TITLE: Binary Quadratic Forms in Characteristic 2
QUESTION [19 upvotes]: One of the reasons why the classical theory of binary quadratic forms
is hardly known anymore is that it is roughly equivalent to the theory 
of ideals in quadratic orders. There is a well known correspondence
which sends the $SL_2({\mathbb Z})$-equivalence class of a form 
$$ (A,B,C) = Ax^2 + Bxy + Cy^2 $$
with discriminant 
$$ \Delta = B^2 - 4AC $$
to the equivalence class of the ideal
$$ \Big(A, \frac{B - \sqrt{\Delta}}2\Big). $$
One then checks that this gives a group isomorphism between the 
primitive forms with discriminant $\Delta$ and the class group 
of ideals coprime to the conductor of the order with discriminant $\Delta$.
If we go from forms with integral coefficients to forms with 
coefficients in a polynomial ring $k[T]$ for some field $k$, 
then everything transforms nicely as long as $k$ has 
characteristic $\ne 2$. In characteristic $2$, there is a well 
known theory of function fields (instead of $Y^2 = f(T)$,
consider equations $Y^2 + Y h(T) = f(T)$), but I have no idea
what the right objects on the forms side are. There should be 
decent objects, since, after all, calculations in the Jacobian
of hyperelliptic curves are performed essentially by composition
and reduction of forms. But I don't thinkg that $Ax^2 + Bxy + Cy^2$
will work; for example, these guys all have square discriminant.
Question: which objects (forms or something else?) correspond to 
ideals in quadratic function fields with characteristic $2$? 
Added: Thank you for the (three at this point) excellent answers.
I am now almost convinced that binary forms over $F_2[T]$ actually
do work; the reason why I thought they would not was not so much the
square discriminant but the fact that the action of SL$_2$ fixed the
middle coefficient $B$ of $(A,B,C)$, so I didn't get much of a 
reduction. But now I see that perhaps this is not so bad after all,
and that some pair $(B,?)$ will be the analog of the discriminant in 
the characteristic 2 case.
I knew about Kneser's work on composition, but was convinced (by a
lengthy article of Towber in Adv. Math.) that classical Gauss composition 
of forms over PID's, which I was interested in, is quite far away from
composition of quadratic spaces over general rings. Special thanks to
Torsten for showing how to go from quadratic spaces back to forms!

REPLY [12 votes]: Classes of binary quadratic forms over any commutative ring R (with no conditions on characteristic, PID, etc.) correspond to certain modules over quadratic extensions of R.  This theory is completely general also in the forms it covers, i.e. covers non-primitive and discriminant zero forms.  We can then restrict our attention to the cases of interest. 
For example, when R is an integral domain and the binary quadratic form has non-zero discriminant, the modules associated to the forms are exactly the ideal classes of the quadratic extension. 
The construction is as follows, simplied for your case of interest.  From a form $ax^2+bxy+cy^2$, we form a quadratic $R$-algebra $Q:=R[\tau]/(\tau^2+b\tau+ac)$ and a module $M =Rx\oplus Ry$ with $\tau x=-cy-bx$ and $\tau y=ax$.  Given a quadratic algebra and ideal, we can pick an $R$ basis $x,y$ of the ideal and then shift a generator $\tau$ of the quadratic algebra as necessary so that $\tau y$ is a multiple of $x$.  Then you can simply read off the inverse map.
What I've given above is the case when the binary quadratic form, quadratic algebra, and module are free over $R$.  For general $R$, they will only be locally free, and the above gives a construction locally on free patches.  There are also global, coordinate-free descriptions of the correspondence between binary quadratic forms and modules over quadratic algebras.  For the details I refer you to my paper
"Gauss composition over an arbitrary base" http://www.sciencedirect.com/science/article/pii/S0001870810003257