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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 ?

Mr Lolo
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    $R_n$ is only a vector space of $n$ dimensions if $R$ is a field. – Thomas Andrews Nov 12 '22 at 20:22
  • It is not true that $\phi$ is an isomorphism in general, even when $R$ is a field. If $A,B$ have a common factor, then every element in the range has that factor, too. – Thomas Andrews Nov 12 '22 at 20:27
  • Thanks, I have changed vector space to module. Isn't it true that $\phi$ is an isomorphism when $\operatorname{Res}(A, B) \neq 0$ ? – Mr Lolo Nov 13 '22 at 09:40
  • No. The matrix is one-to-one if the determinant is non-zero, but it might not be onto. – Thomas Andrews Nov 13 '22 at 13:44
  • Thank you for your answer. Can't we use the fact the domain $R_d \times R_e$ and the codomain $R_{d+e}$ have the same dimension ? – Mr Lolo Nov 14 '22 at 10:21
  • See also https://math.stackexchange.com/questions/4483154/resultants-if-a-b-have-degrees-m-n-why-are-there-polynomials-u-v-of-deg/4484219#4484219 – darij grinberg Nov 21 '22 at 00:14

1 Answers1

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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.

KReiser
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