Exercise 3, Appendix 2

Let $A = A_0\oplus A_1$ be a graded ring of type $\mathbf{Z}/2\mathbf{Z}$. Suppose that $A_0$ is contained in the center of A and that every element of $A_1$ has square zero.

a) Let $A^2_1$ be the $\mathbf{Z}$-submodule of $A_0$ generated by the products of two elements of $A_1$; show that $A^2_1$ is a nil ideal of $A_0$ and that $A^2_1\oplus A_1$ is a two-sided nil ideal of A.

b) Let $p, q$ be integers $\geqslant 0$. Denote by $A^{p|q}$ the free graded right A-module $A^p_d\oplus A_d(1)^q$ (II, §11, No. 2, p. 365, Example 3) and by $\mathbf{G}\mathbf{L}(p|q,A)$ the group of (graded) automorphisms (of degree 0) of $A^{p|q}$. Such an automorphism is represented in the canonical basis of $A^{p|q}$ by a matrix $X = \begin{pmatrix} A & B \\ C & D \end{pmatrix}$ with $A\in \mathbf{M}_p(A_0)$, $B\in \mathbf{M}_{p,q}(A_1),C\in \mathbf{M}_{q,p}(A_1),D\in \mathbf{M}_q(A_0)$. Prove that the matrices A and D are invertible (use a) and Exercise 16 of VIII, p. 168). c) For X as above, set sdetX = det($A-$ BD$^{-1}C$) det $D^{-1}$. Prove that sdet X is an invertible element of $A_0$.

d) Prove that every matrix $X\in \mathbf{G}\mathbf{L}(p|q,A)$ can be written as a product

(Take $R = BD^{-1}$, $P = A-BD^{-1}C$, $Q = D$, $S = D^{-1}C$.)

e) Prove that sdet is a homomorphism from $\mathbf{G}\mathbf{L}(p|q,A)$ to the group of invertible elements of $A_0$. (Use d) to reduce to proving the equality sdet(XY) = sdet X sdet Y when X is of the form $\begin{pmatrix} I & 0 \\ S & I \end{pmatrix}$, where S is an elementary matrix. Observe that for every matrix $R\in \mathbf{M}_{p,q}(A_1)$, the product of two arbitrary entries of the matrix RS (resp. SR) is zero, and deduce the equality $\det (I-RS) = (\det (I-SR))^{-1}$.)

f ) Let $X = \begin{pmatrix} A & B \\ C & D \end{pmatrix}$ be an element of $\mathbf{G}\mathbf{L}(p|q,A)$; set $^{st}X = \begin{pmatrix} {}^tA & {}^tC \\ -{}^tB & {}^tD \end{pmatrix}$ and $^{\pi}X = \begin{pmatrix} D & C \\ B & A \end{pmatrix}$. Let $Y\in \mathbf{G}\mathbf{L}(p|q,A)$; we have $^{st}($XY$) =^{st}Y^{st}X$ and $^{\pi}($XY$) =^{\pi}X^{\pi}Y$. Prove the equalities sdet X = sdet$^{st}X$ = (sdet$^{\pi}X$)$^{-1}$ (use d) to calculate sdet$^{\pi}X$).