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Is $operatorname{Hom}_R(S,R)$ as an $S$-module torsionless?
$begingroup$
Let $R = k[x]$ be a polynomial ring in one indeterminant and $S supset R$ a finite free $R$-module of rank $n$ over $R$ and of Krull-dimension one.
Consider $M = operatorname{Hom}_R(S,R)$ as an $S$-module via
$(scdotphi)(x) = phi(sx)$ for all $s,x in S$ and $phi in M$.
My question is: $quad$ Is $M$ as an $S$-module torsionless?
That is, is the canonical map $M to M^{**}$ from $M$ to its double dual (dual as an $S$-module!) injective? Or equivalently, does for every non-zero $phi in M$ exist some $f in M^*$ such that $f(phi) neq 0$?
I know that the dual of a module is torsionless, that is $M$ considered as an $R$-module is torsionless.
The first ansatz that comes to mind might be the following: For given $R$-linear $phi: S to R$ we need to find an assignment that produces an element in $S$ and is $S$-linear. If $phi neq 0$, there is some $x in S$ with $phi(x) neq 0$. Hence we could try the assignment $f$ that maps $psi mapsto psi(x)$ for every $psi in M$. Then we would have $f(phi) neq 0$. But $f$ is not $S$-linear since $phi$ need not be $S$-linear: $$f(s cdot phi) = (s cdot phi)(x) = phi(sx) neq s phi(x) = s f(phi).$$
commutative-algebra modules
asked Jan 24 at 9:47
windsheafwindsheaf
607312
$endgroup$
add a comment |
$begingroup$
Let $R = k[x]$ be a polynomial ring in one indeterminant and $S supset R$ a finite free $R$-module of rank $n$ over $R$ and of Krull-dimension one.
Consider $M = operatorname{Hom}_R(S,R)$ as an $S$-module via
$(scdotphi)(x) = phi(sx)$ for all $s,x in S$ and $phi in M$.
My question is: $quad$ Is $M$ as an $S$-module torsionless?
That is, is the canonical map $M to M^{**}$ from $M$ to its double dual (dual as an $S$-module!) injective? Or equivalently, does for every non-zero $phi in M$ exist some $f in M^*$ such that $f(phi) neq 0$?
I know that the dual of a module is torsionless, that is $M$ considered as an $R$-module is torsionless.
The first ansatz that comes to mind might be the following: For given $R$-linear $phi: S to R$ we need to find an assignment that produces an element in $S$ and is $S$-linear. If $phi neq 0$, there is some $x in S$ with $phi(x) neq 0$. Hence we could try the assignment $f$ that maps $psi mapsto psi(x)$ for every $psi in M$. Then we would have $f(phi) neq 0$. But $f$ is not $S$-linear since $phi$ need not be $S$-linear: $$f(s cdot phi) = (s cdot phi)(x) = phi(sx) neq s phi(x) = s f(phi).$$
commutative-algebra modules
asked Jan 24 at 9:47
windsheafwindsheaf
607312
$endgroup$
add a comment |
$begingroup$
Let $R = k[x]$ be a polynomial ring in one indeterminant and $S supset R$ a finite free $R$-module of rank $n$ over $R$ and of Krull-dimension one.
Consider $M = operatorname{Hom}_R(S,R)$ as an $S$-module via
$(scdotphi)(x) = phi(sx)$ for all $s,x in S$ and $phi in M$.
My question is: $quad$ Is $M$ as an $S$-module torsionless?
That is, is the canonical map $M to M^{**}$ from $M$ to its double dual (dual as an $S$-module!) injective? Or equivalently, does for every non-zero $phi in M$ exist some $f in M^*$ such that $f(phi) neq 0$?
I know that the dual of a module is torsionless, that is $M$ considered as an $R$-module is torsionless.
The first ansatz that comes to mind might be the following: For given $R$-linear $phi: S to R$ we need to find an assignment that produces an element in $S$ and is $S$-linear. If $phi neq 0$, there is some $x in S$ with $phi(x) neq 0$. Hence we could try the assignment $f$ that maps $psi mapsto psi(x)$ for every $psi in M$. Then we would have $f(phi) neq 0$. But $f$ is not $S$-linear since $phi$ need not be $S$-linear: $$f(s cdot phi) = (s cdot phi)(x) = phi(sx) neq s phi(x) = s f(phi).$$
commutative-algebra modules
asked Jan 24 at 9:47
windsheafwindsheaf
607312
$endgroup$
Let $R = k[x]$ be a polynomial ring in one indeterminant and $S supset R$ a finite free $R$-module of rank $n$ over $R$ and of Krull-dimension one.
Consider $M = operatorname{Hom}_R(S,R)$ as an $S$-module via
$(scdotphi)(x) = phi(sx)$ for all $s,x in S$ and $phi in M$.
My question is: $quad$ Is $M$ as an $S$-module torsionless?
That is, is the canonical map $M to M^{**}$ from $M$ to its double dual (dual as an $S$-module!) injective? Or equivalently, does for every non-zero $phi in M$ exist some $f in M^*$ such that $f(phi) neq 0$?
I know that the dual of a module is torsionless, that is $M$ considered as an $R$-module is torsionless.
The first ansatz that comes to mind might be the following: For given $R$-linear $phi: S to R$ we need to find an assignment that produces an element in $S$ and is $S$-linear. If $phi neq 0$, there is some $x in S$ with $phi(x) neq 0$. Hence we could try the assignment $f$ that maps $psi mapsto psi(x)$ for every $psi in M$. Then we would have $f(phi) neq 0$. But $f$ is not $S$-linear since $phi$ need not be $S$-linear: $$f(s cdot phi) = (s cdot phi)(x) = phi(sx) neq s phi(x) = s f(phi).$$
commutative-algebra modules
commutative-algebra modules
asked Jan 24 at 9:47
windsheafwindsheaf
607312
asked Jan 24 at 9:47
windsheafwindsheaf
607312
edited Jan 24 at 9:53
windsheaf
asked Jan 24 at 9:47
windsheafwindsheaf
607312
asked Jan 24 at 9:47
windsheafwindsheaf
607312
asked Jan 24 at 9:47
windsheafwindsheaf
607312
607312
add a comment |
add a comment |
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