 Degree drop in a specialized resultant

The actual-degree condition in the resultant theorem matters when coefficients depend on a parameter. TakeThe nominal degree-one Sylvester matrix is $\left(\begin{array}{cc}t& 1\\ t& 2\end{array}\right)$, whose determinant is $t$. Its specialization at $t=0$ therefore vanishes. But the specialized polynomials are the nonzero constants $1$ and $2$, which have no common root at all.

For $t\ne 0$, the roots of a polynomial are $-1/t$ and $-2/t$; neither has a finite limit as $t\to 0$. Both leading coefficients vanish at the exceptional parameter. The fixed-size determinant has retained the original degree format even though the specialized polynomial degrees have dropped. Recompute using the actual specialized polynomials and check the original equations before accepting a zero of a parameter-dependent resultant as a solution.