Ore condition and nilpotence

Proof. Assume first that (1) is satisfied. Let \(x \in \pi _n(R/Rs)\); we wish to show that it is annihilated by some element of \(S\). Using the exact sequence \[ \pi _n(R/Rs) \xrightarrow {\phi } \pi _{n-d-1}(R) \xrightarrow {\cdot s} \pi _{n-1}(R), \] we deduce that \(\phi (x) \in \pi _{n-d-1}(R)\) is annihilated by right multiplication by \(s\). Using condition (b) of the left Ore condition, we deduce that there exists an element \(t \in S\) of degree \(d'\) such that \(0 = t\phi (x) = \phi (tx) \in \pi _{n+d'-d-1}(R)\). Using the exactness of the sequence \[ \pi _{n+d'}(R) \xrightarrow {\psi } \pi _{n+d'}(R/Rs) \xrightarrow {\phi } \pi _{n+d'-d-1}(R), \] we conclude that \(tx = \psi (y)\) for some \(y \in \pi _{n+d'}(R)\). Using condition (a) of the left Ore condition, we can find an element \(u \in S\) of degree \(d''\) and an element \(z \in \pi _{n+d'+d''-d}(R)\) such that \(uy=zs\). It follows that the image of \(uy\) in \(\pi _{n+d'+d''}(R/Rs)\) vanishes, so that \((ut)x = u(tx) = u\psi (y) = \psi (uy)=0\). This completes the proof that (1) implies (2).

Now suppose that (2) is satisfied. We will show that \(S\) satisfies conditions (a) and (b) of the left Ore condition. For (a), suppose that \(x \in \pi _*(R)\) is a homogeneous element of degree \(n\) and \(s \in S\) has degree \(d\). We wish to show that there exist \(y \in \pi _*(R)\) and \(t \in S\) such that \(tx=ys\). By \(S\)-nilpotence, the image of \(x\) in \(\pi _n(R/Rs)\) is annihilated by multiplication by some homogeneous element \(t \in S\) of degree \(d'\). It follows that \(tx\) belongs to the kernel of the map \(\pi _{n+d'}(R) \to \pi _{n+d'}(R/Rs)\), and therefore to the image of the map \(\cdot s\colon \pi _{n+d'-d}(R) \to \pi _{n+d'}(R)\) given by right multiplication by \(s\). Thus \(tx=ys\) for some \(y\in \pi _{n+d'-d}(R)\), which proves (a).

We now verify (b). It suffices to show that if \(x \in \pi _n(R)\) is annihilated by right multiplication by some element \(s \in S\) of degree \(d\), then \(tx=0\) for some \(t \in S\). The exactness of the sequence \[ \pi _{n+d+1}(R/Rs) \xrightarrow {\phi } \pi _n(R) \xrightarrow {\cdot s} \pi _{n+d}(R) \] shows that \(x = \phi (y)\) for some \(y \in \pi _{n+d+1}(R/Rs)\). Since \(R/Rs\) is \(S\)-nilpotent, there exists an element \(t \in S\) such that \(ty=0\), from which it follows immediately that \(tx = t\phi (y) = \phi (ty) = 0\). □