Localization of the polynomial ring at a prime ideal modulo maximal ideal is isomorphic to polynomial ring modulo prime ideal.

Question

Let $p \in K[T]$ irreducible, s.t. $\text{LC}(p) = 1$. Then $$ K[T]/(p) \cong K[T]_{(p)}/pK[T]_{(p)}.$$

What I have is: \begin{align*} &K[T] \hookrightarrow K[T]_{(p)} \text{ and } K[T]_{(p)} \twoheadrightarrow K[T]_{(p)}/pK[T]_{(p)}\\ & \implies K[T] \overset{f}\longrightarrow K[T]_{(p)}/pK[T]_{(p)} \end{align*} The kernel of $f$ is $(p)$ so $$ K[T]/(p) \cong \text{im} f = \left\{g + pK[T]_{(p)} \mid g \in K[T] \right\} \overset{?} = K[T]_{(p)}/pK[T]_{(p)}.$$

Answer 1

Let $\overline{\frac{r}{s}} \in K[T]_{(p)}/(p)K[T]_{(p)}$. Then \begin{align*} p \not | s &\implies \exists a,b \in K[T]: ap + bs = 1\\ & \implies bs = 1 - ap\\ & \implies f(bs) = \overline{bs} = 1 + (p)K[T]_{(p)}\\ & \implies f(rb) =f(r) f(b) \equiv \overline{\frac{r}{s}} \end{align*} Hence $$ \text{im } f = K[T]_{(p)}/(p)K[T]_{(p)}.$$