Dimension, Tangent Space and Smoothness Week 4 Mar 30, Apr 1, 3

Home , Smoothness, Zariski tangent space

Dimension, Tangent space and Smoothness Week 4 Mar 30, Apr 1, 3

Here we give the definition of dimension, tangent space and smooth points for an algebraic set. These concepts come from geometry but their definitions (at first) seems unrelated to it. We then show that they have indeed a more intuitive presentation. Here the organization of the notes. - Section1: we recall some facts about field extensions. Apart from the the first two definitions, the rest of the section is just a preparation for the proof of the main result of Section3. - Section2: we give the definition of dimension of an algebraic set Z as the transcendence degree of the field extension k ⊂ k(Z). - Section3: we show that any quasi-projective variety is birational to an hypersurface.

- Section4: we introduce and study the Zariski Tangent Space at a point. The definition will be given in Subsection 4.3, using derivations. Then in Subsection 4.4, we give an alternative description for affine case, which is more explicit and helpful during exercises and computations. In Subsection 4.5, we present an alternative description using the maximal ideal of the local ring of germs of regular functions. - Section5: we explain the relationship between the dimension and the tangent space. And then we define what is a smooth/singular point.

1 Some facts about ﬁeld extensions

The aim of this section is to recall some facts about field extensions and proving a stronger version of Noether Normalisation Lemma for integral domains (see Proposition 10). Most of the effort here is needed for dealing with the separability of a field extension. It is a phenomenon appearing exclusively in positive characteristic. If you feel that characteristic zero is more than enough for you, my suggestion is to skip the parts about separability and go to the next section (and maybe come back later). Here a road map for the characteristic zero case:

• Deﬁnition1 and Example2. • Deﬁnition3 and Example4. • Noether Normalisation Lemma for integral domains, i.e. Proposition 10. The proof in characteristic 0 follows immediately from NNL.

Here K denotes an arbitrary field and K its algebraic closure. Definition 1. Let K ⊂ L be a field extension (i.e. an inclusion of fields). An element α ∈ L is algebraic over K if there exists a polynomial f ∈ K[x] such that f(α) = 0. An element α ∈ L is transcendental over K if for any polynomial f ∈ K[x] we have f(α) 6= 0.

1 Example 2. (a) Let Q ⊂ C. Then i is algebraic, because f(x) = x2 + 1 ∈ Q[x]. Instead, π and e are transcendental. (b) Let K ⊂ K(x), then x is transcendental over K. Definition 3. Let K ⊂ L be a field extension, the transcendence degree of L over K is the maximum number of transcendental elements with respect to K ⊂ L. We denote this number by tr. degK (L). We leave to the reader to check that the transcendence degree is a well-defined number, i.e. given to maximal sets {f1, . . . , fr} and {g1, . . . , gs} of K-algebraically independent elements of L, then r = s.

Example 4. tr. degk(k(x1, . . . , xn)) = n. Let K ⊂ L be a ﬁeld extension and α ∈ L be an algebraic element over K. Then the kernel of the homomorphism of rings π : K[x] → L : f 7→ f(α) is a principal ideal, i.e I := ker(π) = (h) for some h ∈ K[x]. The generator of I is well-deﬁned up to scalar. In particular, there exists a unique monic generator h for I. We call h the minimal polynomial of α with respect to K ⊂ L. Equivalently: The minimal polynomial of α with respect to K ⊂ L is the unique monic polynomial h of least degree among all the polynomial f ∈ K[x] such that f(α) = 0. Note that any α ∈ L algebraic over K admits a minimal polynomial.

Example 5. Assume K = Q, L = C and α = i. Then x2 + 1 ∈ Q[x] is the minimal polynomial of i. Deﬁnition 6. Let K ⊂ L be a ﬁeld extension. (i) A polynomial f ∈ K[x] is separable if it has distinct roots over K. (ii) Given α ∈ L algebraic over K, we say α separable over K if its minimal polynomial is separable.

Example 7. (a) If K have characteristic 0 (e.g. Q, R, C, C(x1, . . . , xn)) then any algebraic element in L is separable. ∼ (b) If K is a finite field (e.g. K = Fp = Z/pZ with p prime), the any algebraic element in L is separable. (c) The interesting cases (where algebraic > separable) appear when K is infinite and of positive characteristic. The standard example is the (algebraic) extension

p K := Fp(t ) ⊂ Fp(t) =: L. The element t ∈ L is algebraic over K with minimal polynomial

h := xp − tp ∈ K[x] ⇒ (X − t)p ∈ K[x].

Hence the polynomial h is not separable and so neither t as an element over K. Definition 8. Let K ⊂ L be a field extension. (i) A field extension K ⊂ L is algebraic if any element in L is algebraic over K.

(ii) A ﬁeld extension K ⊂ L is separable if any element in L is separable over K. Example 9. By the previous example. (a) If K has characteristic 0 any algebraic ﬁeld extension K ⊂ L is separable.

(b) If K is a finite field, any algebraic field extension K ⊂ L is separable.

