Localization of $k[t]$ at a prime ideal is not a finitely generated $k[t]$-algebra?

Question

Let $k$ be a field (infinite and algebraicaly closed), consider the prime ideal $(t)$ in $k[t]$.

Consider the localization of $k[t]$ at $(t)$: $k[t]_{(t)}$.

Now $k[t]_{(t)}$ is a $k[t]$-algebra. I want to show that $k[t]_{(t)}$ is not a finitely generated $k[t]$-algebra. How do I show that? Any help will appreciated.

Answer 1

For any field as a $k$-algebra $$k(t) = k[t,\{\frac{1}{f(t)}, f \in k[t] \text{ monic irreducible}\}]$$

As a $k$-vector space the basis are the $t^k, k \ge 0$ and the $\frac{t^n}{f(t)^m},m \ge 1, n < \deg(f), f \in k[t]$ monic irreducible.