Binary Quadratic Forms in Characteristic 2

Question

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!

Answer 1

Starting, generally, with a commutative ring $R$ and a rank $2$ projective module $P$ given with a quadratic form $\varphi$ we can form its Clifford algebra $C(P)$. Its even part $S:=C^+(P)$ is then a commutative $R$ algebra of rank $2$. If (which we may assume locally) $P=Re_1+Re_2$ we have that $S$ has basis $1,e_1e_2=:h$ and $h^2=e_1e_2e_1e_2=e_1(\langle e_1,e_2\rangle-e_1e_2)e_2=\langle e_1,e_2\rangle h-e_1^2e_2^2=\langle e_1,e_2\rangle h-\varphi(e_1)\varphi(e_2)$, where $\langle-,-\rangle$ is the bilinear form associated to $\varphi$, giving an explicit quadratic algebra over $R$. Furthermore, $C^-(P)=P$ and it thus becomes an $C^+(P)$-module, explicitly $h\cdot e_1=e_1e_2e_1=\langle e_1,e_2\rangle e_1-e_1^2e_2=\langle e_1,e_2\rangle e_1-\varphi(e_1)e_2$. Furthermore, putting $L:=S/R$ we get an isomorphism $\gamma\colon \Lambda^2_RP \to L$ given by $u\land v \mapsto \overline{uv}$ (note that $u^2=\varphi(u)$ which maps to zero in $L$). Note for future use that, putting $[u,v]:=\gamma(u\land v)$, we also have $[[u,v]u,u]=\varphi(u)[u,v]$, where the left hand side is well-defined as $[([u,v]+r)u,u]=[[u,v]u,u]+r[u,u]=[[u,v]u,u]$. We have therefore constructed from $(P,\varphi)$ a triple $(S,P,\gamma)$, where $S$ is a quadratic (i.e., projective of rank $2$) $R$-algebra, $P$ an $S$-module projective of rank $2$ as $R$-module and an isomorphism $\gamma\colon \Lambda^2_RP \to S/R$ of $R$-modules. Conversely, given such a triple $(S,P,\gamma)$ there is a unique quadratic form $\varphi$ on $P$ such that $[[u,v]u,u]=\varphi(u)[u,v]$ (where again the left hand side is well-defined). It is easily verified that these constructions are inverse to each other.

In case $R=\mathbb Z$ I hope this gives the usual association between forms and modules over orders in quadratic number fields (but I admit shamelessly that I haven't checked it). When $R=k[t]$ I again hope this gives what we want.

In terms of algebraic group schemes and torsors we have the following situation. If the quadratic form is perfect, we have that $S$ is an \'etale covering and hence corresponds to an element of $H^1(\mathrm{Spec}R,\mathbb Z/2)$. It is the image under $O_2 \to \mathbb Z/2$ of the torsor in $H^1(\mathrm{Spec}R,O_2)$ corresponding to $\varphi$. As $\mathbb Z/2$ also acts as an automorphism group of $SO_2$ (by conjugation in $O_2$) we can use the torsor in $H^1(\mathrm{Spec}K,\mathbb Z/2)$ to twist $SO_2$ to get $SO(\varphi)$, the connected component of $O(\varphi)$. Torsors over this group corresponds to isomorphism classes of pairs consisting of a rank $2$ quadratic $R$-forms $\phi$ and an $R$-isomorphism $C^+(\phi)\simeq S$. We have an alternative description of $SO(\varphi)$: We can consider the units $T:=S^\ast$ of $S$ as an algebraic group and have a (surjective) norm map $T \to \mathbb G_m$ and then $SO(\varphi)$ is the kernel of this norm map. Using the exact sequence $1\to SO(\varphi)\to T\to\mathbb G_m\to1$ we see that an $SO(\varphi)$-torsor is given by a projective $S$-module $P$ of rank $1$ together with the choice of an isomorphism $\Lambda^2_RP\simeq R$.

[Added] Overcoming some of my laziness I did the calculation in the integral case ($R=\mathbb Z$ although the only thing we use is that $2$ is invertible): Start with the quadratic form $Cx^2+Bxy+Ay^2$ (the switch between $A$ and $C$ is to make my definition come out the same as the formula of the question). We then have $h^2=Bh-AC$ so that $h=\frac{B-\sqrt{\Delta}}{2}$. We have $h\cdot e_1=Be_1-Ce_2$ and $h\cdot e_2=Ae_1$. This is just an abstract module over $S$ but we can make it a fractional ideal by mapping $e_2$ to $A$, then $e_1$ maps to $h$ so that the fractional ideal is indeed $(A,\frac{B-\sqrt{\Delta}}{2})$.

