Let $A$ be a commutative integral domain, with fraction field $K$. Let $T$ be a torsion-free finitely generated $A$ module, so $T \otimes_A K$ is a finite dimensional vector space $V$. Let $T^*$ be the set of $y$ in the dual vector space, $V^*$, such that $\langle x, y \rangle \in A$ for every $x \in T$.

Under what hypotheses on $A$ can I conclude that $T^*$ is a free $A$-module? My current conjecture is that this holds whenever $A$ is a UFD. (Of course, it trivially holds if $A$ is a PID.)

Here are a few ideas of mine. Define the rank of $T$ to be $\dim_K T \otimes_A K$. I can show that, if $A$ is a UFD, then $T^*$ is free for $T$ of rank $1$. For any $T$, we can make a short exact sequence $$0 \to S \to T \to U \to 0$$ where $S$ is rank $1$ and $U$ is torsion free with rank one less than $T$. So we have $$0 \to U^* \to T^* \to S^* \to \mathrm{Ext}^1(U,A) \to \cdots$$. This looks like a good start, but I don't know how to control that Ext group. I suspect that one of you does!

This is motivated by Kevin Buzzard's question about matrix rings.

Let $A$ be a commutative integral domain, with fraction field $K$. Let $T$ be a torsion-free finitely generated $A$ module, so $T \otimes_A K$ is a finite dimensional vector space $V$. Let $T^*$ be the set of $y$ in the dual vector space, $V^*$, such that $\langle x, y \rangle \in A$ for every $x \in T$.

Under what hypotheses on $A$ can I conclude that $T^*$ is a free $A$-module? My current conjecture is that this holds whenever $A$ is a UFD. (Of course, it trivially holds if $A$ is a PID.)

Here are a few ideas of mine. Define the rank of $T$ to be $\dim_K T \otimes_A K$. I can show that, if $A$ is a UFD, then $T^*$ is free for $T$ of rank $1$. For any $T$, we can make a short exact sequence $$0 \to S \to T \to U \to 0$$ where $S$ is rank $1$ and $U$ is torsion free with rank one less than $T$. So we have $$0 \to U^* \to T^* \to S^* \to \mathrm{Ext}^1(U,A) \to \cdots$$. This looks like a good start, but I don't know how to control that Ext group. I suspect that one of you does!

This is motivated by Kevin Buzzard's question about matrix rings.

Let $A$ be a commutative integral domain, with fraction field $K$. Let $T$ be a torsion-free finitely generated $A$ module, so $T \otimes_A K$ is a finite dimensional vector space $V$. Let $T^\*$ be the set of $y$ in the dual vector space, $V^\*$, such that $\langle x, y \rangle \in A$ for every $x \in T$.

Under what hypotheses on $A$ can I conclude that $T^*$ is a free $A$-module? My current conjecture is that this holds whenever $A$ is a UFD. (Of course, it trivially holds if $A$ is a PID.)

Here are a few ideas of mine. Define the rank of $T$ to be $\dim_K T \otimes_A K$. I can show that, if $A$ is a UFD, then $T^\*$ is free for $T$ of rank $1$. For any $T$, we can make a short exact sequence $$0 \to S \to T \to U \to 0$$ where $S$ is rank $1$ and $U$ is torsion free with rank one less than $T$. So we have $$0 \to U^\* \to T^\* \to S^\* \to \mathrm{Ext}^1(U,A) \to \cdots$$. This looks like a good start, but I don't know how to control that Ext group. I suspect that one of you does!

This is motivated by Kevin Buzzard's question about matrix rings.

Let $A$ be a commutative integral domain, with fraction field $K$. Let $T$ be a torsion-free finitely generated $A$ module, so $T \otimes_A K$ is a finite dimensional vector space $V$. Let $T^*$ be the set of $y$ in the dual vector space, $V^*$, such that $\langle x, y \rangle \in A$ for every $x \in T$.

Under what hypotheses on $A$ can I conclude that $T^*$ is a free $A$-module? My current conjecture is that this holds whenever $A$ is a UFD. (Of course, it trivially holds if $A$ is a PID.)

Here are a few ideas of mine. Define the rank of $T$ to be $\dim_K T \otimes_A K$. I can show that, if $A$ is a UFD, then $T^*$ is free for $T$ of rank $1$. For any $T$, we can make a short exact sequence $$0 \to S \to T \to U \to 0$$ where $S$ is rank $1$ and $U$ is torsion free with rank one less than $T$. So we have $$0 \to U^* \to T^* \to S^* \to \mathrm{Ext}^1(U,A) \to \cdots$$. This looks like a good start, but I don't know how to control that Ext group. I suspect that one of you does!

This is motivated by Kevin Buzzard's question about matrix rings.