Bourbaki

Proposition 5

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

Algebra, VIII, p. 343, § 17. Reduced Norms and Traces

For any xx in A, let txt_x be the linear form y↦y\mapsto TrdA/K(xy)_{A/K}(xy) on A.

a) The mapping t:x↦txt:x\mapsto t_x is an isomorphism of (A,A)(A,A)-bimodules from A to its dual HomK(A,K)_K(A,K).

b) Let hh be a linear form on A. The following properties are equivalent:

(i) There exists an element λ\lambda of K such that h(x)=λh(x) =\lambda TrdA/K(x)_{A/K}(x)

for every x∈Ax\in A.

(ii) We have h(xy)=h(yx)h(xy) =h(yx) for all x,yx, y in A.

Recall (II, §1, No. 14, p. 225–226) that the (A(A, A)-bimodule structure on A∗=A^*= HomK(A,K)_K(A,K) is defined by the relation

⟨atb,c⟩=⟨t,bca⟩(43) \langle atb, c\rangle =\langle t, bca\rangle \tag{43}

for a,b,c∈Aa, b, c\in A and t∈A∗t\in A^*. In particular, for every xx in A, we have

⟨atxb,c⟩=⟨tx,bca⟩=\langle at_xb, c\rangle =\langle t_x, bca\rangle = TrdA/K(xbca)_{A/K}(xbca),

⟨taxb,c⟩=\langle t_{axb}, c\rangle = TrdA/K(axbc)_{A/K}(axbc),

and these two elements are equal by formula (37) of VIII, p. 342. We therefore have atxb=taxbat_xb=t_{axb}, which means that tt is a homomorphism of (A(A, A)-bimodules from A to A∗A^*.

We choose an extension L of the field K and an isomorphism θ\theta from the L-algebra A(L)A_{(L)} to the matrix algebra Mn(L)\mathbf{M}_n(L) (VIII, p. 252, Theorem 1). We identify the vector space (A∗)(L)(A^*)_{(L)} with the dual of the vector space A(L)A_{(L)} over the field L. By Proposition 4 of VIII, p. 341, with these conventions, we have

(44) TrdA(L)/L=1L⊗_{A_{(L)}/L}= 1_L\otimes TrdA/K_{A/K}.

Let t(L)t_{(L)} be the L-linear mapping from A(L)A_{(L)} to A(L)∗A^*_{(L)} deduced from tt by extension of scalars; by formula (44) and Corollary 1, we have

(45) ⟨t(L)(x),y⟩=\langle t_{(L)}(x), y\rangle = TrdA(L)/L(xy)=_{A_{(L)}/L}(xy) = Tr(θ(x)θ(y)\theta (x)\theta (y))

for x,yx, y in A(L)A_{(L)}. By Proposition 7 of II, §10, No. 11, p. 358, the mapping t(L)t_{(L)} is bijective; it follows that tt is bijective. We have proved a).

Let hh be in A∗A^*; by the above, there exists an element aa of A such that hh is equal to tat_a. By a), we have

h(xy)−h(yx)=ta(xy−yx)=tax(y)−txa(y) h(xy)-h(yx) =t_a(xy-yx) =t_{ax}(y)-t_{xa}(y)

Consequently, the relation “h(xy)=h(yx)h(xy) =h(yx) for x,yx, y in A” is equivalent to “tax−xa=0t_{ax-xa}= 0 for every x∈x\in A”, and by part a) of the proof, this means that aa belongs to the center K of A. This proves b) in view of formula (36).

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.