Bourbaki

Exercise 25, § 11

Algebra, III, 653

Let C be a cogebra over a ring A. A sub-A-module S of C is called a coideal if c(S) ⊂ Im(S ⊗ C) + Im(C ⊗ S). If C is counital, with counit γ, a coideal S is called conull if γ(x) = 0 for x ∈ S.

(a) A right or left coideal is a coideal. Every sum of coideals is a coideal. (b) If S is a coideal of C, the orthogonal S^0 is a subalgebra of the dual algebra C*. If C is counital and S is conull, S^0 is unital. (c) Suppose that A is a field and C a counital cogebra. Show that if E' is a unital subalgebra of C* with the same unit element as C*, the orthogonal E'^0 in C is a conull coideal. (d) If S is a coideal of C, show that there exists on C/S one and only one cogebra structure such that the canonical mapping p: C → C/S is a cogebra morphism. If C is counital and S is conull, C/S is counital. (e) For every cogebra morphism f: C → C', S = Ker(f) is a coideal of C, S' = Im(f) a subcogebra of C' and, in the canonical factorization

C→pC/S→gS′→jC′ \begin{array}{ccc} C & \xrightarrow{p} & C/S \\ & & \xrightarrow{g} \\ & & S' \\ & & \xrightarrow{j} \\ & & C' \end{array}

of f, g is a cogebra isomorphism.

(26) Let C be a coassociative counital cogebra over a commutative field K.

(a) For every vector subspace E of C which is a left C*-module under the law (u, x) ↦ u ↼ x, the left annihilator of this C*-module is a two-sided ideal J' of the algebra C*. Show that if E is finite-dimensional over K, C*/J' is finite-dimensional. (b) Deduce from (a) that for every element x ∈ C, the subcogebra of C generated by x is finite-dimensional. (Note that the C*-module E generated by x is finite-dimensional and that, if J' is its left annihilator, x ∈ J'^0, the orthogonal of J' in C.)

(27) Let B be an associative unital algebra over a commutative field K and let m: B ⊗_K B → B be the K-linear mapping defining the multiplication on B. The tensor product B* ⊗_K B* (where B* is the dual vector space of the vector space B) is canonically identified with a vector subspace of (B ⊗_K B)* (II, § 7, no. 7, Proposition 16).

(a) Show that for an element w ∈ B* the following conditions are equivalent: (1) ^t m(w) ∈ B* ⊗ B*; (2) the set of x ↼ w, where x runs through B, is a finite-dimensional subspace of B*; (3) the set of w ↼ x, where x runs through B, is a finite-dimensional subspace of B*; (4) the set of x ↼ w ↼ y, where x and y run through B, is contained in a finite-dimensional subspace of B*.

(b) Let B′⊂B∗B' \subset B^* be the set of w∈B∗w \in B^* satisfying the equivalent conditions in (a). Show that tm(B′)⊂B′⊗KB′^t m(B') \subset B' \otimes_K B'. (For w∈B′w \in B', write tm(w)=∑iui⊗vi^t m(w) = \sum_i u_i \otimes v_i, where for example the uiu_i are linearly independent in B∗B^*; show that then vi∈B′v_i \in B' for all ii using the associativity of BB and arguing by reductio ad absurdum. ) B′B' is therefore the largest cogebra contained in B∗B^*, for which tm^t m is the coproduct. B′B' is called the dual cogebra of the algebra BB. When BB is finite-dimensional, B′=B∗B' = B^*.

(c) Show that if BB is a commutative field of infinite dimension over KK, B′={0}B' = \{0\}. (Note that the orthogonal of the set of x⊥wx \perp w, for same w∈B∗w \in B^*, xx running through BB, is an ideal of BB.)

(d) Let B1,B2B_1, B_2 be two KK-algebras and f:B1→B2f : B_1 \to B_2 a KK-homomorphism of algebras. Show that tf(B2′)⊂B1′^t f(B'_2) \subset B'_1 and that tf^t f, restricted to B2′B'_2, is a cogebr*a homomorphism of B2′B'_2 into B1′B'_1.

(e) If CC is a coassociative counital cogebr*a over KK, the canonical injection C→C∗∗C \to C^{**} of the vector space CC into its bidual maps CC onto a subspace of (C∗)′(C^*)', the dual cogebr*a of the algebra C∗C^* dual to CC, and is a cogebr*a homomorphism of CC into (C∗)′(C^*)'.

Solution

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Transcribed from the printed volume. Not translated.