Show that there is a canonical category of pairs $(V,L)$ consisting of a finite-dimensional $K$-vector space $V$ and a $K$-linear endomorphism $L : V \to V$, that is equivalent to the category of torsion finitely generated $K[T]$-modules.
My attempt was to look at the generators. Let $M$ be a finitely generated $K[T]$ module, with basis as $m_1,\dots,m_n$. Then there exists $g_1(T),\dots,g_n(T) \in K[T]$ such that $g_i(T)m_i = 0$ and such that $g_i(T)$ generates the ideal of all polynomials in $K[T]$ that annihilates $m_i$. This is justified, since $K$ being a field implies that $K[T]$ is a principal ideal domain and the set of all polynomials in $K[T]$ that annihilates $m_i$ is easily checked to be an ideal. Let $g$ be the generator of the ideal $(g_1) \cap \dots \cap (g_n)$. Then I claim that the category of objects as pairs $(K^n, g(T))$ and the morphisms as linear transformations between $K^n$ and $K^m$ for $n, m \geq 0$ is the category equivalent to the category of torsion finitely generated $K[T]$ modules. The functor $F$ that sends $(K^n, g(T))$ to $K^n$ whose element $e_i$ is annihilated by $g(T)$ is fully faithful and essentially surjective. Indeed, it's easy to see that it's full and essentially surjective, and By construction of $g$ we may see that any two modules $M$ and $N$ whose corresponding $g_M$ and $g_N$ have a $K[T]$-linear function that is bijective.
I was asking someone about this question and he said that my solution does not work because "it's not functorial." I wasn't sure what he meant by that. I know that my claim that the functor is fully faithful and essentially surjective is not correct; however, I am failing to see how one can solve this in another way.