Can a polynomial over a division ring (skew-field) have more roots than its degree?
Solution
Yes. Over a noncommutative division ring, a polynomial can have more roots than its degree.
Take the Hamilton quaternions \(\mathbb H\) and the polynomial \[ p(x)=x^2+1. \] For every pure imaginary unit quaternion \[ q=bi+cj+dk, \qquad b^2+c^2+d^2=1, \] one has \[ q^2 =b^2i^2+c^2j^2+d^2k^2 +bc(ij+ji)+bd(ik+ki)+cd(jk+kj). \] Using \[ i^2=j^2=k^2=-1,\qquad ij=-ji,\quad ik=-ki,\quad jk=-kj, \] this becomes \[ q^2=-(b^2+c^2+d^2)=-1. \] Thus \[ p(q)=q^2+1=0. \] The unit sphere \[ \{bi+cj+dk:b^2+c^2+d^2=1\} \] therefore consists entirely of roots of the quadratic polynomial \(x^2+1\). Hence a degree-\(2\) polynomial over the division ring \(\mathbb H\) has infinitely many roots.