{p}/m_{p}^{2} ≅ (k[x_{1}, ..., x_{n}]/I_{lin})_{1} since Sym^{∗}(m_{p}/m_{p}^{2}) is a polynomial ring and k[x_{1}, ..., x_{n}]/I_{lin} is generated by its degree one part, and also a polynomial ring. This also explains why the two k-algebra homomorphisms from Sym^{∗}(m_{p}/m_{p}^{2}) to k[x_{1}, ..., x_{n}]/I_{init}, which agree in degree one, must agree in all degrees.

Finally, notice that if V is a vector space over k of dimension n, then Sym^{∗}(V) is naturally the coordinate ring of the dual space Hom_{k}(V, k) viewed as affine n-space. Thus V (ker(ρ)) may be viewed as a closed cone in the Zariski tangent space Hom_{k}(m_{p}/m_{p}^{2}, k) when thought of as affine space.

Remark: In the proof, I have had to dance around the statement I would have liked to make, namely: “V (ker(ρ)) is isomorphic to C_{p}V ”. If C_{p}V is reducible, we could finesse this by saying that the components are isomorphic. But even when C_{p}V is irreducible, I actually have more information than the mere isomorphism of affine varieties, since I know that the defining ideals are isomorphic (think of the cusp example above). With the theory of schemes, one can more simply define the tangent cone as the affine scheme associated to the ring ⊕_{d=0}^{∞}m_{p}^{d}/m_{p}^{d+1}, which remembers the isomorphism of ideals.

The tangent cone allows us to define types of singularities. E.g.

Definition: A singular point p ∈ X is called a hypersurface singularity if dim_{k}(m_{p}/m_{p}^{2}) = dim(X) + 1. A hypersurface singularity of a curve is called a plane curve singularity. A plane curve singularity is ordinary if ker(ρ) (which is a homogeneous ideal in k[x, y]) is principal, generated by a quadratic polynomial. It is an:

(a) ordinary node if the generator is a product of distinct linear factors.

(b) ordinary cusp if the generator is a square.

I want to prove that the tangent cone to X at p always has the same dimension as X (unlike the Zariski tangent space, which jumps in dimension). The most convenient way to do this is to introduce the notion of blowing up.

Blowing up a point in affine space: Consider the rational map:

Φ : A_{k}^{n} − − > P_{k}^{n−1} given by Φ(p_{1}, ..., p_{n}) = (p_{1} : ... : p_{n})

in other words, Φ maps a point p ∈ A_{k}^{n} (other than the origin) to the point in projective space corresponding to the line through p and the origin. Let

Γ = {((p_{1}, ..., p_{n}), (p_{1} : ... : p_{n})) | (p_{1}, ..., p_{n}) ≠ (0, ..., 0)} ⊂ A_{k}^{n} × P_{k}^{n−1}

i.e. Γ is the graph of Φ, and let Ã_{k}^{n} := \overline{Γ} be its closure in A_{k}^{n} × P_{k}^{n−1}.

Consider the polynomial ring k[x_{1}, ..., x_{n}, y_{1}, ..., y_{n}] graded by degree in the y-variables as in the proof of Theorem 14.11, so that in particular, the closed subsets of A_{k}^{n} × P_{k}^{n−1} are all of the form V (I) for homogeneous ideals I by the claim in that theorem. Then:

Theorem 16.8: (a) The first projection restricts to a birational regular map:

σ : Ã_{k}^{n} → A_{k}^{n}

(σ is variously called a blow-down, σ-process or monoidal transformation) More precisely, σ^{−1}(0, ..., 0) = {(0, ..., 0)} × P_{k}^{n−1} and σ is an isomorphism elsewhere.

(b) Ã_{k}^{n} = V (⟨x_{i}y_{j} − x_{j}y_{i}⟩_{i,j=1}^{n})

(c) The open sets U_{i} := Ã_{k}^{n} − V (y_{i}) are isomorphic to A_{k}^{n} with natural isomorphisms k[U_{i}] ≅ k[\frac{y_{1}}{y_{i}} , ..., \frac{y_{i−1}}{y_{i}} , x_{i}, \frac{y_{i+1}}{y_{i}} , ..., \frac{y_{n}}{y_{i}}].

