Post #1815096

@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.