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Introduction to Linear Algebraic Groups Stephen Donkin University of York November 19, 2009 Stephen Donkin (University of York) Introduction to Linear Algebraic Groups November 19, 2009 1 / 21 Lecture 6 Stephen Donkin (University of York) Introduction to Linear Algebraic Groups November 19, 2009 2 / 21 Lecture 6 Hilbert’s Nullstellensatz (i) Let A be a ﬁnitely generated k -algebra. Then every maximal ideal of A has codimension 1. (ii) Let (V , A) be an afﬁne variety. Let I be an ideal of A with I = A. Then V(I) = ∅ (i.e. the ideal I has a “zero point"). Stephen Donkin (University of York) Introduction to Linear Algebraic Groups November 19, 2009 2 / 21 Lecture 6 Hilbert’s Nullstellensatz (i) Let A be a ﬁnitely generated k -algebra. Then every maximal ideal of A has codimension 1. (ii) Let (V , A) be an afﬁne variety. Let I be an ideal of A with I = A. Then V(I) = ∅ (i.e. the ideal I has a “zero point"). By a maximal ideal of A I mean an ideal I such that I = A and such that the only ideal of A properly containing I is A itself. Stephen Donkin (University of York) Introduction to Linear Algebraic Groups November 19, 2009 2 / 21 Lecture 6 Hilbert’s Nullstellensatz (i) Let A be a ﬁnitely generated k -algebra. Then every maximal ideal of A has codimension 1. (ii) Let (V , A) be an afﬁne variety. Let I be an ideal of A with I = A. Then V(I) = ∅ (i.e. the ideal I has a “zero point"). By a maximal ideal of A I mean an ideal I such that I = A and such that the only ideal of A properly containing I is A itself. If I say that f is a regular function on an afﬁne variety V I just mean that f belongs to the coordinate algebra k [V ]. Stephen Donkin (University of York) Introduction to Linear Algebraic Groups November 19, 2009 2 / 21 Lecture 6 Hilbert’s Nullstellensatz (i) Let A be a ﬁnitely generated k -algebra. Then every maximal ideal of A has codimension 1. (ii) Let (V , A) be an afﬁne variety. Let I be an ideal of A with I = A. Then V(I) = ∅ (i.e. the ideal I has a “zero point"). By a maximal ideal of A I mean an ideal I such that I = A and such that the only ideal of A properly containing I is A itself. If I say that f is a regular function on an afﬁne variety V I just mean that f belongs to the coordinate algebra k [V ]. Application Suppose that V is an afﬁne variety and f is a regular function on V which is never zero. Then the function 1/f is also regular. To see this take I = k [V ] · f . Then V(I) = ∅ so that I = k [V ], i.e. f is invertible. Stephen Donkin (University of York) Introduction to Linear Algebraic Groups November 19, 2009 2 / 21 From the Nullstellensatz one may deduce the strong Nullstellensatz. First some additional terminology. Let A be a commutative ring. An element f ∈ A is nilpotent if f n = 0 for some positive integer n. It is an exercise to show that the set of all nilpotent elements forms an ideal. This is called the nilradical of A and we denote it N(A). Stephen Donkin (University of York) Introduction to Linear Algebraic Groups November 19, 2009 3 / 21 From the Nullstellensatz one may deduce the strong Nullstellensatz. First some additional terminology. Let A be a commutative ring. An element f ∈ A is nilpotent if f n = 0 for some positive integer n. It is an exercise to show that the set of all nilpotent elements forms an ideal. This is called the nilradical of A and we denote it N(A). The ring A is called reduced if N(A) = 0. For example there are no non-zero nilpotent elements in Z so N(Z) = {0}, i.e. Z is reduced. Stephen Donkin (University of York) Introduction to Linear Algebraic Groups November 19, 2009 3 / 21 From the Nullstellensatz one may deduce the strong Nullstellensatz. First some additional terminology. Let A be a commutative ring. An element f ∈ A is nilpotent if f n = 0 for some positive integer n. It is an exercise to