Post #1815095

2026-04-07 21:48 UTC

@hallasurvivor I vaguely remembered something about all unit vectors being roots of -1 in the quaternions, and I only just remembered a fun way to see it, which is that _every_ quaterion q is a root of a quadratic polynomial over 𝐑, namely χ=X²-tr(q)X+det(q), where tr(q)=2Re(q)=2a is twice the real part of q=a+ib+jc+kd and det(q)=a²+b²+c²+d² is the square of the absolute value of q (χ is the characteristic polynomial à la https://arxiv.org/abs/math/0203276), so taking a=0 gives you an S² worth of roots. 𝐑 is the center of 𝐇, so the map 𝐑[X]→𝐇 that evaluates in q is a ring homomorphism, and its kernel (the ideal of polynomials with q as a root) is generated by χ for q∉𝐑, which is a (roundabout) way of seeing that that only the q∉𝐑 with Re(q)=0 and |q|=1 make the cut. Incidentally χ factors over the (commutative) subring 𝐑[q]≅𝐂 generated by 𝐑 and q as χ=(X-q)(X-q*), since q*=2Re(q)-q is in there and we have tr(q)=q+q* and det(q)=qq*=q*q.

Replies (1)

@at@mathstodon.xyz 2026-04-07 22:01

@hallasurvivor The notes I learnt algebra from (https://websites.math.leidenuniv.nl/algebra/algebra2.pdf#page=22) actually prove division with remainder in R[X] (by polynomials of lower degree with invertible leading coefficient) for arbitrary coefficient rings R, but the usual factorization in terms of roots only for integral domains R, and it’s sort of swept under the rug what stops you for more general R (it’s the failure of the evaluation map R[X]→R to be a homomorphism). You can still factor f=q⋅(X-a) if a∈R is a root of f∈R[X], but you can’t induct further because the other roots of f needn’t be roots of q, and the previous toot gives you an example: we have X²+1=(X-i)(X+i), but j also satisfies j²=-1, yet we have (j-i)²=(j-i)tr(j-i)-det(j-i)=-2.

Open ##1815096