The resultant of two homogeneous polynomials is homogeneous I haven't been able to find a proof for this theorem in the literature:

Let $f,g\in k[x_0,\dots,x_k]$ be homogeneous polynomials, of degree $m$ and $n$ respectively. Then $R_{x_0}(f,g)$ is homogeneous of degree $mn$, where $R_{x_0}(f,g)$ stands for the resultant of $f$ and $g$ viewed as elements of $k[x_1,\dots,x_k][x_0]$.

 A: This theorem is not true. Take
$$f=x_0x_1x_2 - x_1x_2x_3 ~\text{ and }~ g=x_0x_3x_4-x_0x_4x_5.$$ These polynomials are both of homogeneous degree 3, so according to the theorem the resultant $Res_{x_0}(f,g)$ should be of degree $9$, however
$$Res_{x_0}(f,g)=-x_1x_2x_3^2x_4+x_1x_2x_3x_4x_5$$
is of homogeneous degree $5$.
The mistake in the proof in the comments under the question is in determining the homogeneous degree of the $(i,j)$-the entry of the Sylvester matrix, specifically in this part

the $(i,j)$-th entry of the Sylvester matrix is homogeneous of degree $m−j+i$ if $i\leq n$, and $−j+i$ otherwise

as it doesn't take into account that certain entries of the matrix are $0$, e.g. if $m=2$ and $n=3$ then at position $(0,3)$ the entry is $0$, but $m+i-j=2+0-3=-1$.
Edit: I believe that the correct homogeneous degree is
$$m\cdot\deg_{x_0}g+(n-\deg_{x_0}g)\cdot\deg_{x_0}f$$
because if $f=a_m+a_{m-1}x+\cdots+a_{m-k}x^k$ and $g=b_n+b_{n-1}x+\cdots+b_{n-l}x^l$ (the coefficients are indexed like so because then the homogeneous degree of $a_i$ and $b_i$ is exactly $i$) then the term $a_m^lb_{n-l}^k$ generically appears in the resultant, up to sign. It may not appear for particular instances of $f$ and $g$, e.g. as in my example above if we switch around $f$ and $g$, but all other terms that do appear have the same degree anyway.
