Consider the following statement on resultants, given for example on the Wikipedia page for resultants
Let $A, B$ be polynomials in $R[X]$ where $R$ is a commutative ring. Let $A, B$ have degrees $d, e$ respectively. There exists a polynomial $P$ of degree less than $e$ and a polynomial $Q$ of degree less than $d$ such that $ \operatorname {res} (A,B)=AP+BQ.$
The definition of the resultant is given as follows:
Let $R_n$ be the $n$-dimensional module of polynomials in $R[X]$ of degree strictly less than $n$. Then the resultant of $A, B \in R[X]$ with degrees $d, e$ is the determinant of the matrix representing the following linear map $\phi$:
$$\phi:R_e \times R_d \rightarrow R_{d+e}
$$
$$\phi(P, Q) = AP + BQ
$$
My confusion
It seems to me that if $\operatorname{Res}(A, B) = 0$, the proposition holds trivially, while if $ \operatorname{Res}(A, B) \neq 0$, $\phi$ is an isomorphism and so $P, Q$ can be chose such that $AP + BQ$ is any element of $R_{d+e}$.
Is this correct ? I ask because because the above proposition feels like an odd restriction of a more general result. Is this simply a convenient form for use in other proofs ?
Your observations apply in the case that $R$ is a field. Unfortunately, they don't work for general rings. Here is a counterexample: take $R=\Bbb Z$, $A=x^2+2$, $B=x$. Then the resultant of these two polynomials is $2$, and the map $\phi: R_1\times R_2\to R_3$ has image all polynomials with even constant term and therefore is not surjective and cannot be an isomorphism.
This result is stated this way because it's legitimately useful - it leads to fun elimination-theoretic properties, and those are things we want to have over a general base ring, not just a field.