Proof: We start with the affines U_{y_{i}} := A_{k}^{n} × P_{k}^{n−1} − V (y_{i}) ≅ A_{k}^{n+(n−1)} with coordinate rings k[x_{1}, ..., x_{n}, \frac{y_{1}}{y_{i}} , ..., \frac{y_{n}}{y_{i}}]. The equations x_{i}y_{j} = x_{j}y_{i} may be rewritten as x_{j} = x_{i}(\frac{y_{j}}{y_{i}}) on k[U_{y_{i}}], so we obtain the isomorphism:

U_{y_{i}} ∩ V (⟨x_{i}y_{j} − x_{j}y_{i}⟩) ≅ A_{k}^{n}, with k[A_{k}^{n}] = k[\frac{y_{1}}{y_{i}} , ..., \frac{y_{i−1}}{y_{i}} , x_{i}, \frac{y_{i+1}}{y_{i}} , ..., \frac{y_{n}}{y_{i}}]

(in particular, check that the other equations x_{k}y_{l} − x_{l}y_{k} are redundent.)

Let W_{i} = U_{y_{i}} ∩ V (⟨x_{i}y_{j} − x_{j}y_{i}⟩_{i,j=1}^{n}) for the moment. Since the U_{y_{i}} cover A_{k}^{n} × P_{k}^{n−1}, since each of the W_{i} is irreducible, and finally since each intersection W_{i} ∩ W_{j} is nonempty (as you may easily check), it follows that V (⟨x_{i}y_{j} − x_{j}y_{i}⟩) = ∪_{i=1}^{n}W_{i} is irreducible.

If (p_{1}, ..., p_{n}) ≠ (0, ..., 0), then p_{i}q_{j} = p_{j}q_{i} for all i and j if and only if (p_{1} : ... : p_{n}) = (q_{1} : ... : q_{n}), so Γ = V (⟨x_{i}y_{j} − x_{j}y_{i}⟩) − {(0, ..., 0)} × P_{k}^{n−1}, hence Γ ⊂ V (⟨x_{i}y_{j} − x_{i}y_{i}⟩) is open, which allows us to conclude (b) and (c) where now U_{i} = W_{i}.

One checks directly that σ^{−1}(0, ..., 0) = (0, ..., 0) × P_{k}^{n−1} using (b). The first projection retricts on U_{i} to σ : U_{i} → A_{k}^{n} corresponding to the inclusion:

ι : k[x_{1}, ..., x_{n}] ↪ k[\frac{y_{1}}{y_{i}} , ..., \frac{y_{i−1}}{y_{i}} , x_{i}, \frac{y_{i+1}}{y_{i}} , ..., \frac{y_{n}}{y_{i}}] sending x_{j} ↦ x_{i} (\frac{y_{j}}{y_{i}} ) .

Notice that σ^{−1}(0, ..., 0) ∩ U_{i} = V (x_{i}). On the coordinate rings of the open complements of V (x_{i}) ⊂ U_{i} and V (x_{i}) ⊂ A_{k}^{n} the map ι localizes to:

ι_{x_{i}} : k[\frac{x_{1}}{x_{i}} , ..., \frac{x_{n}}{x_{i}}] → k[\frac{y_{1}}{y_{i}} , ..., \frac{y_{n}}{y_{i}}] sending \frac{x_{j}}{x_{i}} ↦ \frac{y_{j}}{y_{i}} ,

and it follows that σ becomes an isomorphism from each U_{i} − σ^{−1}(0, ..., 0) = U_{i} − V (x_{i}) to its image. You check now that (id, Φ) furnishes an inverse to σ : Ã_{k}^{n} − σ^{−1}(0, ..., 0) → A_{k}^{n} − (0, ..., 0) which gives the rest of (a).

Example: We have already encountered a blow-down! If n = 2, then the restriction σ : U_{2} → A_{k}^{2} corresponds to the inclusion k[x, y] ↪ k[x, \frac{x}{y}] which means that this restriction of σ is none other than the map (x, y) ↦ (x, xy) we have considered before.

Remark: Notice that Ã_{k}^{n} is nonsingular, since it is covered by affine spaces(!) Notice also that the “exceptional” fiber σ^{−1}(0, ..., 0) is locally given by one equation (namely the equations x_{i} = 0 on each U_{i}). Thus blowing up replaced the origin in A_{k}^{n} with a copy of P_{k}^{n−1} (which you should think of as the collection of lines through the origin), which is locally described by a single equation. This becomes quite important in the next theorem.

Suppose V ⊂ A_{k}^{n} is a closed, irreducible subset passing through the origin. Let Ṽ ⊂ A_{k}^{n} × P_{k}^{n−1} be the closure of the graph of Φ restricted to V, that is, the closure of the graph Γ_{V} := {((p_{1}, ..., p_{n}), (p_{1} : ... : p_{n})) | (p_{1}, ..., p_{n}) ∈ V }.

