Let $k$ be a field of characteristic $ \neq 2$, and let $f \in k[T]$ be a polynomial of degree $\ge 1$ which is square free. Let $K$ be the quadratic extension $k(T)(\sqrt{f})$ of $k(T)$. I know that the integral closure of $k[T]$ in $K$ is $k[T, \sqrt{f}]$. My question is as follows.

Is a prime ideal $(g)$ of $k[T]$, where $g$ is an irreducible polynomial, ramifies in the extension $K/k(T)$ if and only if $g$ divides $f$?

Let $k$ be a field of characteristic $ \neq 2$, and let $f \in k[T]$ be a polynomial of degree $\ge 1$ which is square free. Let $K$ be the quadratic extension $k(T)(\sqrt{f})$ of $k(T)$. I know that the integral closure of $k[T]$ in $K$ is $k[T, \sqrt{f}]$. My question is as follows.

Does a prime ideal $(g)$ of $k[T]$, where $g$ is an irreducible polynomial, ramify in the extension $K/k(T)$ if and only if $g$ divides $f$?