Bourbaki

Example 1

Tag 00DL, permanent. Cite it as [Bourbaki, Tag 00DL].

Algebra, VIII, p. 198, § 11. Grothendieck Groups

Let K be a commutative field. Let G be a group, and let K[G] be the algebra of the group G. We identify G with its canonical image in K[G] (III, §2, No. 6, p. 446).

We endow K[G] with the structure of a bigebra with coproduct cc and counit γ\gamma given by

c(g)=g⊗g,γ(g)=1(g∈G)(15) c(g) =g\otimes g ,\gamma (g) = 1(g\in G) \tag{15}

Let E and F be K[G]-modules; by formula (12), the K[G]-module structure on E⊗FE\otimes F is given by

g(x⊗y)=gx⊗gy(g∈G,x∈E,y∈F)(16) g(x\otimes y) =gx\otimes gy(g\in G, x\in E, y\in F) \tag{16}

The K[G]-module KγK_{\gamma} is the vector space K endowed with the action of G defined by gλ=λg\lambda =\lambda for g∈Gg\in G and λ∈K\lambda \in K.

The ring RK(K[G])R_K(K[G]) is also denoted by RK(G)R_K(G). It is commutative; its multiplication is given by [E][F]=[E⊗KF][E] [F] = [E\otimes_KF], and the unit element of RK(G)R_K(G) is [Kγ][K_{\gamma}].

Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.

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Transcribed from the printed edition, 2023, Springer Nature. Not translated.