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

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

$$$$p(y)=y^2+(m^2-n^2f)$$$$

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.

## 1 Answer

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.

• 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 $f'(x)$. – Cornelius Dec 7 '18 at 18:08