2 (c) If K is infinite and of positive characteristic, then it may be possible that algebraic6=separable. Indeed, p the algebraic field extension Fp(t ) ⊂ Fp(t) is not separable. We now show a stronger version of the Noether Normalisation Lemma for integral domains. Proposition 10. Assume k algebraically closed field. Let A be a finitely generated k-algebra with n generators. Assume that A is an integral domain with field of fractions F . Then there exists m ≤ n and y1, . . . , ym ∈ A such that

(i) y1, . . . , ym are algebraically independent over k, i.e. it does not exists f ∈ k[x1, . . . , xm] such that f(y1, . . . , ym) = 0.

(ii) A is a ﬁnite k[y1, . . . ym]-algebra.

(iii) k(y1, . . . , ym) ⊂ F is a separable extension. Proof. Let p the characteristic of the ﬁeld k. We distinguish two cases. p = 0 By Noether Normalisation Lemma points (i) and (ii) hold. In particular, point (ii) implies that

the field extension k(y1, . . . , ym) ⊂ F must be algebraic. Since p = 0, then it must be separable (see Example 9). p > 0 The plan is to repeat the proof of NSS with some modifications. We declare the following Claim: there exists a subring A0 ⊂ A such that: (a) A0 is a finitely generated k-algebra with n − 1 generators, (b) A is a finite A0-algebra, (c) the extension of fields of fractions Q(A0) ⊂ F is separable. 0 Assume the claim holds, by induction there exists y1, . . . , ym ∈ A ⊂ A with m ≤ n − 1 satisfying Thm(i). 0 Furthermore, A is a finite k[y1, . . . , ym]-algebra. We recall that if A ⊂ B and B ⊂ C are finite algebras, then A ⊂ C is a finite algebra. Point (b) in the claim implies that A is a finite k[y1, . . . , ym]-algebra (see 0 Prop. 26 Week 1), i.e. Thm(ii). Again by induction, k(y1, . . . , ym) ⊂ Q(A ) is a separable extension. We remark that if K ⊂ L and L ⊂ Q are separable extensions, then K ⊂ Q is a separable extension (exercise). Hence, by point (c) in the claim, we get Thm(iii). It remains to show the claim. By assumptions there exists a surjective homomorphism

π : k[x1, . . . , xn] → A, with ker(π) =: I prime ideal. Assume I 6= 0, otherwise the proposition is obvious. Fix a non-zero irreducible (it is possible because I prime) polynomial f ∈ I. For any i, we can write d f = adxi + ... + a0, ai ∈ Ki

where Ki is the subring of k[x1, . . . , xn] of those polynomials with the degree zero with respect to xi (e.g. if i = 0 then K0 = k[x2, . . . , xn]). Let Fi be the ﬁeld of fractions of Ki. Then one (and only one) of the following two cases happens:

1. f is separable in Fi[xi], p 2. f ∈ Fi[xi ] ⊂ Fi[x]. We want to show that there exists a choice of i, such that 1. holds. We argue by contradiction. Assume that f is not separable for any i, then p p f ∈ k[x1, . . . , xn]. Then using the identity (a + b)p = ap + bp in k, one can show that f = gp for some polynomial g. But this is in contradiction with the irreducibility of f. Hence, f is separable for some i, say i = n. Arguing as in the proof of NNL (see proof of Theorem 25 Week 1), we get that 0 0 f(x1 + c1xn, . . . , xn−1 + cn−1xn, xn) 0 0 0 gives a (monic) separable relation for π(xn) over A = π(k[x1, . . . , xn−1]). Proving the claim.

3 2 Dimension

We now ﬁnally deﬁne the dimension of a quasi-projective algebraic set.

n Deﬁnition 11. Let Z ⊂ Pk be a quasi-projective algebraic set. If Z is irreducible, the dimension of Z is the transcendence degree of the extension of the ﬁeld k(Z) of rational functions of Z over k. We set

dim Z := tr. degk(k(Z)).

For a general Z with Z1,...,Zr irreducible components, we set

dim Z := maxi{dim Zi}.

The dimension of Z at p is the number

dimp Z := max{dim Zi s.t. p ∈ Zi}.

n n Example 12. If Z = P or Ak , then dim Z = n. Lemma 13. The following hold (i) If Z and W irreducible and birational, then dim Z = dim W . (ii) If Z irreducible and W ⊂ Z proper closed subset then dim W < dim Z.

(iii) Let Z ⊂ An be an irreducible algebraic subset, then Z have dimension n − 1 if and only it is an irreducible hypersurface, i.e. Z = V (f) for some f ∈ k[x1, . . . , xn] irreducible.

Proof. Point (i) We have seen W and Z are birational if and only if k(Z) =∼ k(W ). Point (ii) . We saw that any quasi-projectve algebraic set admits a cover by aﬃne varieties. By previous

point, we may assume Z and W affine and irreducible. Fix f1, . . . , fd ∈ k(W ) trascendence basis. Since W is affine, the field of rational functions is the field of fractions of the integral domain O(W ) = A(W ). Without loss of generality, we may assume that fi ∈ A(W ). Since W and Z are affine and W ⊂ Z, the restriction map O(Z) = k[x1, . . . , xn]/I(Z) → k[x1, . . . , xn]/I(W ) = O(W )

is surjective. Choose a lift F1,...,Fd ∈ A(Z) for f1, . . . , fd ∈ A(W ) and take an element 0 6= g ∈ A(Z) s.t. g(q) = 0 for any q ∈ W (it exists because Z \ W 6= ∅). For proving the point, it is enough to show that F1,...Fd, g are algebraically independent on k. We argue by contradiction. Suppose there exists

k k−1 P = Y Pk(X) + Y Pk−1 + ... + P0(X),Pi(X) ∈ k[X1,...,Xd]

d1 dr such that P (F, g) = 0 ∈ A(Z) ⊂ k(Z). Let P = Q1 ··· Qr be the irreducible decomposition of P , then by assumptions d1 dr P (F, g) = 0 ∈ A(Z) ⇔ P (F, g) = Q1 (F, g) ··· Qr (F, g) ∈ I(Z).