show that the set of all nilpotent elements forms an ideal. This is called the nilradical of A and we denote it N(A). The ring A is called reduced if N(A) = 0. For example there are no non-zero nilpotent elements in Z so N(Z) = {0}, i.e. Z is reduced. √ More generally, if I is an ideal of A we write I for the set of all f ∈ A such that f n √ I for some positive integer √ From the deﬁnitions it ∈ n. follows that I is the ideal of A such that I/I = N(A/I). Stephen Donkin (University of York) Introduction to Linear Algebraic Groups November 19, 2009 3 / 21 Stephen Donkin (University of York) Introduction to Linear Algebraic Groups November 19, 2009 4 / 21 Now, given a closed set Z in an afﬁne variety V we write N (Z ) for the ideal of all functions f ∈ k [V ] that vanish on Z . Stephen Donkin (University of York) Introduction to Linear Algebraic Groups November 19, 2009 4 / 21 Now, given a closed set Z in an afﬁne variety V we write N (Z ) for the ideal of all functions f ∈ k [V ] that vanish on Z . What happens if we start with an ideal I of k [V ] then we take V(I), the closed set consisting of the points at which all elements of I vanish, and then we take N (V(I))? Clearly this ideal contains I but how much bigger can it be? Stephen Donkin (University of York) Introduction to Linear Algebraic Groups November 19, 2009 4 / 21 Now, given a closed set Z in an afﬁne variety V we write N (Z ) for the ideal of all functions f ∈ k [V ] that vanish on Z . What happens if we start with an ideal I of k [V ] then we take V(I), the closed set consisting of the points at which all elements of I vanish, and then we take N (V(I))? Clearly this ideal contains I but how much bigger can it be? The Strong Nullstellensatz (i) Let A be a ﬁnitely generated k -algebra. Then the intersection of the maximal ideals of A is the nilradical of A. (ii) Let (V , A) be an afﬁne variety and let I be an ideal of A. Then √ N (V(I)) is the radical I = {f ∈ A | f r ∈ I for some r ≥ 0}. Stephen Donkin (University of York) Introduction to Linear Algebraic Groups November 19, 2009 4 / 21 This effectively produces a complete dictionary between ﬁnitely generated reduced algebras over k and afﬁne algebraic varieties. Stephen Donkin (University of York) Introduction to Linear Algebraic Groups November 19, 2009 5 / 21 This effectively produces a complete dictionary between ﬁnitely generated reduced algebras over k and afﬁne algebraic varieties. On the one hand, if A = k [V ], for some variety V then A is reduced (if f n = 0 then for x ∈ V we have f n (x) = f (x)n = 0, so f (x) = 0. Hence f = 0. Stephen Donkin (University of York) Introduction to Linear Algebraic Groups November 19, 2009 5 / 21 This effectively produces a complete dictionary between ﬁnitely generated reduced algebras over k and afﬁne algebraic varieties. On the one hand, if A = k [V ], for some variety V then A is reduced (if f n = 0 then for x ∈ V we have f n (x) = f (x)n = 0, so f (x) = 0. Hence f = 0. One the other hand suppose A is just some ﬁnitely generated reduced commutative k -algebra. Suppose A has generators a1 , a2 , . . . , an . Consider V = An . Then k [V ] = k [X1 , . . . , Xn ], the free polynomial algebra in the coordinate functions X1 , . . . , Xn . Stephen Donkin (University of York) Introduction to Linear Algebraic Groups November 19, 2009 5 / 21 This effectively produces a complete dictionary between ﬁnitely generated reduced algebras over k and afﬁne algebraic varieties. On the one hand, if A = k [V ], for some variety V then A is reduced (if f n = 0 then for x ∈ V we have f n (x) = f (x)n = 0, so f (x) = 0. Hence f = 0. One the other hand suppose A is just some ﬁnitely generated reduced commutative k -algebra. Suppose A has generators a1 , a2 , . . . , an . Consider V = An . Then k [V ] = k [X1 , . . . , Xn ], the free polynomial algebra in the coordinate functions