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 $\alpha\in\mathbb{F}$. 



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> If $\mathbb{F}$ is algebraically closed, is it true that the irreducible elements are exactly those of degree one? 
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What about the special case when $\mathbb{F}$ has characteristic $p>0$, and $\sigma$ is the Frobenius automorphism?



Irreducible skew polynomials over an algebraically closed field
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.


