Homework 1 for Algebraic Groups, Spring 2007 Solutions All group schemes and algebras are de¯ned over a ¯xed ¯eld k. Problem 1. State and prove the Yoneda Lemma. De¯nition 1. Let A be a k-vector space. A is called a coalgebra if there exist maps ¢A : A ! A ­ A (coproduct) and ²A : A ! k (counit) such that the following diagrams commute: ¢ A / A ­ A A J tt JJ 1­id tt JJid­1 tt JJ tt JJ ztt J$ ¢ id­¢ k ­ AAJd ¢ : ­ k JJ tt JJ tt ² ² JJ tt ¢­id ²­id JJ tt id­² A ­ A / A ­ A ­ A ² t A ­ A De¯nition 2. Let A, B be coalgebras. A map Á : A ! B is a coalgebra homomorphism if it commutes with coproduct. Explicitly, (Á ­ Á) ± ¢A = ¢B ± Á: Problem 2. Let A be a Hopf algebra. Show that multiplication in A is a coalgebra map. Note: In fact, in the de¯nition of a Hopf algebra one can choose between the two equivalent conditions: (1) coproduct and counit are homomorphisms of algebras; (2) product and unit are homomorphisms of coalgebras. Solution. (Zsolt Patakfalvi). We want to prove, that £A is coalgebra morphism, i. e.: (1) (£A ­ £A) ± ¢A­A = ¢A ± £A This holds, because of the following: ¢A is an algebra morphism. That is, (2) £A­A ± (¢A ­ ¢A) = ¢A ± £A However after performing the adequate identi¯cations both £A ­ £A and £A­A are the map: A ­ A ­ A ­ A ! A ­ A x ­ y ­ z ­ v 7! xz ­ yv and both ¢A­A and ¢A ­ ¢A are: A ­ A ! A ­ A ­ A ­ A x ­ y 7! ¢(x) ­ ¢(y) So, (2) is just the same equation as (1). Problem 3. Does there exist a group scheme G such that G(R) ' Z=2 for any k-algebra R? 1 2 Solution 1. (Jeremy Berquist). There does not even exist an a±ne scheme XA such that XA(R) is a set with 2 elements for any such R. For, given two algebras R and S, their direct product » R £ S is also a k-algebra, and then we have XA(R £ S) = Homk¡alg(A; R £ S) = Homk¡alg(A; R) £ Homk¡alg(A; S) = XA(R) £ XA(S). Then if XA(R) and XA(S) each have two elements, XA(R £ S) must have 4. Solution 2. (Ariana Dundon). Suppose that G is a group scheme such that G(R) = Z=2Z for any k-algebra R. Let A = k[G]. We know that A 6= k, since Homk(k; k) = fidg 6' Z=2Z. By hypothesis, G(k) = Homk(A; k) = Z=2Z. Now, since there's an injection ® : k ! A, and Homk(A; ¡) is a left-exact, covariant functor, we have that ®¤ : Homk(A; k) ! Homk(A; A) is also injective. But since ®¤ is an injection from G(k) = Z=2Z to G(A) = Z=2Z, it must be an isomorphism. However, id : A ! A 2 Homk(A; A) is not in the image of ®¤, since it can't be written as ® ± Á for some Á 2 Homk(A; k) because ® is not surjective (k 6= A). So ®¤ is not surjective, which is a contradiction. Thus, such a group scheme cannot exist. Problem 4. Let G be a group scheme. Assume that G(K) is a trivial group for any ¯eld extension K=k. Is it true that G ' pt as a group scheme? Solution. (Antonion Kirson). Let k be the ¯eld with two elements, so that the characteristic of k is 2. In class it k[x] was shown that G = Ga(1) is a group scheme represented by (x2) . k[x] For any ¯eld extension K=k we have that Hom( (x2) ;K) =< e > since x must get sent to an element whose square is 0 in K. i.e. it must get sent to 0. But G is not isomorphic to pt: Problem 5. Let GLn be a group scheme de¯ned by £ GLn(R) = f(aij 2 Matn(R) j det(aij) 2 R g 1 (1) Show that the coproduct in k[GLn] = k[Xij ; ] is given by the formula det(Xij ) Xn ¢(Xij ) = Xik ­ Xkj k=1 (2) Find and prove the formula for the coniverse σ : k[GLn] ! k[GLn]. Solution. (Andrey Novoseltsev). (1). Let A = k[GLn]. Let f; g 2 GLn(A ­ A) = Homk-alg(A; A ­ A) be given by f(a) = a ­ 1 and g(a) = 1 ­ a. Then f corresponds to the matrix F = (fij), fij = Xij ­ 1, and g corresponds to the matrix G = (gij ), gij = 1 ­ Xij . Their product Pn Pn F ¢ G = H = (hij) has entries hij = k=1(Xik ­ 1) ¢ (1 ­ Xkj) = k=1 Xik ­ Xkj , Pn thus H corresponds to the map h given on generators by h(Xij) = k=1 Xik ­Xkj , as required. (2) Let A = k[GLn]. Then the identity map IdA 2 GLn(A) = Homk-alg(A; A) ¡1 corresponds to the matrix X = (Xij ). Its inverse is given by Cramer's Rule X = (Xeij = det(Xij )), where Xeij are cofactors. Thus the coinverse σ 2 Homk-alg(A; A) is given by σ(Xij) = Xeij= det(Xij )..