Since Z irreducible, I(Z) prime. Hence, at least one irreducible polynomial Qi(F, g) is in I(Z). Up to replacing P with Qi, we can assume P irreducible. The restriction of P (F, g) to W gives

P (f, 0) = P0(f) = 0.

Since fi are algebraically independent over k, then P0 = 0. Since P is irreducibile, then P = cY for some c ∈ k. Then g = 0. Contradiction. Point (iii) Assume W = V (f) with f irreducible polynomial. Up to reorder the coordinates, we can write d d−1 f = adxn + ad−1xn + ... + a0,

4 with ai ∈ k[x1, . . . , xn−1], d > 0 and ad 6= 0. Hence, we have

k(W ) = k(x1, . . . , xn−1)[xn]/(f)

So, x1, . . . , xn−1 is transcendence basis for k(W ). Hence, dim W = n − 1. n On the other hand, suppose dim Z = n − 1. The dimension implies Z is not dense in Ak . Then, there exists 0 6= f ∈ I(Z) ⊂ k[x1, . . . , xn]. Since Z irreducible, we can assume f irreducible (use the fact that I(Z) is prime). The inclusion Z ⊂ V (f) must be an equality, otherwise by (ii), we must have dim Z < n − 1.

3 Reduction to an hypersurface

This section is devoted to show the following.

n Proposition 14. Let Z ⊂ Pk be an irreducible quasi-projective algebraic set of dimension l. Then, it is l+1 birational to an hypersurface V ⊂ Ak . m+1 ∼ It is enough to construct an hypersurface V ⊂ Ak such that k(V ) = k(Z). Without loss of generality, n we may assume that Z is an affine algebraic subset of Ak . The ring of regular function A(Z) is an integral domain, a finitely generated k algebra with n generators and its field of fractions is exactly k(Z). By Prop. 10, there exists y1, . . . , ym ∈ A(Z) such that

(i) y1, . . . , ym are algebraically independent over k.

(ii) A(Z) is a ﬁnite k[y1, . . . ym]-algebra.

(iii) k(y1, . . . , ym) ⊂ k(Z) is a separable extension.

By hypothesis tr. degk(A(Z)) = l, hence y1, . . . , ym is a maximal set of transcendental elements, i.e. m = l. We now need the following Theorem 15 (Primitive Element Theorem). Let K be an infinite field, and K ⊂ L be a finite separable field extension; then there exists α ∈ L, which generates L as an extension over k, i.e. for any x ∈ L, then

m x = c0 + c1α + . . . cmα , ci ∈ k

Furthermore, if L is generated over K by elements β1, . . . , βr, then the element α can be chosen to be a linear combination α = a1β1 + ... + arβr

We apply the Theorem to the extension K := k(y1, . . . , yl) ⊂ k(Z). The ﬁrst part of the Theorem tell ∼ us that there exists an element α ∈ k(Z) such that y1, . . . , ym, α generates k(Z) over k. Since A(Z) = k[x1, . . . , xn]/I, the monomials x1, . . . , xn generates A(Z) (and so k(Z)) over k. So the second part of the Theorem tell us that α can be chosen in A(Z). Since K ⊂ k(Z) is a separable extension, there exists a separable polynomial for α:

d ad−1 d−1 a0 K[x] 3 f = x + x ... + , ai, bi ∈, k[y1, . . . , ym] bd−1 b0 such that f(α) = 0. We then obtain the polynomial

d d−1 k[y1, . . . , yl][x] ∈ Q := (b1 ··· bd−1)f = cdx + cd−1x ... + c0, ci ∈ k[y1, . . . , yl],

Up to divide by an element in k[y1, . . . , yl], we may assume that c0, . . . , cd are pairwise relatively prime. We claim that Q is irreducible as polynomial in k[y1, . . . , yl, x]. We argue by contradiction, assume that Q = Q1Q2 with Qi ∈ k[y1, . . . , yl, x] irreducible. Since f is separable, it is irreducible in K[x] := k(y1, . . . , yl)[x], otherwise it is not minimal among the polynomials

5 vanishing at α. In particular, also the polynomial Q is irreducible in K[x]. Then, either Q1 or Q2 are unit in K[x], say Q1. The conditions Q1 ∈ k[y1, . . . , yl, x] and unit in K[x] imply that

Q1 ∈ k[y1, . . . , yl] \{0}.

Hence Q1 divides Q if and only if it divides all the coeﬃcients c1, . . . , cd−1. Since they are pairwise relatively l+1 primes, Q1 must be a unit in k[y1, . . . , yl]. Then Q is irreducible. By setting V := V (Q) ∈ Ak , we get ∼ ∼ k(Z) = k(y1, . . . yl)[x]/Q(x) = k(V ), concluding the proof.

