Let $\mathbb{Q}[t,t^{-1}]$ be the ring of Laurent polynomials. I view this as the localization of $\mathbb{Q}[t]$ by $S= \{ t^i \mid i \in \mathbb{N} \cup \{0\} \}$. I want to understand the decomposition of finitely generated modules over a PID in this case, since $\mathbb{Q}[t,t^{-1}]$ is a PID. More concretely, I would like to understand that if $(q_i)$ is a non-zero primary ideal, then $Q[t,t^{-1}]/(q_i)$ is a finite dimensional vector space.
For that, I should understand the quotient $\mathbb{Q}[t,t^{-1}]/ (q_i)$ where $(q_i)$ is a primary ideal. I have started with an example and I am already confused.
Assume that I take $(q_i)$ to be $< (t-1)^2>_S$, that is, the localization of the ideal $<(t-1)^2>$ in $\mathbb{Q}[t]$. Then, $$\mathbb{Q}[t]_S / < (t-1)^2>_S= (\mathbb{Q}[t]/<(t-1)^2>)_{\bar{S}}$$ where $\bar{S}$ is the image of $S$ in $\mathbb{Q}[t]/<(t-1)^2>$.
I would like to understand why $(\mathbb{Q}[t]/<(t-1)^2>)_{\bar{S}}$ is a finite dimensional $\mathbb{Q}$-vector space.
I understand that $\mathbb{Q}[t]/<(t-1)^2>$ is a $2$-dimensional $\mathbb{Q}$-vector space because $t^2= 2t-1$, so $\{\bar{1}, \bar{t} \}$ is a $\mathbb{Q}$-basis for $\mathbb{Q}[t]/<(t-1)^2>$.
Nevertheless, I don't get why $(\mathbb{Q}[t]/<(t-1)^2>)_{\bar{S}}$ is a finite dimensional $\mathbb{Q}$-vector space. I don't see which would be a $\mathbb{Q}$-basis, since $$\bar{S}= \{ \bar{t}^i \mid i\in \mathbb{N} \cup \{0\} \},$$ so for example $\bar{t}^4= (2t-1)(2t-1)= 4t^2-4t+1= 4(2t-1)-4t+1= 4\bar{t}-3$ should be in $\bar{S}$. Hence, in $(\mathbb{Q}[t]/<(t-1)^2>)_{\bar{S}}$ I would have for instance the element $$ \frac{\bar{t}}{4\bar{t}-3},$$ right?
Thanks in advance!
EDIT: I think that I get it, I just want someone to tell me that I'm correct/incorrect.
From $t^2= 2t-1$ we get that $t^{-1}= 2-t$, $t^{-2}= 2t^{-1}-1= 2(2-t)-1= 3-2t$ and so on. So, for instance, $$\frac{1+\bar{t}}{\bar{t}^2}= (1+\bar{t})(3-2t)$$
Firstly, as a $\mathbb{Q}$-vector space, $\mathbb{Q}[t]/(t^2-2t+1)=\mathbb{Q}\oplus \mathbb{Q}t$. If you are inverting all powers of $t$, we claim you get as a $\mathbb{Q}$-vector space, $\mathbb{Q}\oplus \mathbb{Q}\bar{t}$. Higher powers of $\bar{t}$ can be reduced to lower ones by using $\bar{t}^2=2\bar{t}-1$.
Conversely, a negative power of $\bar{t}$ can be brought to positive powers by using $\frac{1}{\bar{t}^k}=\frac{2\bar{t}-\bar{t}^2}{\bar{t}^k}$.