# Isomorphism between a vector space and its dual as $k[t]-$modules, for a given operator $T$

Let $k$ be a field and $V$ a finite-dimensional vector field over it. Let $T$ be a linear endomorphism of $V$, and denote by $[V,T]$ the $k[t]$-module where $t$ acts through $T$.

Let $V^*$ be the dual space of $V$, and $T^*:V^*\to V^*$ the adjoint operator $T(\psi)=\psi(T\cdot)$. Show that $[V,T]$ and $[V^*, T^*]$ are isomorphic as $k[t]-$modules.

My solution:

We know that $V$ and $V^*$ are isomorphic as vector spaces ($k-$modules). Denote this isomorphism as $v\mapsto v^*$. Therefore, the $k-$modules $k[t]\otimes_{k[t]}V$ and $k[t]\otimes_{k[t]}V^*$ are isomorphic by putting $\psi:p(t)\otimes v\mapsto p(t)\otimes v^*$, and this is actually a $k[t]-$modules isomorphism too, since additivity, injectivity and surjectivity are clear, and $$q(t)(p(t)\otimes v)=q(t)p(t)\otimes v\mapsto q(t)p(t)\otimes v^*=q(t)(p(t)\otimes v^*).$$

Furthermore, the maps $$\phi:k[t]\otimes V\to V$$ $$p(t)\otimes v\mapsto p(t) v$$ and $$k[t]\otimes V^*\to V$$ $$\chi:p(t)\otimes v^*\mapsto p(t) v^*$$ are $k[t]-$ isomorphisms because of the definition of the action of $t$. So composing $\chi\circ \psi\circ\phi^{-1}$ we have an isomorphism from $V$ to $V^*$ as $k[t]-$modules.

Is this correct? Is there a simpler way?

This is not correct: how do you know your map $\psi$ is well-defined? For instance, in $k[t]\otimes_{k[t]}V$, $t\otimes v=1\otimes tv$, so in order for $\psi$ to be well-defined you would need to have $t\otimes v^*=1\otimes(tv)^*$, which may not be true.
In fact, you can't take just any isomorphism between $V$ and $V^*$ as $k$-modules and get from it an isomorphism of $k[t]$-modules. Instead, you will need to make a very special (and non-canonical) choice of such an isomorphism. I would suggest to first consider the case that $V$ is a cyclic $k[t]$-module, and then reduce to that case using the classification of finitely generated $k[t]$-modules.