4 The Tangent Space

Consider the circle Z = V (x2 + y2 − 1) ⊂ 2 . You probably know that the (complex analytic) tangent space AC at a point (p1, p2) is the line T Z = {(x , x ) ∈ 2 s.t. 2p x + 2p x = 0} ⊂ A2 p 1 2 AC 1 1 2 2 C In other words, the aﬃne line

p + T Z = {(x , x ) ∈ 2 s.t. 2p (x − p ) + 2p (x − p ) = 0} p 1 2 AC 1 1 1 2 2 2

is the unique line containing p and tangent at Z in p. More in general, if we have Z = V (f1, . . . , fr), we would like to describe TpZ by r-linear polynomial equations, given by the derivatives of the polynomials fi. In this section, we aim to do this formally. We first give a definition of the tangent space in terms of derivations which works in the more general context of quasi-projective algebraic sets. And than later we will show that in the affine case coincides with the construction sketched above.

4.1 Localisation First, we need to introduce the concept of germ of a regular function. Roughly speaking, if Z is an algebraic subset and p ∈ Z, the germ of a regular function at p is a function regular in a neighbourhood of p. Two germs are equal if, locally at p, they take the same values. This concept comes from topology and probably you have already used implicitly many times. For instance, when you describe a differentiable function locally using the Taylor series. As for the regular and rational functions, we introduce a geometric definition and an algebraic one and then we show that they agree in the affine case. Let’s start with the geometric approach. Fix a quasi- n projective algebraic subset Z ⊂ Pk and point p ∈ Z. Consider the set of pairs (U, ϕ), where • U ⊂ Z (not-necessarily dense) open containing p ∈ Z,

1 • ϕ : U → Ak regular function. 0 0 0 0 We say (U, ϕ) ∼ (U , ϕ ) if there exists an open V ⊂ Z containing p such that V ⊂ U ∩ U and ϕV = ϕV . We leave to the reader to check that is an equivalence relation. Given two equivalence classes [(U, ϕ)] and [(V, ψ)], we deﬁne the following operations (Sum)[( U, ϕ)] + [(V, ψ)] := [(U ∩ V, ϕ + ψ)], (Product)[( U, ϕ)] · [(V, ψ)] := [(U ∩ V, ϕψ)] Deﬁnition 16. Let Z be a quasi-projective variety p ∈ Z. The ring of the germs of regular functions on Z near p is the set of equivalence classes OZ,p := {[(U, ϕ)]} 1 with the operations described as above (the identity is given by the class of Z → Ak : q 7→ 1).

6 Remark 17. A local ring is a ring with a unique maximal ideal. The set mp of classes in OZ,p vanishing at p is an ideal. Furthermore, it is the unique maximal ideal of OZ,p, in particular OZ,p is a local ring. It can be −1 showed by observing that if [(U, ϕ)] ∈/ mp then it is a unit. Indeed, [(U \ V (ϕ), ϕ )] is a well-defined class and it is the (multiplicative) inverse for [(U, ϕ)]. Remark 18. Note that there are some similarities with the rational functions. Indeed, in a suitable way one can see that the field of rational functions is the ”local ring at the generic point of Z”. ∼ Remark 19. If U ⊂ Z is an open subset containing p, then the natural restriction OZ,p = OU,p is an isomorphism of rings. This is will be very useful during the proofs and computations, because we can always restrict to the case when Z is affine. We now introduce the algebraic counterpart. Let R be a ring and I ∈ R be a prime ideal. We define an equivalence relation on R × (R \ I) as it follows (r, p) ∼ (r0, p0) if ∃u∈ / I s.t. u(rp0 − r0p) = 0 ∈ R We denote the localization of R at I as

RI := R × (R \ I)/ ∼ r An equivalence class [(r, p)] will be denoted with the fraction p . It has a natural ring structure: r r0 rp0 + r0p r r0 rr0 + := , · := p p0 pp0 p p0 pp0

Note that, when R is an integral domain (0) is prime and the localization R(0) is exactly the ﬁeld of fractions Q(R) (confront with Remark 18).

Lemma 20. The localization RI is a local ring, i.e. it has a unique maximal ideal, namely IRI . Proof. Exercise. ∼ Example 21. (a) Z and p ∈ Z prime, then Z(p) = {r/q | r, q ∈ Z and q∈ / (p)}.

(b) k[x1, . . . , xn] and f ∈ k[x1, . . . , xn] irreducible, then

k[x1, . . . , xn](f) = {h/g | h, g ∈ k[x1, . . . , xn] and f does not divide g}.

k[x1, . . . , xn]mp = {f/g | f, g ∈ k[x1, . . . , xn] and g(p1, . . . , pn) 6= 0}.

n (d) Let Z ⊂ Ak be quasi-aﬃne. Let mp ⊂ A(Z) the (maximal) ideal of polynomial functions vanishing at p ∈ Z. Then ∼ f A(Z)mp = s.t. f, g ∈ k[x1, . . . , xn] and g(p) 6= 0 . g ∼

f f 0 0 0 where g ∼ g0 if and only if there exists u ∈ k[x1, . . . , xn] such that u(p) 6= 0 and u(fg − f g) ∈ I(Z). When Z irreducible (i.e. I(Z) prime), we may choose u = 1. Furthermore, we have the following n Proposition 22. If Z ⊂ Ak is a quasi-aﬃne algebraic subset and p ∈ Z point, then there is an isomorphism ∼ OZ,p = A(Z)mp of local rings. Proof. Using the isomorphism of Examples 21(d), wee get a homomorphism of rings

A(Z)mp → OZ,p f h f i g 7→ Z \ V (g), g We leave to the reader to check that it is well-deﬁned and bijective.

