Bourbaki

Exercise 4, § 11

Algebra, III, 647

Let E be a commutative bigebra over A, mm the product E⊗E→EE \otimes E \to E, cc the coproduct E→E⊗EE \to E \otimes E, ee the unit element and γ\gamma the counit of E. (a) Show that for every commutative A-algebra B, the law of composition which associates with every ordered pair

(u,v)∈HomA−alg.(E,B)×HomA−alg.(E,B) (u, v) \in \mathrm{Hom}_{A-\mathrm{alg.}}(E, B) \times \mathrm{Hom}_{A-\mathrm{alg.}}(E, B)
the A-algebra homomorphism
E→cE⊗E→u⊗vB⊗B→mBB E \xrightarrow{c} E \otimes E \xrightarrow{u \otimes v} B \otimes B \xrightarrow{m_B} B
(where mBm_B is the product in B), defines a monoid structure on HomA−alg.(E,B)\mathrm{Hom}_{A-\mathrm{alg.}}(E, B). (b) In order that, for every commutative A-algebra B, HomA−alg.(E,B)\mathrm{Hom}_{A-\mathrm{alg.}}(E, B) be a group, it is necessary and sufficient that there exist an A-algebra homomorphism i:E→Ei: E \to E such that i(e)=ei(e) = e and the composite mappings m∘(1E⊗i)∘cm \circ (1_E \otimes i) \circ c and m∘(i⊗1E)∘cm \circ (i \otimes 1_E) \circ c are both equal to the mapping x↦γ(x)ex \mapsto \gamma(x)e. ii is then called an inversion in E; it is an isomorphism of E onto the opposite bigebra E0E^0.

Solution

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Transcribed from the printed volume. Not translated.