Given a two-dimensional $\mathbb{Q}$-module, $V$ with basis $\{e_1,e_2\}$, define two $\mathbb{Q}[t]$-module structures as follows: $$T_1:= \begin{cases} te_1 \mapsto e_2\\ te_2 \mapsto e_1 \\ \end{cases} $$ $$ T_2:=\begin{cases} te_1 \mapsto e_2\\ te_2 \mapsto e_1+e_2 \\ \end{cases} $$ Attempt: I claim $V_{T_2}$ and $V_{T_2}$ are not isomorphic. Isomorphism classes of $k[t]$ modules are in a one-to-one correspondence with similarity classes of the underlying matrix representations of the endomorphism that gives a $k$-vector space its $k[t]$-module structure. Consider the two matrices representing the two $k[t]$-modules, $$M_{T_1}=A=\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$$ $$M_{T_2}=B=\begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}$$ The corresponding characteristic polynomials are $P_A(t)=t^2-1$ and $P_B(t)=t^2-t-1$, note that this implies that $A$ has two eigenvalues while $B$ has no eigenvalues. Similar matrices have the same eigenvalues $\implies A$ and $B$ are not similar, $\implies V_{T_1} \not\cong V_{T_2}$ as $\mathbb{Q}[t]$-modules.
I wanted to check if this reasoning works, because on my final last semester I was told to be careful about using the characteristic polynomial. I made a statement along the lines of, "They have different characteristic polynomials therefore they're not similar" which I was shown wasn't necessarily true. But does this eigenvalue argument work? My linear algebra really needs work before my algebra qualifier coming up...
Two matrices with different characteristic polynomials cannot be similar. Indeed, $\det(A - xI) = \det(QAQ^{-1} - x I)$ for any $Q$, since $QAQ^{-1} - x I = Q(A - x I)Q^{-1}$.