7 4.2 Derivations Here we introduce the derivations. They are the algebraic version of derivations coming from calculus, they are defined using the usual properties of classical derivations. In this section, R is a ring. We recall the definition of a module. Definition 23. An R-module M is an abelian group together with an operation · : R × M → M such that for any r, s ∈ R and x, y ∈ M, • r · (x + y) = r · x + r · y, • (r + s) · x = r · x + s · s, • (rs) · x = r · (s · x), • 1 · x = x. A homomorphism ϕ : M → N of R-modules is a homomorphism of abelian groups such that ϕ(r·x) = r·ϕ(x). Because of the compatibility, we will use the notation rx instead of r · x during the notes. Example 24. Here a (hopefully) complete list of all the R-modules that we are going to meet during the lectures. So, if you have trouble remember the above definition, just memorize these examples. (a) If R = k is a field, then the definitions of k-module and k-vector space agree. (b) R has a natural structure of R-module given by the multiplication on itself. (c) An abelian subgroup I ⊂ R is an R-module (via multiplication) if and only if I is an ideal. When I = (1) we get the previous example. (d) Any quotient R/I carries a natural structure of R-module given by the multiplication, i.e. r · [x] := [rx]. ∼ (e) A fundamental case of the previous example is the following. Consider the ring OZ,p, then OZ,p/mp = k and the OZ,p-module structure of k is given by f · λ := f(p)λ for λ ∈ k. We denote by kp the field k with the OZ,p-module structure described here.

(f) The homomorphism of rings A(Z) → OZ,p makes the OZ,p-module kp an A(Z)-module by composition. With abuse of notation, we will denote with the same symbol kp the ﬁeld k with this A(Z)-module structure. Deﬁnition 25. Let R be a k-algebra (i.e. k ⊂ R) and M be an R-module. A k-derivation from R to M is a map D : R → M such that: (i) D(λ) = 0 if λ ∈ k,

(ii) D(f + g) = D(f) + D(g) for any f, g ∈ k[x1, . . . , xn] (Linearity),

(iii) D(fg) = fD(g) + D(f)g for any f, g ∈ k[x1, . . . , xn] (Leibniz rule).

Example 26. The standard example is R = M = k[x1, . . . , xn] and ∂f k[x1, . . . , xn] → k[x1, . . . , xn]: f 7→ ∂xi

It is called formal partial derivative with respect to xi. It can be described as the unique k-derivation such that ( ∂x 1 if i = j j = ∂xi 0 if i 6= j ∂ When the variables are clear from the context, we set ∂i := . ∂xi

8 Observe that the set of k-derivations from R to M

Derk(R,M) is an R-module. Indeed • (D + D0)(f) := D(f) + D0(f),

• f · D(g) := fD(g) for f, g ∈ R and D ∈ Derk(R,M). And because k is contained in R, it is also k-vector space.

4.3 The Zariski Tangent Space n Let Z ⊂ Pk be an algebraic set, then observe that the evaluation homomorphism

OZ,p → k : [(U, ϕ)] 7→ ϕ(p)

makes k an OZ,p-module, we denote the module by kp (see also Example 24(e)). We finally give the definition n Definition 27. Let Z ⊂ Pk be a quasi-projective algebraic set. The Zariski tangent space of Z at p is the k-vector space TpZ := Derk(OZ,p, kp)

where kp is the OZ,p-module described above. The deﬁnition seems obscure at ﬁrst, but it has the advantage of being clearly functorial, i.e. given a regular morphism of algebraic set, we have

∗ ϕ : Z → W =⇒ ϕp : OW,ϕ(p) → OZ,p. And then we get a k-linear homomorphism

∗ dϕp : TpZ → Tϕ(p)W : D 7→ D ◦ ϕp of k-vector spaces. Furthermore, given another regular morphism ψ : W → Y , then

d(ψ ◦ ϕ)p = d(ψ)ϕ(p) ◦ d(ϕ)p : TpZ → Tψ(ϕ(p))Y. In particular, if ϕ is an isomorphism then ∼ d(ϕ)p : TpZ = Tϕ(p)W. Remark 28. The deﬁnition as a counterpart in diﬀerential geometry. Usually the tangent space of a (smooth) manifold M at a point p is described by the set of equivalence classes of curves passing through p, i.e. {γ :[−1, 1] → M s.t. γ(0) = p}/ ∼,

ϕ where γ ∼ η if and only if there exists one (and so for any) chart M ⊃ U −→ Rn such that p ∈ U and (ϕ ◦ γ)0(0) = (ϕ ◦ η)0(0). Here (−)0 is the derivative of a function [−1, 1] → Rn : t 7→ f(t). Each equivalence class [γ] defines an R-derivation ∞ Dγ : C (M) → R as it follows, for any differentiable function f : M → R we set 0 Dγ (f) := (f ◦ γ) (0), (note that f ◦ γ :[−1, 1] → R) And conversely, given a derivation D : C∞(M) → R, it is possible to reconstruct the curve. One last remark: in the differential world a derivation takes values in the ring of functions defined everywhere in M. For technical reasons (in a nutshell we do not have the partition of unity), in our setting we need to work with germs of functions but the idea is the same. However, as next lemma shows, when Z is quasi-affine we can replace the germs with the functions defined globally.

9 n Lemma 29. Let Z ⊂ Ak be a quasi-aﬃne algebraic set and p ∈ Z a point. Then, the natural homomorphism f A(Z) → A(Z) = O : f 7→ mp Z,p 1 induces a k-linear isomorphism of vector spaces

∼= π : TpZ := Derk(OZ,p, kp) −→ Derk(A(Z), kp). Proof. The homomorphism π is well-deﬁned and k-linear. Injectivity Assume π(D) = 0, i.e. D(f/1) = 0 for any f ∈ A(Z). Then for any f/g ∈ A(Z)mp , we have D(f/g) = f(p)D(1/g) + D(f)(1/g(p)) = f(p)D(1/g). In particular, 0 = D(1) = D(g/g) = g(p)D(1/g). Since g(p) 6= 0, then D(1/g) = 0 and so D(f/g) = 0.

Hence D = 0 and π is injective. Surjectivity Given a derivation D : A(Z) → kp, we deﬁne a map

De : OZ,p = A(Z)mp → kp, as it follows: for any f, g ∈ A(Z) such that g(p) 6= 0, set f D(g) 1 De := −f(p) + D(f) . g g2(p) g(p)

We leave to the reader to check that it is a k-derivation. Obviously, π(De) = D proving the surjectivity.

4.4 The Embedded Tangent Space. For explicit computations the definition of Zariski Tangent Space is not very helpful. In this section, we introduce the embedded tangent space for a quasi-affine algebraic sets (which essentially is the one we sketched at the beginning of the Section), which is easier to handle. Since any algebraic set is covered by affine sets, this description can be used also for computing the tangent space of a quasi-projective set. Then we will show that it agrees with the definition of tangent space. We recall the formal partial derivatives of Example 26. The formal partial derivative with respect to xi is the (unique) function ∂i : k[x1, . . . , xn] → k[x1, . . . , xn], such that:

(i) ∂i(λ) = 0 if λ ∈ k,

(ii) ∂i(f + g) = ∂i(f) + ∂i(g) for any f, g ∈ k[x1, . . . , xn] (Linearity),

(iii) ∂i(fg) = f∂i(g) + ∂i(f)g for any f, g ∈ k[x1, . . . , xn] (Leibniz rule), ( 1 if i = j (iv) ∂i(xj) = , 0 if i 6= j

n n−1 In particular, ∂i(xi ) = n∂i(xi ). Using the formal derivatives, we get the following n Definition 30. Let Z ⊂ A be a quasi-affine algebraic subset and fix a point p = (p1, . . . , pn) ∈ Z. We define the embedded tangent space of Z at p by ( ) e n X n Tp Z = x ∈ Ak s.t. ∂if(p)xi = 0 ∀f ∈ I(Z) ⊂ Ak i and the affine tangent space to V at p by ( ) e n X n p + Tp Z = x ∈ Ak s.t. ∂if(p)(xi − pi) = 0 ∀f ∈ I(Z) ⊂ Ak . i

10 n From the deﬁnition, If Z is the closure of of Z in Ak , then e e e e Tp Z = Tp Z, and p + Tp Z = p + Tp Z

for any p ∈ Z ⊂ Z. Furthermore, if f1, . . . , fm are the generators of I(Z), then ! e X X n Tp Z = V ∂if1(p)xi,..., ∂ifm(p)xi ⊂ Ak i i e In particular, Tp Z is an aﬃne algebraic subset. Actually, it is a linear subspace since all the polynomials e are homogeneous and linear. There is a similar description for p + Tp Z (replace xi with xi − pi). n e n n n Example 31. (a) if Z = Ak , then Tp Ak = Ak for any p ∈ Ak . 3 2 e 2 (b) if Z = V (x − y ), then Tp Z = V (3a x − 2by) where p := (a, b) ∈ Z. Observe that ( 2 if p = (0, 0), dim T eZ = p 1 if p ∈ Z \{(0, 0)}

2 Since Z is an hypersurface of Ak, we have dim Z = 1. As you can see, the dimension of the variety and the dimension of the tangent space agree on a dense open subset of Z. We will see later that this always true for any algebraic set, providing a good criterion for computing the dimension of an algebraic set.

n (c) More in general, if Z = V (f) ⊂ Ak , we have ( n if p ∈ V (f, ∂ f, . . . , ∂ f), dim T eZ = 1 n p n − 1 otherwise Furthemore, we have the following. Proposition 32. The function Z → Z e p 7→ dim Tp Z is upper-semicontinuous (for the Zariski topology), i.e. for any integer m the subset e S(m) := {p ∈ Z s.t. dim Tp (Z) ≥ m} ⊂ Z is closed in Z.

Proof. Let f1, . . . , fr be the generators of the ideal I(Z). Then, we have ! e X X Tp (Z) = V ∂if1(p)xi,..., ∂ifm(p)xi . i i Using this interpretation, by linear algebra we get that     ∂1f1(p) ··· ∂1fm(p)    . .. .  S(m) = p ∈ Z s.t. rank  . . .  ≤ n − m    ∂nf1(p) ··· ∂nfm(p)  So, S(m) is described by the vanishing of the (n − m − 1) × (n − m − 1)-minors of the matrix   ∂1f1 ··· ∂1fm  . .. .   . . .  ∈ Mn×m(k[x1, . . . , xn]). ∂nf1 ··· ∂nfm The minors are clearly polynomials, since they are determinants of a matrix of polynomials. So S(m) ⊂ Z is a closed subset.

11 e We now show that the deﬁnition agrees with the previous one. Take a vector v := (v1, . . . , vn) ∈ Tp Z ⊂ e n n Tp Ak , then we may deﬁne the derivation in Derk(k[x1, . . . , xn], kp) = TpAk . X Dv : k[x1, . . . , xn] → kp : f 7→ Dv(f) := ∂if(p)vi i

e Note that since v ∈ TZ we get that Dv(f) = 0 for any f ∈ I(Z). In particular, Dv factors along A(Z). In other words, we have deﬁned a k-linear homomorphism

e ∼ β : Tp Z → Derk(A(Z), kp) = Derk(OZ,p, kp) =: TpZ v 7→ Dv

n Proposition 33. Let Z ⊂ Ak be a quasi-aﬃne algebraic set and p ∈ Z be a point. The homomorphism β e ∼ above is an isomorphism of k-vector space, i.e. Tp Z = TpZ.

n Proof. Assume Z = Ak Consider the derivations

Di : k[x1, . . . , xn] → kp : f 7→ ∂if(p)

We then have β(v1, . . . , vn) = D1v1 + ... + Dnvn. In particular, β(v)(xi) = vi. So β must be injective. On the other hand, given a derivation D : k[x1, . . . , xn] → kp, we must have D = D(x1)D1 + ... + D(xn)Dn. Hence, β(D(x1),...,D(xn)) = D and β is surjective. n Assume Z ⊂ Ak quasi-aﬃne By construction, β is the restriction of the previous isomorphism

e n ∼ n Tp Ak = TpAk

Hence β is clearly injective. Take a derivation D ∈ Derk(A(Z), kp) ⊂ Derk(k[x1, . . . , xn], kp). Hence, D can be seen as a derivation in k[x1, . . . , xn] such that D(f) = 0 for any f ∈ I(Z). By previous point, for any f ∈ I(Z) 0 = D(f) = D(x1)D1(f) + ...D(xn)Dn(f) = ∂1f(p)D(x1) + ... + ∂nf(p)D(xn). e By deﬁnition of embedded tangent space, we have (D(x1),...D(xn)) ∈ Tp Z. Hence, β(D(x1),...D(xn)) = D and β is surjective.

4.5 Another description of the Zariski Tangent space. We now present an alternative presentation of the Zariski tangent space. We ﬁrst give the description for the embedded tangent space case and then we sketch how to generalize to the (non-embedded) case. Let Z be a quasi-aﬃne algebraic variety and p ∈ Z a point. Let mp ⊂ A(Z) be the maximal ideal of polynomial functions vanishing at p. Consider the quotient of abelian groups

2 mp/mp. It carries a canonical structure of k-vector space. Indeed, take f ∈ A(Z) and deﬁne

2 f.[x] := [fx] for any [m] ∈ mp/mp.

2 ∼ But then α.[x] = 0 if α ∈ mp ⊂ A(Z). Hence the multiplication above factors through A(Z)/mp = k. Hence, 2 mp/mp is a k-vector space. n Proposition 34. Let Z ⊂ Ak be a quasi-aﬃne algebraic set and p ∈ Z be a point. There exists a natural isomorphism of vector spaces e 2 ∨ Tp Z = (mp/mp) , where (−)∨ denotes the dual of a vector space.

12 Proof. Let f ∈ A(Z) = k[x1, . . . , xn]/I(Z) and choose two lifts fb1, fb2 ∈ k[x1, . . . , xn]. Then, by deﬁnition of tangent space X X ∂ifb1(p)vi − ∂ifb2(p)vi = 0 for any (v1, . . . , vn) ∈ TpZ. i i In particular, we have a well-deﬁned map of k-vector spaces ∨ A(Z) → (TpZ) f 7→ df(p),

where df(p): TpZ → k is the linear function (v1, . . . , vn) 7→ ∂1f(p)v1 +. . . ∂nf(p)vn. Consider the restriction to the maximal ideal ∨ δ : mp → (TpZ) . 2 The map is clearly linear (by the linearity of ∂i) and surjective. Because of Leibniz rule, δ(mp) = 0. We then have a surjective homomorphism of k-vector spaces 2 ∨ δe : mp/mp → (TpZ) . n It remains to show that it is injective. Assume ﬁrst that Z = A and δe(f) = 0, i.e. ∂i(f)(p) = 0 for any i. By deﬁnition, we can write

f = (x1 − p1)g1 + ... + (xn − pn)gn gi ∈ k[x1 − p1, . . . , xn − pn] 2 n Then, ∂i(f)(p) = 0 ⇔ f has quadratic terms in (x1 − p1),..., (xn − pn) ⇔ f ∈ mp. Assume now Z ⊂ Ak and δe(f) = 0, i.e. df(p)(v) = 0 for any v ∈ TpZ

Let fb ∈ k[x1, . . . , xn] such that [fb] = f ∈ A(Z), then n n dfb(p)(v) = 0 for any v ∈ TpZ ⊂ TpAk = Ak ⇒ df(p) ∈ I(TpZ) ⊂ k[x1, . . . , xn]

By deﬁnition of TpZ, there exists g ∈ I(Z) such that dg(p) = dfb(p). In particular, we have • [fb− g] = f, • (fb− g)(p) = 0, • d(fb− g)(p) = 0. n 2 By second, third point and the case Z = Ak , we get fb− g ∈ (x1 − p1, . . . xn − pn) ⊂ k[x1, . . . , xn]. By ﬁrst 2 2 point, f ∈ mp = (x1 − p1, . . . xn − pn) + I(Z). The above proposition has a counterpart for the Zariski Tangent space. This allows us to generalize the previous result to the more general case of quasi-projective varieties. The main idea is to replace the embedded tangent space with the Zariski tangent space and the ring of polynomial functions with the ring of germs of regular functions near p. More precisely, let Z be a quasi- projective algebraic variety and p ∈ Z be a point. Let mp ⊂ OZ,p be the maximal ideal of germs of regular functions vanishing at p. Consider the quotient of abelian groups 2 mp/mp.

It carries a canonical structure of k-vector space. Indeed, take f ∈ OZ,p and deﬁne 2 f.[m] := [fm] for any [m] ∈ mp/mp. 2 ∼ But then f.[m] = 0 if f ∈ mp ⊂ OZ,p. Hence the multiplication above factors through OZ,p/mp = k. So, 2 mp/mp is a k-vector space. n Theorem 35. Let Z ⊂ Pk be a quasi-projective algebraic set and p ∈ Z be a point. There exists a natural isomorphism of vector spaces 2 ∨ TpZ = (mp/mp) , where (−)∨ denotes the dual of a vector space.

13 5 Smooth and Singular points

The next theorem explains the relationships between the dimension of an algebraic set and the Zariski tangent space. n Theorem 36. Let Z ⊂ Pk be a quasi-projective algebraic set. For any p ∈ Z, we have dimp Z ≤ dim TpZ. Moreover, there exists an open not-empty dense subset Zsm ⊂ Z where the equality holds.

n Proof. Without loss of generality, we may assume Z irreducible. We distinguish three cases. Z = Ak n obvious. Z = V (f) ⊂ Ak aﬃne hypersurface Then we already know that dim Z = n − 1. Furthermore

n ! X TpZ = V ∂if(p)xi i=1 In particular, ( n if p ∈ V (f, ∂1f, . . . , ∂nf), dim TpZ = n − 1 otherwise Pn We need to show that Z \ V (f, i=1 ∂if(p)xi) is dense in Z. We argue by contradiction. Assume

Z = V (f, ∂1, . . . , ∂nf),

then ∂if ∈ (f), i.e. ∂if = af for some polynomial a. But the degree of ∂if at xi is less than degree of f at xi. Hence ∂if = 0 for any i. In characteristic 0, it implies immediately that f is constant which is in contradiction with the fact that Z is an hypersurface. In characteristic p > 0, it happens only when p p p f ∈ k[x1, . . . , xn] then again f = g for some polynomial g contradicting the irreducibility of f. n Z ⊂ Pk general By Proposition 14, Z is birational to an hypersurface. Since the dimension is a birational invariant, by previous point there exists a dense subset U where dim Z = dim TpZ for any p ∈ U. Assume there exists q ∈ Z such that dim Z > dim TqZ. By proposition 32 and the fact that Z is covered by affine varieties, the locus V of points q such that dim Z > dim TqZ is open in Z. Since Z is irreducible, we would get V ∩ U 6= ∅ giving a contradiction. The advantage of this theorem is that it gives a useful criterion to compute the dimension of an algebraic set by simply computing the dimension of the tangent space in a dense subset. Furthermore, we can finally define the smooth/singular points for an algebraic set.

Deﬁnition 37. We say that Z is smooth at p if dim TpZ = dimp Z. The set of smooth points is denoted by Zsm. When Zsm = Z, we say Z is smooth. We say Z is singular at p if Z is not smooth at p. The closed subset of singular points is denoted by Zsing. n n Example 38. (a) Ak and Pk are clearly smooth. (b) If Z = V (x3 − y2), then we saw that Zsm = Z \ V (x, y) and V (x, y) = Zsing.

sing (c) More in general, if Z = V (f) then Z = V (f, ∂1f, . . . , ∂nf). n (d) if Z ⊂ Pk is quasi-projective, then one can restrict to the aﬃne pieces and use that embedded Zariski tangent space for studying the smoothness of Z. (e) The singular/smooth locus is preserved by the isomorphism class of the algebraic set. In particular, it gives a criterion for checking if two algebraic sets are isomorphic. As an example, the bijective regular morphism of Exercise 2 Week 2

1 3 2 2 3 A → Z := V (x − y ): t → (t , t ) is not an isomorphism because the source is smooth and target has a singular point.