Exercise 4, Appendix 2

Keep the assumptions of Exercise 3. Let $M = M_0\oplus M_1$ be a finitely generated free graded right A-module; there exist unique integers $p, q$ such that M is isomorphic to $A^{p|q}$. We say that a basis of M is adapted if its first $p$ prime vectors are homogeneous of degree 0 and the last $q$ are homogeneous of degree 1.

a) Let $u$ be an automorphism of the graded module M, and let $M_{\mathscr{B}}(u)$ be its matrix with respect to an adapted basis of $\mathscr{B}$. Prove that the element sdet $M_{\mathscr{B}}(u)$ does not depend on the choice of $\mathscr{B}$. We denote it by sdet $u$.

b) Denote by $u(1)$ the endomorphism $u$ viewed as an automorphism of the graded module M(1). If $\mathscr{B}$ is an adapted basis of M, then the basis $^{\pi}\mathscr{B}$ obtained by interchanging the first $p$ vectors of $\mathscr{B}$ and the last $q$ is an adapted basis of M(1), and we have $M_{_{\pi}\mathscr{B}}(u(1)) =^{\pi}M_{\mathscr{B}}(u)$ (Exercise 3, f)). Deduce that we have sdet($u(1)$) $=$ (sdet $u$)$^{-1}$.

c) On every graded left A-module N, we define a canonical graded right A-module structure by setting $ax= (-1)^{ij}xa$ for $a\in A_i,x\in N_j$. In particular, we view the graded dual $M^{*gr}$ of M as a graded right A-module, also isomorphic to $A^{p|q}$. Let $u$ be an automorphism of M. We denote by $^{st}u$ the automorphism of $M^{*gr}$ characterized by the formula $\langle^{st}u(x^*), x\rangle = (-1)^{rs}\langle x^*, u(x)\rangle$ for $x^*\in (M^{*gr})_r,x\in M_s$. If $\mathscr{B}$ is an adapted basis of M, then the dual basis $\mathscr{B}^*$ is an adapted basis of $M^{*gr}$, and we have $M_{\mathscr{B}^*}(^{st}u) =^{st}M_{\mathscr{B}}(u)$ (Exercise 3, f)). We have sdet($^{st}u$) $=$ sdet($u$).