X1 , . . . , Xn . Stephen Donkin (University of York) Introduction to Linear Algebraic Groups November 19, 2009 5 / 21 We deﬁne θ : k [X1 , . . . , Xn ] → A to be the k -algebra homomorphism taking Xi to ai . Then θ is surjective with kernel I, say. Thus θ induces an isomorphism from k [V ]/I to A. Let Z = V(I), the closed set in V deﬁned by I. Now Z is an afﬁne variety with coordinate algebra k [V ]/J, where J is the set of f ∈ √ ] such that f |Z = 0. This means that √ k [V J = N (V(I)) = I. Now I/I = N(k [V ]/I) = 0 (because k [V ]/I is isomorphic to A hence reduced). So J = I. Hence k [Z ] is isomorphic to k [V ]/I and hence to A. So every commutative ﬁnitely generated reduced k -algebra is (isomorphic to) the coordinate algebra of some afﬁne variety. Stephen Donkin (University of York) Introduction to Linear Algebraic Groups November 19, 2009 6 / 21 One may make this correspondence a bit more canonical and precise as follows. Stephen Donkin (University of York) Introduction to Linear Algebraic Groups November 19, 2009 7 / 21 One may make this correspondence a bit more canonical and precise as follows. Remark Let A be a ﬁnitely generated k -algebra which is reduced (i.e. 0 is the only nilpotent element). We write Var(A) for the set Homk −alg (A, k ). We have a natural map θ : A → Map(Var(A), k ), given by θ(a)(α) = α(a), for α ∈ Var(A). By the Nullstellensatz, this map is injective. Identifying A with an algebra of functions on Var(A) via θ we thus obtain an afﬁne algebraic variety (Var(A), A). It follows now from Exercise 6 of Chapter 1 that the category of afﬁne varieties is anti-equivalent to the category of ﬁnitely generated reduced k -algebras. Stephen Donkin (University of York) Introduction to Linear Algebraic Groups November 19, 2009 7 / 21 Dominant Morphisms Stephen Donkin (University of York) Introduction to Linear Algebraic Groups November 19, 2009 8 / 21 Dominant Morphisms A continuous map φ : X → Y is called dominant if Im(f ) is dense in Y . Stephen Donkin (University of York) Introduction to Linear Algebraic Groups November 19, 2009 8 / 21 Dominant Morphisms A continuous map φ : X → Y is called dominant if Im(f ) is dense in Y . Exercise Show that a morphism of afﬁne varieties φ : X → Y is dominant if and only if the comorphism φ : k [Y ] → k [X ] is injective. Stephen Donkin (University of York) Introduction to Linear Algebraic Groups November 19, 2009 8 / 21 Dominant Morphisms A continuous map φ : X → Y is called dominant if Im(f ) is dense in Y . Exercise Show that a morphism of afﬁne varieties φ : X → Y is dominant if and only if the comorphism φ : k [Y ] → k [X ] is injective. We will call a dominant morphism of varieties φ : X → Y ﬁnite if k [X ] is an integral extension of Im(φ ). Stephen Donkin (University of York) Introduction to Linear Algebraic Groups November 19, 2009 8 / 21 Dominant Morphisms A continuous map φ : X → Y is called dominant if Im(f ) is dense in Y . Exercise Show that a morphism of afﬁne varieties φ : X → Y is dominant if and only if the comorphism φ : k [Y ] → k [X ] is injective. We will call a dominant morphism of varieties φ : X → Y ﬁnite if k [X ] is an integral extension of Im(φ ). In the prove of the following Lemma we shall use the result (proved in the appendix) that if A is a subring of B with B integral over A then for each prime ideal P of A there exists a prime ideal Q of B “over P", meaning that A Q = P. Stephen Donkin (University of York) Introduction to Linear Algebraic Groups November 19, 2009 8 / 21 We will also use the following observation. Let V be a variety and x ∈ V . Then N = {f ∈ k [V ] | f (x) = 0} is an ideal of codimension 1 (i.e. dim k [V ]/N = 1), in fact N = Ker( x ) and x : k [V ] → k induces an isomorphism k [V ]/N → k . Conversely, given an ideal M of k [V ] of codimension 1, we have k [V ] = M ⊕ k and we may deﬁne a homomorphism