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.



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> 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 $f$?
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Prime ideal ramified in extension if and only if certain polynomial divides another one?
Set $A=k[T,\sqrt f]$. In fact, $A=k[T,U]/(U^2-f)$.  

Then $A/gA\simeq k[T,U]/(g,U^2-f)$. If we set $L=k[T]/(g)$, then $A/gA\simeq L[U]/(U^2-\bar f)$.  

If $g$ ramifies in $A$ then there is $h\in k[T]$ such that $U^2-\bar f=(U-\bar h)^2$, so $g\mid h$ and $g\mid f-h^2$ hence $g\mid f$.


