Homework 1 for Algebraic Groups, Spring 2007 Solutions All Group

Homework 1 for Algebraic Groups, Spring 2007 Solutions

All group schemes and algebras are defined over a fixed field k.

Problem 1. State and prove the Yoneda Lemma.

Deﬁnition 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 1⊗id tt JJid⊗1 tt JJ tt JJ ztt J$ ∆ id⊗∆ k ⊗ AAJd ∆ : ⊗ k JJ tt JJ tt JJ tt ∆⊗id ²⊗id JJ tt id⊗² A ⊗ A / A ⊗ A ⊗ A t A ⊗ A

Deﬁnition 2. Let A, B be coalgebras. A map φ : A → B is a coalgebra homomorphism if it commutes with coproduct. Explicitly,

(φ ⊗ φ) ◦ ∆A = ∆B ◦ φ.

Problem 2. Let A be a Hopf algebra. Show that multiplication in A is a coalgebra map. Note: In fact, in the deﬁnition 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) ◦ ∆A⊗A = ∆A ◦ ×A

This holds, because of the following: ∆A is an algebra morphism. That is,

(2) ×A⊗A ◦ (∆A ⊗ ∆A) = ∆A ◦ ×A

However after performing the adequate identiﬁcations both ×A ⊗ ×A and ×A⊗A are the map:

A ⊗ A ⊗ A ⊗ A → A ⊗ A x ⊗ y ⊗ z ⊗ v 7→ xz ⊗ yv

and both ∆A⊗A 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 aﬃne 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) = {id} 6' Z/2Z. By hypothesis, G(k) = Homk(A, k) = Z/2Z. Now, since there’s an injection α : k → A, and Homk(A, −) is a left-exact, covariant functor, we have that α∗ : Homk(A, k) → Homk(A, A) is also injective. But since α∗ is an injection from G(k) = Z/2Z to G(A) = Z/2Z, it must be an isomorphism. However, id : A → A ∈ Homk(A, A) is not in the image of α∗, since it can’t be written as α ◦ φ for some φ ∈ Homk(A, k) because α is not surjective (k 6= A). So α∗ 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 ﬁeld extension K/k. Is it true that G ' pt as a group scheme? Solution. (Antonion Kirson). Let k be the ﬁeld 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 ﬁeld 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 deﬁned by × GLn(R) = {(aij ∈ Matn(R) | det(aij) ∈ R } 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 ∈ 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 ∈ 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 ∈ Homk-alg(A, A) is given by σ(Xij) = Xeij/ det(Xij ).