Irreducible skew polynomials over an algebraically closed field

Question

Let $\mathbb{F}$ be a field, and denote with $\mathbb{F}[t,\sigma]$ the skew-polynomial ring, where $\sigma$ is an automorphim of $\mathbb{F}$. Recall that the multiplication of this ring is defined by the rule $t\cdot \alpha =\sigma(\alpha) t$ for every $\alpha\in\mathbb{F}$.

If $\mathbb{F}$ is algebraically closed, is it true that the irreducible elements are exactly those of degree one?

What about the special case when $\mathbb{F}$ has characteristic $p>0$, and $\sigma$ is the Frobenius automorphism?

Answer 1

There is no compelling reason for this proprty to be true in general, but it holds for quadratic polynomials in characteristic $p$ and the Frobenius automorphism.

Let us consider the special case of a monic reciprocal quadratic polynomial $p(t)=t^2-ct+1,\,$ to be factored as $(t-a)(t-b).\,$ Equating the coefficients, $a+\sigma(b)=c$ and $ab=1$, so $a=b^{-1}$ and $b^{-1}+\sigma(b)=c$. The resulting equation for $b$ is not algebraic in general, and need not have solutions: if ${\Bbb F}={\Bbb C}$ and $\sigma$ is the complex conjugation, $c=0$ would mean that $|b|=-1$, which is impossible; thus $p(t)=t^2+1$ is irreducible. On the other hand, if $\sigma(b)=b^p$ is the Frobenius automorphism in characteritic $p $ then the equation is algebraic, has a root by the algebraic closedness assumption, and such a factorization exists. (This argument easily extends to general quadratics.)

There is extensive literature on skew-polynomial rings and their generalizations, and I recommend consulting it for further information on this and other basic questions about their properties. For example the monograph of McConnell and Robson "Noncommutative Noetherian Rings" has a chapter devoted to these classes of rings.