Finding the integral closure of $k[x]$ in $k(x)(\sqrt{f})$, where $f(x)=x^6+tx^5+t^2x^3+t\in k[x]$.

Question

This is an exercise (1.15) of Dino Lorenzini' s An Invitation to Arithmetic Geometry.

Let $F$ be a field of characteristic $2$. Let $k:=F(t)$, the field of rational functions in the variable $t$. Let $f(x)=x^6+tx^5+t^2x^3+t\in k[x]$.

We want to show that the integral closure of $k[x]$ in $k(x)(\sqrt{f})$ is $k[x][\sqrt{f}]$.

So let $\alpha=m+n\sqrt{f}\in k(x)(\sqrt{f})$, where $m,n\in k(x)$. Since $F(t)$ is a field, $k[x]$ is a PID and therefore $\alpha$ is integral over $k[x]$ if and only if its minimal polynomial in $k(x)$ has coefficients in $k[x]$. Let's then find the minimal polynomial of $\alpha$.

\begin{align} p(y)&=(y-(m+n\sqrt{f}))(y-(m-n\sqrt{f})) \\ &=y^2-2my+(m^2-n^2f) \end{align} and since $m\in k(x)=F(t)k(x)$ and the characteristic of $F$ equals $2$ we get that $2my=0$. Hence

\begin{equation} p(y)=y^2+(m^2-n^2f) \end{equation}

Thus, $\alpha=m+n\sqrt{f}$ is integral over $k[x]$ iff $m^2-n^2f\in k[x]$. I' m pretty sure that I'm on the right course but I don' t know how to continue.

Answer 1

Denote as $B$ the integral closure of $k[x]$ in $k(x)(\sqrt{f}) $. The inclusion $k[x][\sqrt{f}]\subseteq B$ follows immediately from the fact that the coefficients of the minimal polynomial of an arbitrary element of $k[x][\sqrt{f}]$ lay in $k[x]$. For the reverse inclusion you have to use exercise 1.14. First observe that $f$ is squarefree in $\overline{k}[x]$ and then that $f'(x)=5tx^4+3t^2x^2=x^2(5tx^2 +3t^2)$. Now, let $l=a+b\sqrt{f}/c\in B$ where $a,b,c \in k[x]$ and gcd$(a,b,c)=1$. Note that since $l$ is integral we have $\left (a+b\sqrt{f}/c \right )^2\in k[x]$ which implies that $c^2|a^2 +b^2f$. From exercise 1.14 we have that $c^2$ divides $f'$ and since $5tx^2 +3t^2$ is squarefree(why?) we conclude that $c$ divides $x^2$ and hence $c=1$ or $c=x$. If $c=1$ we are done. Suppose now for a contradiction that $c=x$. Also note that when $a\in k[x]$, since $k$ is of characteristic 2, we have that $a^2$ has only powers divisible by $2$ and its constant term squared. Since $c=x$ we have that $x^2h=a^2 +b^2f$ for some $h\in k[x]$ thus, the constant term of $a^2 +b^2f$ i.e $a_0^2 +b_0^2t$ must be zero. Finally, \begin{equation*} a_0^2 +b_0^2t=0\Rightarrow a_0^2=b_0^2t\Rightarrow a_0^2b_0^{-2}=t\Rightarrow(a_0b_0^{-1})^2=t \end{equation*} which implies that $\sqrt{t} \in F(t)$ a contradiction and you get the desired result.