Bourbaki

Exercise 22, § 8

Algebra, III, 643

Bourbaki marks this one as harder than the rest.

Let F be a mapping of Mn(K)M_n(K) into a set E, where K is a commutative field, such that, for every triple of matrices X,Y,ZX, Y, Z in Mn(K)M_n(K),

F(XYZ)=F(XZY). F(XYZ) = F(XZY).

We except the case M2(F2)M_2(F_2); show then that there exists a mapping Φ\Phi of K into E such that F(X)=Φ(det⁡(X))F(X) = \Phi(\det(X)). (Using the Corollary to Proposition 17 (no. 9), show first that, for U∈SLn(K)U \in SL_n(K), F(X)=F(XU)F(X) = F(XU); deduce that, in GLn(K)GL_n(K), F(X)F(X) depends only on det⁡(X)\det(X). On the other hand, if, for r≤n−1r \leq n - 1, we write

Xr=(Ir000) X_r = \begin{pmatrix} I_r & 0 \\ 0 & 0 \end{pmatrix}

show that there exists a matrix YY of rank rr such that Xr−1=YXrX_{r-1} = YX_r and Y=XrYY = X_rY and deduce that F(X)F(X) has the same value for all matrices of rank <n< n.)

Solution

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Transcribed from the printed volume. Not translated.