Post #1815094

@hallasurvivor The real-analytic proof given on wikipedia shows that a higher degree polynomial with real coefficients 'can always be divided by some quadratic polynomial with real coefficients'. Together with the complex conjugate theorem, it proves the result over the complex numbers. However, for quaternions it's simply not true by identity: There are infinitely many square roots of -1, given by combination of the i/j/k basis vectors on a unit sphere instead of just 2. Rather than ±i, it's the set of i,j,k such that ||i+j²+k²|| = 1 I see that this doesn't exactly respond to your question about "where is the use of commutativity". To pursue that thread I guess I'd look down at the quadratic formula. When it gets to the step where you have to take the square root of a negative number, you just have to use the rule for quaternions, not complex numbers: you have to pick pars of numbers that are like specific kinds of conjugates, of which ±i, ±j and ±k would be the prototypical examples. You can say that 1j and 1k are both roots of x²+1 in some sense but unlike when you say -i and +i are roots of the same equation it is NOT the case that jk+1=0 (or kj+1=0) So, I think in that sense it's also true that the conjugate rule holds, you still have to use conjugate pairs of roots that multiply together to satisfy the original quadratic equation. When it comes to polynomials with quaternion coefficients, sorry, my head explodes.