Graded modules over an algebra with an invertible odd element

Write $u^{-1}=x_0+x_1$ with $x_i\in A^i$. From $1=u{u}^{-1}=u{x}_0+u{x}_1$, the odd component $u{x}_0$ and even component $u{x}_1$ must be $0$ and $1$, respectively. Since $u$ is invertible, $x_0=0$ and $u^{-1}=x_1\in A^1$. Consequently $A^1=u{A}^0$: inclusion follows from [F1], and for $a\in A^1$, $u^{-1}a\in A^0$ gives $a=u(u^{-1}a)$.

For any graded left $A$-module $M$, multiplication by $u$ maps $M^0$ bijectively onto $M^1$: it maps into $M^1$ by [F1], and its inverse $u^{-1}$ maps $M^1$ into $M^0$. Thus every homogeneous component of $M$ is generated from $M^0$ by the $A$-action.

Define $\mu :A{\otimes }_{A^0}{M}^0\to M$ by $\mu (a\otimes m)=am$. The multiplication map is balanced over $A^0$, so [F3] makes $\mu$ well defined; associativity makes it $A$-linear. Its even part $A^0{\otimes }_{A^0}{M}^0\to M^0$ is an isomorphism with inverse $m\mapsto 1\otimes m$, by the balancing relation. By step 1.1, the map $M^0\to A^1{\otimes }_{A^0}{M}^0$, $m\mapsto u\otimes m$, is an isomorphism: its inverse is $a_1\otimes m\mapsto (u^{-1}{a}_1)m$, which is well defined because $u^{-1}{a}_1\in A^0$ and respects balancing. Under this identification, the odd part of $\mu$ is $m\mapsto um$, bijective by step 2.1. Hence $\mu$ is a graded $A$-module isomorphism.

If $L\subseteq M$ is a graded submodule, then $L^1=u{L}^0$ by invertibility of $u$ and the grading. Conversely, for any $A^0$-submodule $N\subseteq M^0$, the subspace $N\oplus uN$ is graded and is stable under $A$: for $a\in A^0$, $u^{-1}au\in A^0$, so $a(un)=u(u^{-1}au)n\in uN$; for $b\in A^1$, step 1.1 writes $b=u{a}_0$ with $a_0\in A^0$, whence $bN\subseteq uN$ and $b(uN)\subseteq N$ because $u{a}_{0}u\in A^0$. Thus it is an $A$-submodule. By the definition of graded-simple in [F2], $M$ has a proper nonzero graded submodule exactly when $M^0$ has a proper nonzero $A^0$-submodule.

For any left $A^0$-module $N$, the induced module $A{\otimes }_{A^0}N$ splits as $A^0{\otimes }_{A^0}N\oplus A^1{\otimes }_{A^0}N$: the map $(a_0+a_1)\otimes n\mapsto (a_0\otimes n,a_1\otimes n)$ is well defined by balancing and has inverse given by addition. This is a grading because left multiplication respects the parity of $A$, while the balancing relations use only even elements. Its degree-zero part is isomorphic to $N$ by $a_0\otimes n\mapsto a_{0}n$, with inverse $n\mapsto 1\otimes n$. If $N$ is simple, step 3.2 makes the induced module graded-simple. Conversely, step 3.1 identifies any graded $M$ with the module induced from $M^0$, and step 3.2 says $M$ is graded-simple exactly when $M^0$ is simple. Restriction to degree zero and induction are therefore inverse on isomorphism classes: even $A$-isomorphisms restrict to $A^0$-isomorphisms, and $A^0$-isomorphisms induce even $A$-isomorphisms.