[Added later] One could say that the association of the ideal $(A,h)$ to the quadratic form $Cx^2+Bxy+Ay^2$ over the ring $R[h]/(h^2-Bh+AC)$ is an answer to the question which workds in all classical cases ($R=\mathbb Z$ and
$R=k[T]$). The reason that this looks simpler than the traditional ($R=\mathbb Z$) answer is that we let the presentation of the quadratic order depend on the quadratic form itself. Usually we have fixed the quadratic order and want to
consider all forms with this fixed quadratic order as associated order. This
means that we should fix some normal form for the order and then express the
ideal in terms of this normal. When $R=\mathbb Z$ orders are in bijection with their discriminants $\Delta$ which are actual integers (as the discriminant is well-determined modulo squares of units). A normal form for the order is
$\mathbb Z+\mathbb Z\sqrt{\Delta}$ or $\mathbb Z+\mathbb Z(\sqrt{\Delta}+1)/2$ depending on whether $\Delta$ is even or odd. A curious feature of the form
$(A,\frac{B-\sqrt{\Delta}}{2})$ of the ideal is that the distinction between the odd
and even case is not apparent. However, the crucial thing is that it expresses a
$\mathbb Z$-basis for the ideal in terms of the canonical form of the quadratic
order.

The case of $R=k[T]$ for $k$ a field of odd characteristic is somewhat deceptive as the discriminant $\Delta=B^2-4AC$ is not quite an invariant of the quadratic order as there are units different from $1$ that are squares (except when $k=\mathbb Z/3$!). Hence the formula $(A,\frac{B-\sqrt{\Delta}}{2})$ is not quite of the same
nature as for the $\mathbb Z$ case as there is a very slight dependence of $\Delta$
on the form (and not just on the order). We could fix that by choosing coset
representatives for the squares as a subgroup of $k^\ast$ and then the formula for
the ideal would would take the form $(A,\frac{B-\lambda\sqrt{\Delta}}{2})$ where $\Delta$ now has been normalised so as to have its top degree coefficient to be a coset
representative.

The case when $k$ has characteristic $2$ is more complicated. We get and order of the form $k[T][h]/(h^2+gh+f)$ but the question of a normal form is trickier. The discriminant of the order (in the sense of the discriminant of the trace form) is equal to $g^2$ and as all elements of $k$ are squares we can normalise things so that $g$ is monic. However, the order is not determined by its discriminant. This can be seen already in the unramified case when $g=1$ when a normal form for $f$ is that it contain no monomials of even degree and the constant term is one of a chosen set of coset representives for the subgroup $\{\lambda^2+\lambda\}$ of $k$. We can decide to rather arbitrarily fix a generator $H$ for every order with $H^2=GH+F$ where one sensible first normalisation is that $G$ be monic (for which we have to assume that $k$ is perfect). Then we have that $h=H+a$, with $a\in k[T]$, and the ideal would be $(A,H+a)$.

There is a particular (arguably canonical) choice of $H$: We assume $G$ is monic and then can write uniquely $G=G_1G_2\cdots G_n$ with $G_i$ monic and $G_{i+1}|G_i$. The normal form is then that $F$ have the form $F=G_1F_1+G_1^2G_2F_2+\cdots+G_1^2G_2^2\cdots G_nF_n+G^2F'$ where $\deg F_i<\deg G_i$ and $F'$ contain no square monomials and its constant term belongs to a chosen set of coset representatives for $\{\lambda^2+\lambda\}$ in $k$.

One further comment relating to the classical formulas. When passing from a fractional ideal to a quadratic form one classically divides by the norm of the ideal (as is done in KConrad's reply). This means that the constructed form is primitive, i.e., the ideal generated by its values is the unit ideal. Hence if one starts with a quadratic form, passes to the ideal and then back to a quadratic form one does not end up with the same form if the form is not primitive. Rather the end form is the "primitivisation" where one has divided the form by a generator for its ideal of values. This of course only makes sense if the base ring is a PID. Even for a general Dedekind ring if one wants to work with primitive forms one has to accept quadratic forms that take values in general rank $1$ modules (i.e., fractional ideals).

The approach above makes another choice. It deals only with $R$-valued forms but accepts non-primitive ones. This would seem to lead to a contradiction as the classical construction leads to a bijection between modules and primitive forms and the above leads to a bijection between modules and arbitrary forms. There is no contradiction however (phew!) as the above construction leads to smaller orders than the classical one in the non-primitive case.

Classically what one really works with (when $R$ is a PID) are lattices in $L$ (the fraction field of $S$), where a lattice $M$ is a finitely generated $R$-submodule of $K$ containing a $K$-basis ($K$ the fraction field of $R$) for $L$. The order one associates to $M$ is the subring of $L$ stabilising $M$. When the quadratic form $\varphi$ is non-primitive $C^+(P)$ is not equal to this canonically associated order but is a proper suborder by it. Dividing by a generator for the ideal of values gives us a primitive form for which $C^+(P)$ is equal to the canonical order.

Finally there is a particular miracle that occurs for lattices in quadratic extensions (of the fraction field of a Dedekind ring $R$), $M$ as a module over its stabilisation order is projective. This is why classes of primitive forms with fixed order are in bijection with the class group of the order.

