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

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Dimension, Tangent Space and Smoothness Week 4 Mar 30, Apr 1, 3 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 2 L is algebraic over K if there exists a polynomial f 2 K[x] such that f(α) = 0. An element α 2 L is transcendental over K if for any polynomial f 2 K[x] we have f(α) 6= 0. 1 Example 2. (a) Let Q ⊂ C. Then i is algebraic, because f(x) = x2 + 1 2 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 ff1; : : : ; frg and fg1; : : : ; gsg 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 2 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 2 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 2 K[x] such that f(α) = 0. Note that any α 2 L algebraic over K admits a minimal polynomial. Example 5. Assume K = Q, L = C and α = i. Then x2 + 1 2 Q[x] is the minimal polynomial of i. Definition 6. Let K ⊂ L be a field extension. (i) A polynomial f 2 K[x] is separable if it has distinct roots over K. (ii) Given α 2 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 character- istic. The standard example is the (algebraic) extension p K := Fp(t ) ⊂ Fp(t) =: L: The element t 2 L is algebraic over K with minimal polynomial h := xp − tp 2 K[x] ) (X − t)p 2 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 gen- erators. Assume that A is an integral domain with field of fractions F . Then there exists m ≤ n and y1; : : : ; ym 2 A such that (i) y1; : : : ; ym are algebraically independent over k, i.e. it does not exists f 2 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 2 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 2 I. For any i, we can write d f = adxi + ::: + a0; ai 2 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 2 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 2 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 := maxifdim Zig: The dimension of Z at p is the number dimp Z := maxfdim Zi s.t.
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