Let P(C_{p}(V)) := V (I_{init}) ⊂ P_{k}^{n−1} be the “projectivized tangent cone”. Of course, it may have many components. But one consequence of the next theorem is that each component has the same dimension.

Let I = I(V). For each f ∈ I(V), let mult_{p}(f) be the degree of the first homogeneous term in f. Notice that if mult_{p}(f) = e, then

\frac{f}{x_{i}^{e}} = f_{e}(\frac{x_{1}}{x_{i}} , ..., \frac{x_{n}}{x_{i}}) + x_{i}f_{e+1}(\frac{x_{1}}{x_{i}} , ..., \frac{x_{n}}{x_{i}}) + ... + x_{i}^{d−e}f_{d}(\frac{x_{1}}{x_{i}} , ..., \frac{x_{n}}{x_{i}})

Substituting \frac{y_{j}}{y_{i}} for each \frac{x_{j}}{x_{i}} gives an element of k[U_{i}] from part (c) of the previous theorem, which we will be sloppy and denote by \frac{f}{x_{i}^{e}}. For each

i = 1, ..., n let Ĩ_{i} ⊂ k[U_{i}] be the ideal generated by the \frac{f}{x_{i}^{e}} for each f ∈ I.

Exercise: g ∈ Ĩ_{i} iff g = x_{i}^{m}(\frac{f}{x_{i}^{e}}) for some f ∈ I, m ≥ 0.

Theorem 16.9: (a) The first projection restricts to a birational regular map:

σ : Ṽ → V

such that σ^{−1}(0, ..., 0) = {(0, ..., 0)} × P(C_{p}(V)) and σ is an isomorphism elsewhere.

(b) Each Ĩ_{i} ⊂ k[U_{i}] is a prime ideal, and Ṽ ∩ U_{i} = V (Ĩ_{i}) ⊂ U_{i}.

(c) Each intersection σ^{−1}(0, ..., 0) ∩ U_{i} ⊂ Ṽ ∩ U_{i} is the zero locus of the (non-zero!) image of x_{i} in k[V (Ĩ_{i})] = O_{V}(Ṽ ∩ U_{i}).

Proof: Our first goal is to prove that ∪_{i=1}^{n}V (Ĩ_{i}) is a closed, irreducible subset of Ã_{k}^{n} = ∪_{i=1}^{n}U_{i}. To see that it is closed, it suffices to check that for each i, ∪_{i=1}^{n}V (Ĩ_{i}) ∩ U_{i} is closed. In fact, from the identities:

\frac{f}{x_{i}^{e}} = \frac{f}{x_{j}^{e}} (\frac{x_{j}}{x_{i}} )^{e} and x_{j} = x_{i} (\frac{y_{j}}{y_{i}} ) .

it follows that each V (Ĩ_{j}) ∩ U_{i} ⊆ V (Ĩ_{i}), hence in fact ∪_{i=1}^{n}V (Ĩ_{i}) ∩ U_{i} = V (Ĩ_{i}).

To see irreducibility, observe that from the exercise above it follows that if x_{i}h ∈ Ĩ_{i}, then h ∈ Ĩ_{i}, so x_{i} is not a zero divisor in k[U_{i}]/Ĩ_{i}. Next, we have natural isomorphisms (k[U_{i}]/Ĩ_{i})_{x_{i}} ≅ (k[V]/I(V))_{x_{i}}, as you may easily check, so in particular, (k[U_{i}]/Ĩ_{i})_{x_{i}} is an integral domain. But these two facts together imply k[U_{i}]/Ĩ_{i} is an integral domain, as well (why?). Thus Ĩ_{i} is a prime ideal, and as in the previous theorem, the irreducibility of each V (Ĩ_{i}) and nonempty intersections implies that the union is irreducible.

One checks as in the previous theorem that Γ_{V} ⊂ ∪_{i=1}^{n}V (Ĩ_{i}) is the complement of σ^{−1}(0, ..., 0), hence Ṽ is equal to this union. Also, as in the previous theorem, the isomorphisms (k[U_{i}]/Ĩ_{i})_{x_{i}} ≅ (k[V]/I(V))_{x_{i}} correspond to isomorphisms V (Ĩ_{i}) − σ^{−1}(0, ..., 0) = V (Ĩ_{i}) − V (x_{i}) →^{∼} V − V (x_{i}) which taken together imply that (id, Φ) : V − {(0, ..., 0)} → Ṽ − σ^{−1}(0, ..., 0) is an isomorphism, with σ as its inverse. Notice also that as in the previous theorem, σ^{−1}(0, ..., 0) ∩ U_{i} ⊂ Ṽ ∩ U_{i} is given by one equation, namely V (x_{i}).

