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.


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.

  • $\begingroup$ Exercise 1.14 states that$\colon$ Let $k$ be a field of characteristic $2$. Let $f\in k[x]$ be squarefree in $\overline{k}[x]$. Then if $(a+b\sqrt{f})/c$ is integral over $k[x]$, with $a,b,c\in k[x]$ and $\gcd(a,b,c)=1$, we have that $c^2$ divides $f'(x)$. $\endgroup$ – Cornelius Dec 7 '18 at 18:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.