α : k [V ] → k , by α(m + λ) = λ. Since V is a variety, we have α = x , for some x ∈ V and it follows that M = Ker( x ) = {f ∈ k [V ] | f (x) = 0}. Stephen Donkin (University of York) Introduction to Linear Algebraic Groups November 19, 2009 9 / 21 We will also use the following observation. Let V be a variety and x ∈ V . Then N = {f ∈ k [V ] | f (x) = 0} is an ideal of codimension 1 (i.e. dim k [V ]/N = 1), in fact N = Ker( x ) and x : k [V ] → k induces an isomorphism k [V ]/N → k . Conversely, given an ideal M of k [V ] of codimension 1, we have k [V ] = M ⊕ k and we may deﬁne a homomorphism α : k [V ] → k , by α(m + λ) = λ. Since V is a variety, we have α = x , for some x ∈ V and it follows that M = Ker( x ) = {f ∈ k [V ] | f (x) = 0}. Stephen Donkin (University of York) Introduction to Linear Algebraic Groups November 19, 2009 9 / 21 Lemma Let φ : V → W be a dominant ﬁnite morphism of afﬁne varieties. Then φ is surjective and takes closed sets to closed sets. Stephen Donkin (University of York) Introduction to Linear Algebraic Groups November 19, 2009 10 / 21 Lemma Let φ : V → W be a dominant ﬁnite morphism of afﬁne varieties. Then φ is surjective and takes closed sets to closed sets. Proof First we will check that φ(V ) = W . Let A = k [V ] and let B = Im(φ ). Let y ∈ W and let N = {f ∈ k [W ] | f (y ) = 0}. Note that V(N) = {y }. Then φ (N) is an ideal of codimension 1 in B, in particular φ (N) is a prime ideal of B. Hence there exists a prime ideal P of A such that P B = φ (N). Stephen Donkin (University of York) Introduction to Linear Algebraic Groups November 19, 2009 10 / 21 Lemma Let φ : V → W be a dominant ﬁnite morphism of afﬁne varieties. Then φ is surjective and takes closed sets to closed sets. Proof First we will check that φ(V ) = W . Let A = k [V ] and let B = Im(φ ). Let y ∈ W and let N = {f ∈ k [W ] | f (y ) = 0}. Note that V(N) = {y }. Then φ (N) is an ideal of codimension 1 in B, in particular φ (N) is a prime ideal of B. Hence there exists a prime ideal P of A such that P B = φ (N). Let M be a maximal ideal of A containing P. Then M has codimension 1 in A (by the Nullstellensatz) and φ (N) ≤ M. Stephen Donkin (University of York) Introduction to Linear Algebraic Groups November 19, 2009 10 / 21 Let x ∈ V be the element such that Ker( x ) = M. Then we have n(φ(x)) = φ (n)(x) = x (φ (n)) = 0, for n ∈ N. Hence φ(x) ∈ V(N) = {y }. Hence φ(x) = y , and φ is onto. Stephen Donkin (University of York) Introduction to Linear Algebraic Groups November 19, 2009 11 / 21 Let x ∈ V be the element such that Ker( x ) = M. Then we have n(φ(x)) = φ (n)(x) = x (φ (n)) = 0, for n ∈ N. Hence φ(x) ∈ V(N) = {y }. Hence φ(x) = y , and φ is onto. Now if Z is a closed set in X and let T be the closure of φ(Z ) in Y . Then one checks that the restriction φ0 : Z → T is dominant and ﬁnite. So by the above φ0 maps Z onto T , i.e., φ(Z ) is closed. So φ takes closed sets to closed sets. Stephen Donkin (University of York) Introduction to Linear Algebraic Groups November 19, 2009 11 / 21 The next result shows that a dominant map is “approximately a ﬁnite map". Stephen Donkin (University of York) Introduction to Linear Algebraic Groups November 19, 2009 12 / 21 The next result shows that a dominant map is “approximately a ﬁnite map". Proposition Let φ : V → W be a dominant morphism of irreducible afﬁne varieties. Then there exists 0 = g ∈ k [W ] such that the restriction Vφ (g) → Wg factorizes as Vφ (g) → Wg × An → Wg , for some n ≥ 0, where the ﬁrst map is a ﬁnite dominant morphism and the second map is projection. Stephen Donkin (University of York) Introduction to Linear Algebraic Groups November 19, 2009 12 / 21 The next result shows that a dominant map is “approximately a ﬁnite map". Proposition Let φ : V → W be a dominant morphism of irreducible afﬁne varieties. Then there exists 0 = g ∈ k [W ] such that the