The formula $R(fg,h)=R(f,h)R(g,h)$ follows easily if you express each resultant in terms of the roots of the polynomials involved. It can also be obtained by relating $R(f,h)$ to the determinant of multiplication by $f$ on the space $K[x]/(h)$. But is there a way to prove the above formula by extending the three Sylvester matrices involved by cleverly chosen blocks so that the determinants don't change, the three matrices become square matrices of equal size, and one matrix is the product of the other two?
Asked
Active
Viewed 227 times
4
-
I remember seeing a proof that writes $R(f,g)$ and $R(f,h)$ as determinants of block matrices and multiplies them to obtain the Sylvester matrix of $(f,gh)$. Unfortunately I cannto find this proof right now. In the worst case, it was in Apéry's Elimination book, which doesn't seem to have found its way on the internet... – darij grinberg Jul 01 '22 at 12:59