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It is well-known( see Goldschmidt book: Algebraic Functions and Projective Curves) that for $q$ a power of $2$ a quadratic separable extension of $\mathbb F_q(T)$ can be written as: $$K:=\mathbb F_q(T)[y]\text{ with }y^2+y=f(T)\text{ and } f(T)\in\mathbb F_q(T).$$ Moreover $f(T)=\frac{R(T)}{P_1(T)^{2d_1-1}\cdots P_s(T)^{2d_s-1}}$ with $P_i$ irreducibles of $\mathbb F_q[T]$, $d_i$ positive integers, and $R(T)\in\mathbb F_q[T]$ such for all $i$ $P_i\nmid R$. I am looking for describing the integral closure $\mathcal O_K$ of $\mathbb F_q[T]$ in $K$.

Let $x=a+by\in\mathcal O_K$ ($a,b\in\mathbb F_q(T)$). One has $\mathrm{Tr}(x)=b\in\mathbb F_q[T]$ and $N(x)=(a+by)(a+b(y+1))\in\mathbb F_q[T]$. Hence, $a^2+b^2f(T)+ab\in\mathbb F_q[T]$

I am stuck here. I did not manage to obtain conditions on $a$. Thanks in advance for any hint or any solution that you could give me.

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  • $\begingroup$ Work locally, considering pole-order possibilities for $a$ initially at places of $\mathbf{F}_q[T]$ away from the $P_j$'s and then at some $P_j$ (the latter handled using oddness of the pole-order of $f$ at $P_j$ and the known integrality of $b$ there). In this way you'll get $a\in \mathbf{F}_q[T]$; this really works over any field of characteristic 2, nothing special about finite fields. $\endgroup$
    – user74230
    Commented Jan 31, 2015 at 20:52
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    $\begingroup$ I should have said that this works over any perfect coefficient field of characteristic 2 (not necessarily finite), as perfectness is used to adjust $f$ to be in the form given at the outset. $\endgroup$
    – user74230
    Commented Jan 31, 2015 at 21:11
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    $\begingroup$ You can handle integrality of $a$ at places of ${\mathbf F}_q[T]$ other than irreducible factors in the denominator of $f$ by looking at the trace of both $x$ and $xy$ (note $y$ is integral at such places). The trace of $x$ is $b$ and the trace of $xy$ is $a+b$. Thus $xy - x$ has trace $a$, so $a$ is integral at irreducibles in ${\mathbf F}_q[T]$ not appearing in the denominator of $f$. $\endgroup$
    – KConrad
    Commented Feb 1, 2015 at 3:53
  • $\begingroup$ Thanks everybody for your solution. That helped me.For an Arti-Schreier extension in odd characteristic, does there exist a formula giving the ring of integers of the extension? $\endgroup$
    – joaopa
    Commented Feb 1, 2015 at 19:36
  • $\begingroup$ With the hints, I found that $\mathcal O_K$ is $\mathbb F_q[T]\oplus y\prod_{i=1}^sP_i^{d_i}\mathbb F_q[T]=\mathbb F_q[T][y\prod_{i=1}^sP_i^{d_i}]$. Is it correct? I am suspicious, because the different would be $\mathcal O_K$ since $(y^2+y)'=1$. $\endgroup$
    – joaopa
    Commented Feb 2, 2015 at 15:34

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