I am trying to understand the following proof on P. Gille's and T. Szamuely's Central simple algebras and Galois cohomology.
The minimal left ideal $I_1$ cited in the proof is the set of all matrices which are $0$ away from the first column.
I don't understand two things:
- How can we assume that $\lambda(I_1)=I_1$ and
- Why the endomorphism of $K^n$ defined in the standard basis by $\lambda(M)$ has matrix $CMC^{-1}$?
Thank you.
For 1), $\lambda(I_1)$ must be one of the other $I_k$ because they are the minimal left ideals. Say $\lambda(I_1)=I_2$. Take the matrix $A$ determined by $Ae_1=e_2$, $Ae_2=e_1$ and $Ae_k=e_k$ for $k>2$. Then $A^{-1}I_2A=I_1$, so one can look at $X \mapsto A^{-1}\lambda(X)A$ instead of $\lambda$.
For 2), I have the following argument, but it seems actually overcomplicated. Identifying $I_1$ and $K^n$ by $M \mapsto Me_1$, then $C : K^n \to K^n$ can be seen as $Me_1 \mapsto CMe_1$ which is by definition of $C$, $Me_1 \mapsto \lambda(M)e_1$.
So let's calculate $$\lambda(N) : K^n \to K^n, \; x \mapsto \lambda(N)x,$$ or accordingly to the previous identification, $Me_1 \mapsto \lambda(N)Me_1$. The composite $CNC^{-1}$ is $$Me_1 \mapsto \lambda^{-1}(M)e_1=C^{-1}Me_1 \mapsto NC^{-1}Me_1 \mapsto CNC^{-1}Me_1 .$$ But $CNC^{-1}Me_1=\lambda(NC^{-1}M)e_1$ (because $I_1$ is a left ideal and by definition of $C$), and since $\lambda$ is an algebra isomorphism, $\lambda(NC^{-1}M)=\lambda(N)\lambda(C^{-1}M)$, so $$CNC^{-1}Me_1=\lambda(N)\lambda(C^{-1}M)e_1=\lambda(N) CC^{-1}Me_1=\lambda(N)Me_1.$$ Finally $\lambda(N)=CNC^{-1}$.