Determinant of $xP+yQ$ is resultant of $P$ and $Q$

Question

In my textbook, the resultant $R$ of two polynomials $P$ and $Q$ in $K(X)[Y]$, where $K$ is a field, is defined as the monic generator of the ideal $(P,Q) \cap K[X]$. Is it still true that $ R = \det φ$, where $$ φ \colon (r,s) \mapsto rP+sQ$$ is a linear map from the space of polynomials in $K(X)[Y]$ of degree no more than $\deg Q$ times the space of polynomials in $K(X)[Y]$ of degree no more than $\deg P$ to the space of polynomials in $K(X)[Y]$ of degree no more than $\deg P + \deg Q$?

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I can see that if one between $R$ and $\det φ$ is zero, the other is zero too. I wanted to show that they have the same degree with respect to each of the coefficients of $P$ and $Q$, but I don't know how to do it with my definition.

Answer 1

I'll start with an example using the GAP computer algebra system:

gap> P := (x+1)*y^2+(2*x-1)*y+(x-3);;
gap> Q := (2*x+1)*y+(x-4);;
gap> M := sylvester(P,Q);
[ [ x+1, 2*x-1, x-3 ], [ 2*x+1, x-4, 0 ], [ 0, 2*x+1, x-4 ] ]
gap> p := determinant(M);
x^3+x^2-2*x+9

On the other hand :

 gap> GroebnerBasis([P,Q], MonomialLexOrdering([y,x]));
 [ x*y^2+2*x*y+y^2+x-y-3, 2*x*y+x+y-4, -y^2-1/2*x-1/2*y, -1/2*x^2-1/4*x-
 9/4*y,1/4*x^3+1/4*x^2-1/2*x+9/4 ]

This seems to point out that Buchberger's algorithm in this special case with a base of only two polynomials modifies the Sylvester matrix in such a way that at each step a one appears on the diagonal preceded by zeroes, except in the last case where the univariate determinant pops up. Since at each step only sums of terms of the form $Py^i$ and $Qy^j$ appear the rows represent polynomials of the ideal, in casu the last row.