restriction Vφ (g) → Wg factorizes as Vφ (g) → Wg × An → Wg , for some n ≥ 0, where the ﬁrst map is a ﬁnite dominant morphism and the second map is projection. To prove this one puts B = φ (k [W ]). Then make the ring of fractions of A with denominators non-zero elements of B. This produces a ﬁnitely generated algebra over F , the ﬁeld of fractions of B. Then one uses Noether Normalization. Stephen Donkin (University of York) Introduction to Linear Algebraic Groups November 19, 2009 12 / 21 It has an important corollary. Stephen Donkin (University of York) Introduction to Linear Algebraic Groups November 19, 2009 13 / 21 It has an important corollary. Corollary (Chevalley’s Theorem) Let φ : V → W be a dominant morphism of irreducible varieties. Then Im(φ) contains a non-empty open set of W . Stephen Donkin (University of York) Introduction to Linear Algebraic Groups November 19, 2009 13 / 21 It has an important corollary. Corollary (Chevalley’s Theorem) Let φ : V → W be a dominant morphism of irreducible varieties. Then Im(φ) contains a non-empty open set of W . Proof We factorize φ on a principal open set Vf → Wg × An → Wg , as above. The ﬁrst map is ﬁnite so onto by the Lemma above. The second map (projection) is clearly onto. Hence Wg = φ(Vf ) ⊆ Im(φ), so Im(φ) contains the open set Wg of W . Stephen Donkin (University of York) Introduction to Linear Algebraic Groups November 19, 2009 13 / 21 We have an application to algebraic groups. Stephen Donkin (University of York) Introduction to Linear Algebraic Groups November 19, 2009 14 / 21 We have an application to algebraic groups. Proposition 2 Let φ : G → H be a morphism of algebraic groups. Then the image of φ is closed. Stephen Donkin (University of York) Introduction to Linear Algebraic Groups November 19, 2009 14 / 21 We have an application to algebraic groups. Proposition 2 Let φ : G → H be a morphism of algebraic groups. Then the image of φ is closed. n n Proof We can write G = xi G0 . So φ(G) = φ(xi )φ(G0 ). So its i=1 i=1 enough to prove that φ(G0 ) is closed, i.e., we may assume that G is connected. Replacing H by the closure of φ(G) we can assume that φ is dominant. Now φ(G) contains an open set V , say. Stephen Donkin (University of York) Introduction to Linear Algebraic Groups November 19, 2009 14 / 21 But then φ(G) = φ(g)V is open, i.e., φ(G) is open. Hence g∈G hφ(G) is open. So the complement of φ(G) is open and φ(G) h∈H\φ(G) is closed. Stephen Donkin (University of York) Introduction to Linear Algebraic Groups November 19, 2009 15 / 21 But then φ(G) = φ(g)V is open, i.e., φ(G) is open. Hence g∈G hφ(G) is open. So the complement of φ(G) is open and φ(G) h∈H\φ(G) is closed. Stephen Donkin (University of York) Introduction to Linear Algebraic Groups November 19, 2009 15 / 21 But then φ(G) = φ(g)V is open, i.e., φ(G) is open. Hence g∈G hφ(G) is open. So the complement of φ(G) is open and φ(G) h∈H\φ(G) is closed. Dimension and Tangent Spaces Stephen Donkin (University of York) Introduction to Linear Algebraic Groups November 19, 2009 15 / 21 But then φ(G) = φ(g)V is open, i.e., φ(G) is open. Hence g∈G hφ(G) is open. So the complement of φ(G) is open and φ(G) h∈H\φ(G) is closed. Dimension and Tangent Spaces A topological space X has dimension n (see A5) if there is a strictly ascending chain of closed irreducible sets X0 ⊂ X1 ⊂ · · · ⊂ Xn , of length n, and none longer. Stephen Donkin (University of York) Introduction to Linear Algebraic Groups November 19, 2009 15 / 21 If X is an afﬁne variety then a strictly ascending chain of closed irreducible subsets X0 ⊂ X1 · · · ⊂ · · · ⊂ Xr gives rise to a strictly ascending chain of prime ideals N (Xr ) ⊂ · · · ⊂ N (X1 ) ⊂ N (X0 ) of k [X ] and, conversely, a strictly chain of prime ideals P0 ⊂ P1 ⊂ · · · ⊂ Pr gives rise to a strictly ascending chain of closed irreducible subsets V(Pr ) ⊂ · · · ⊂ V(P1 ) ⊂ V(P0 ). Stephen Donkin (University of York) Introduction to Linear Algebraic Groups November 19, 2009 16 / 21 The dimension of X is therefore the dimension dim k [X ] of the ring k [X ], as discussed in A5. In particular the dimension is ﬁnite and, if X is irreducible, is equal to the transcendence degree of the ﬁeld of fractions of k [X ] over k , see A5 Proposition 1. Of course, from the deﬁnition, if X is irreducible and Y is a proper closed subset of X then we have dim Y < dim X . This property will be frequently used in what follows. Stephen Donkin (University of York) Introduction to Linear Algebraic Groups November 19, 2009 17 / 21 There is also a local notion of dimension, that of the dimension of the tangent space Tx (V ), of a point x in an afﬁne variety V , as a k vector space. In this section we see how these different notions are related. Stephen Donkin (University of York) Introduction to Linear Algebraic Groups November 19, 2009 18 / 21 There is also a local notion of dimension, that of the dimension of the tangent space Tx (V ), of a point x in an afﬁne variety V , as a k vector space. In this section we see how these different notions are related. Let (V , A) be an afﬁne variety. For x ∈ V we deﬁne Tx (V ) to be the space of all linear maps α : A → k such that α(fg) = f (x)α(g) + α(f )g(x), for all f , g ∈ A. We get α(1) = α(12 ) = α(1) + α(1) so that α(1) = 0 and, by linearity, α(c.1) = 0 for all c ∈ k . Let g1 , . . . , gr be k -algebra generators of A and let Z be the k -span of g1 , . . . , gr . Now if α ∈ Tx (V ) vanishes on f , g ∈ A then clearly α vanishes on fg. Stephen Donkin (University of York) Introduction to Linear Algebraic Groups November 19, 2009 18 / 21 Thus we get that α vanishes on the algebra generated by g1 , . . . , gr , which is A, and hence α = 0. Thus restriction Tx (V ) → Z ∗ is injective, where Z ∗ = Homk (Z , k ) is the dual of Z . Thus we get that dim Tx (X ) is ﬁnite dimensional in fact we have: Stephen Donkin (University of York) Introduction to Linear Algebraic Groups November 19, 2009 19 / 21 Thus we get that α vanishes on the algebra generated by g1 , . . . , gr , which is A, and hence α = 0. Thus restriction Tx (V ) → Z ∗ is injective, where Z ∗ = Homk (Z , k ) is the dual of Z . Thus we get that dim Tx (X ) is ﬁnite dimensional in fact we have: (1) If k [V ] may be generated by r elements then we have dim Tx (V ) ≤ r for all x ∈ V . Stephen Donkin (University of York) Introduction to Linear Algebraic Groups November 19, 2009 19 / 21 Example We take V = An so that k [V ] = k [X1 , . . . , Xn ]. Let ∂f a = (a1 , . . . , an ) ∈ V . We deﬁne αi : k [V ] → k by αi (f ) = |a , the ∂Xi formal partial derivative of f = f (X1 , . . . , Xn ) evaluated at the point a ∈ An , for 1 ≤ i ≤ n. It is easy to see that the functions α1 , . . . , αn are linearly independent elements of Ta (An ) and hence form a basis. Stephen Donkin (University of York) Introduction to Linear Algebraic Groups November 19, 2009 20 / 21 Suppose that φ : V → W is a morphism of varieties and that x ∈ V . For each α ∈ Tx (V ) the composite α ◦ φ is an element of Tφ(x) (V ). In this way we get a linear map dφx : Tx (V ) → Tφ(x) (W ), called the differential of φ, deﬁned by dφx (α) = α ◦ φ . Stephen Donkin (University of York) Introduction to Linear Algebraic Groups November 19, 2009 21 / 21 Suppose that φ : V → W is a morphism of varieties and that x ∈ V . For each α ∈ Tx (V ) the composite α ◦ φ is an element of Tφ(x) (V ). In this way we get a linear map dφx : Tx (V ) → Tφ(x) (W ), called the differential of φ, deﬁned by dφx (α) = α ◦ φ . Exercise 4. Let V be an afﬁne variety and let U be a principal open set. Show that, for x ∈ U, the inclusion map U → V induces an isomorphism Tx (U) → Tx (V ). Stephen Donkin (University of York) Introduction to Linear Algebraic Groups November 19, 2009 21 / 21