1
$\begingroup$

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 $f$?

Improve this question
$\endgroup$
1
  • 2
    $\begingroup$ Yes; consider that the discriminant of $k(T, \sqrt{f}) / k(T)$ is $-4f$. $\endgroup$
    – Jeff Yelton
    Commented Mar 29, 2016 at 10:01

1 Answer 1

Reset to default
2
$\begingroup$

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 $g\mid f$.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .