Let $\sigma$ be an endomorphism of a skew field $D$ and consider the skew polynomial ring $$R= D[X,\sigma]$$
which is a ring of left polynomials $\sum_{i} a_i X^i$ with twisted multiplication given by
$$Xa = \sigma(a) X$$
Consider the $R$-quotient module
$$M_d:= R/R(X-d)$$
Note that $D$ is an $R$-module via the action $$Xr= \sigma(r) d, r \in D$$
I'm trying to show that
$$D \cong M_d$$ as $R$-modules (it will then follow that $M_d$ is simple).
I defined the map
$$\psi: R\to D: f \mapsto f(d)$$
and want to apply the isomorphism theorem on this map. But I have trouble showing this map is $R$-linear. I can see it suffices to show that
$$\psi(Xf) = X\psi(f)$$
but when I calculate this with a generic element they are not equal?
Any help will be appreciated!
Best Answer
The issue is that you should not be sending $X^n$ to $d^n$ but rather to $$\sigma^{n-1}(d)\cdots \sigma(d)d.$$
For instance, with your definition $\psi(X^2)$ is sent to $d^2$ but $X\psi(X)$ is sent to $X\cdot d = \sigma(d)d$.