It only remains to show that σ^{−1}(0, ..., 0) = {(0, ..., 0)} × P(C_{p}V). But on each V (Ĩ_{i}) = Ṽ ∩ U_{i}, we have σ^{−1}(0, ..., 0) ∩ V (Ĩ_{i}) = V (x_{i}) ∩ V (Ĩ_{i}), so just by examining the equations defining V (Ĩ_{i}) ⊂ U_{y_{i}}, we see that

σ^{−1}(0, ..., 0) ∩ V (Ĩ_{i}) = V (x_{1}, ..., x_{n}, \{f_{e}(\frac{y_{1}}{y_{i}} , ..., \frac{y_{n}}{y_{i}}) | f ∈ I\})

which precisely gives (0, ..., 0) × V (I_{init}) − V (y_{i}) ⊂ (0, ..., 0) × P_{k}^{n−1} − V (y_{i})(!)

Corollary 16.10: The dimension of every component of tangent cone to X at p is equal to dim(X). The point p ∈ X is nonsingular if and only if the tangent cone at p is equal to the Zariski tangent space.

Proof: By 16.9(a),(c), and Krull’s theorem, we see that every component of the projectivized tangent cone has dimension equal to dim(X)−1. By the first exercise in §15, we get the dimension result. Since every component of the tangent cone has dimension equal to dim(X), and since every component is a closed, irreducible subset of the Zariski tangent space, it follows from Krull’s theorem that the dimensions are the same if and only if the tangent cone equals the Zariski tangent space.

Combining several results of this section, we get the following:

Definition: The local ring O_{X,p} associated to a point in a variety over an algebraically closed field k is regular if, equivalently:

(a) dim_{k}(m_{p}/m_{p}^{2}) = dim(X)

(b) Ω_{O_{X,p}} is a free O_{X,p}-module.

(c) The tangent cone at p is equal to the Zariski tangent space.

Remark: There is a fourth characterization using completions, which allows one to conclude to conclude that a regular local ring is always a UFD. Since a UFD is integrally closed, this would tell us that nonsingular implies normal.

Eleventh Homework Assignment

1. Easy properties of modules. Let A be a commutative ring with 1: (a) If M is generated by m_{1}, ..., m_{n} as an A-module, show that M_{S} is

generated by \frac{m_{1}}{1} , ..., \frac{m_{n}}{1} as an A_{S}-module for any multiplicative set S ⊂ A.

(b) If S ⊂ A is a multiplicative subset and I ⊂ A is an ideal, let I_{S} ⊂ A_{S} be the corresponding ideal. Then exhibit an isomorphism:

(A/IA)_{S} ≅ A_{S}/I_{S}A_{S}

and more generally, given an A-module M, an isomorphism of A_{S}-modules:

(M/IM)_{S} →^{∼} M_{S}/I_{S}M_{S}.

If S = A − m for a maximal ideal m, note that A/mA ≅ A_{S}/m_{S}A_{S} and more generally, prove that M/mM ≅ M_{S}/m_{S}M_{S}. Finally, if I, J ⊂ A are ideals, show that (IJ)_{S} = I_{S}J_{S} and use this to prove that:

m^{d}/m^{d+1} ≅ m_{S}^{d}/m_{S}^{d+1}

for any integer d ≥ 0.

(c) If N ⊂ M are modules over A and I ⊂ A is an ideal, show that N/IN is a submodule of M/IM and (M/N)/I(M/N) ≅ (M/IM)/(N/IN)

(d) Prove Proposition 16.3

(e) If I ⊂ A is an ideal and M, N are A-modules, show that:

Hom_{A}(M, N/IN) = Hom_{A/I}(M/IM, N/IN)

2. (a) Prove Euler’s formula: If F ∈ k[x_{0}, ..., x_{n}]_{e}, then:

eF = \frac{∂F}{∂x_{0}} x_{0} + ... + \frac{∂F}{∂x_{n}} x_{n}

(b) Prove that a hypersurface V (F) ⊂ P_{k}^{n} is nonsingular iff V (∇F) = ∅.

3. (a) If d : A → Ω_{A} is a universal derivation, show that Ω_{A} is generated as an A-module by the image of d.

(b) Given an example to show that d in (a) need not be surjective.