Answer 2

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

Answer 3

Classically, to a nonzero ideal $\mathfrak a$ in the ring of integers $R$ of a quadratic field, you pick a $\mathbf Z$-basis $e_1,e_2$ of $R$ and decide a proper way to order the basis. Once such a choice is made, for integers $x$ and $y$ define $$ Q_{\mathfrak a}(x,y) = \frac{{\mathrm N}(xe_1 + ye_2)}{\mathrm N \mathfrak a}, $$ where $x$ and $y$ are in $\mathbf Z$, the $\mathrm N$ upstairs is the ring norm from $R$ down to $\mathbf Z$ and the $\mathrm N$ downstairs is the ideal norm (taking values in the positive integers). This two-variable function $Q_{\mathfrak a}(x,y)$ is a primitive integral quadratic form, blah blah blah. The same thing works if $R$ is only an order in a quadratic field provided we limit our attention to invertible ideals in $R$.

We can try to set up something similar in characteristic 2. Say $K$ is a quadratic Galois extension of $k(T)$, where $k$ is a finite field of characteristic 2. Let $R$ be the integral closure of $k[T]$ in $K$. For any nonzero ideal $\mathfrak a$ in $R$, it is free of rank 2 over $k[T]$. Pick a $k[T]$-basis $e_1, e_2$ of $\mathfrak a$ as $k[T]$-module. For $x$ and $y$ in $k[T]$, define $$ Q_{\mathfrak a}(x,y) = \frac{{\mathrm N}(xe_1 + ye_2)}{\mathrm N \mathfrak a}, $$ where upstairs the $\mathrm N$ is the ring norm from $R$ down to $k[T]$ and downstairs the $\mathrm N$ is the $k[T]$-ideal norm: it is not the size of $R/\mathfrak a$ in the combinatorial sense, but rather the $k[T]$-cardinality of $R/\mathfrak a$: for any finitely generated torsion module $M$ over a PID $A$ (like $A=\mathbf Z$ or $A=k[T]$), write $M$ as a direct sum of cyclic modules $A/a_iA$ and declare the $A$-cardinality of $M$ to be the product of the ideals $a_iA$. (If $A$ is Dedekind but not a PID, the same concept makes sense but the ideals won't be principal in general.) Since $R/\mathfrak a$ is a torsion module over $k[T]$, $\mathrm{card}_{k[T]}(R/\mathfrak a)$ is a principal ideal and its unique monic generator is what $\mathrm N \mathfrak a$ means. Anyway, there you have a quadratic form associated to an ideal, it will have coefficients in $k[T]$, and you could explore how the quadratic forms associated to other ideals equivalent to $\mathfrak a$ are intrinsically related to one another. Such exploration would probably also guide you to a correct notion of ordered basis, analogous to the use of oriented bases in the classical theory.

Incidentally, up to this point I haven't actually used anything about the characteristic of $k$ (except for insisting the quadratic extension of $k(T)$ is Galois). It would be nice to find out where the characteristic being 2 or not will have a real effect on things.

Answer 4

I don't think the discriminant being square is an issue. This seems best understood by avoiding the "quadratic formula" expression and identifying the cohomological explanation for the appearance of the discriminant square class in the classical case over fields away from characteristic 2, as follows.

For non-degenerate quadratic spaces of rank $n$ over a field $K$ (e.g., $k(T)$ above), the discriminant makes explicit a map in cohomology if ${\rm{char}}(K) \ne 2$ but it plays no such role otherwise. More specifically, if $n$ is even or ${\rm{char}}(K) \ne 2$ the $K$-group scheme ${\rm{O}}_ n$ (automorphisms preserving the Bureau of Standards non-degenerate quadratic form in $n$ variables) is smooth and disconnected with identity component ${\rm{SO}}_ n$ and component group $\mathbf{Z}/2\mathbf{Z}$. This yields a map ${\rm{H}}^1(K, {\rm{O}}_ n) \rightarrow {\rm{H}}^1(K, \mathbf{Z}/2\mathbf{Z})$. If ${\rm{char}}(K) \ne 2$ then $\mathbf{Z}/2\mathbf{Z} = \mu_2$ over $K$, so the target of this map is identified with $K^{\times}/(K^{\times})^2$. The cohomology set ${\rm{H}}^1(K, {\rm{O}}_ n)$ classifies isomorphism classes of rank-$n$ non-degenerate quadratic spaces $(V,q)$ over $K$ for any field $K$ (as all such spaces become isomorphic over separable closure), and away from characteristic 2 the computation using $\mu_2$ thereby assigns to the isomorphism class of $(V,q)$ the class ${\rm{disc}}(q) \in K^{\times}/(K^{\times})^2$.

But in characteristic 2 with $n$ even this calculation doesn't work out explicitly in this way at all. Instead, we use Artin-Schreier theory in place of Kummer theory, and the resulting element in $K/\wp(K)$ is the Dickson invariant. In particular, the discriminant being a square or not has no impact at all like in the case of other characteristics (e.g., it does not imply that the quadratic form arises from ${\rm{H}}^1(K, {\rm{SO}}_ n)$).

Coming back to the case $n = 2$, it seems the natural thing to do is to not focus much on the "quadratic formula" aspect, and instead focus on the separable quadratic splitting field of the quadratic form. (The separability expresses the non-vanishing of the discriminant, which is the non-degeneracy condition since $n$ is even.) Then the discriminant sort of fizzles away, and things may become more uniform.