# Existence of orthogonal base for finite Galois extension over characteristic 2

Let $$K$$ be a field of characteristic $$2$$ and $$L$$ be a finite Galois extension of $$K$$. Considering the trace $$Tr_{L/K}: L \to K$$ and $$L$$ as a finite dimensional $$K$$-vectorspace we know, that $$Tr_{L/K} \neq 0$$ hence we get a non-degenerate $$K$$-bilinear map

$$Tr_{L/K}: L \times L \to K$$

$$(x,y) \mapsto Tr_{L/K}(xy)$$

which gives us an isomorphism $$L \to L^*$$.

Does there exist an orthogonal resp. selfdual basis $$\{x_1, \dots, x_n\}$$ of $$L$$, i.e. we have $$Tr_{L/K}(x_i x_j)=0 \iff i \neq j$$.

If the characteristic is not 2, it is easy to construct such a basis inductively. Namely let $$\langle x_1, \dots, x_k\rangle$$ be as wanted (orthogonal and their square has non-zero trace) and by Gram–Schmidt we have a basis of $$\langle x_1, \dots, x_k\rangle^\perp$$. Since $$Tr_{L/K}$$ is non-degenerate we find $$y,z \in \langle x_1, \dots, x_k\rangle^\perp$$ with $$Tr_{L/K}(yz) \neq 0$$ and hence the square of either $$y,z$$ or $$y+z$$ are non-zero (char $$\neq 2$$). For the char=2 case I found only examples where it works so far.

• It's been a while since I needed this piece of information. If $K$ is the prime field then I think such a basis always exists. One argument is that If you start with any basis $\{x_1,\ldots, x_n\}$, then the matrix $A=tr(x_ix_j)$ is symmetric and has full rank. Also at least one of the diagonal entries of $A$ must be non-zero. A result of Seroussi and Lempel then says that $A=MM^T$ for some $n\times n$ matrix $M$. Using $M$ as a change of basis matrix gives you the claim. – Jyrki Lahtonen Feb 4 at 21:25
• IIRC an inductive proof of the Seroussi-Lempel result is not too difficult. Their result covers the construction of a minimal size matrix $M$ for all symmetric matrices $A$ over $\Bbb{F}_2$. And the case where $A$ has all zeros along the diagonal is the tricky one,and there $M$ has $r(A)+1$ columns, otherwise $r(A)$ columns suffices. I am not sure about the more general case. As you observed self-duality is impossible to achieve if $L/K$ is not separable. For then the trace function vanishes altogether. – Jyrki Lahtonen Feb 4 at 21:32
• I need to consult my bible of Finite Fields (in my office) for more, so it will have to wait. – Jyrki Lahtonen Feb 4 at 21:33

The following more general result is true in general.

Thm. Let $$b:E\times E\to F$$ a non degenerate nonalternating symmetric bilinear form over a field $$F$$ of characteristic $$2$$ (where $$E$$ is a finite dimensional $$F$$-vector space. Then $$E$$ has a $$b$$-orthogonal basis.

Let $$f_b: x\in E\to b(x,x)\in F$$. Since we are in characteristic $$2$$, this map is additive. This will come handy for computations.

We say that $$b$$ is alternating if $$f_b$$ is the zero map, and non alternating otherwise.

Note that in your situation, your bilinear form is non alternating, because $$Tr_{L/K}(x^2)=(Tr_{L/K}(x))^2$$, and the trace is a nonzero map since a Galois extension is separable (thus your result will be true more generally for finite separable extensions.)

Proof.

Claim. There exists $$e_1,e_2\in E$$ such that $$b(e_1,e_1)\neq 0$$, $$b(e_1,e_2)=0$$ and $$b(e_2,e_2)\neq 0.$$

Assume the claim is proved. Then, since $$b(e_1,e_1)\neq 0$$,the restriction of $$b$$ to $$Fe_1$$ is non degenerate, so $$E=Fe_1\oplus (Fe_1)^\perp$$. Now the restriction of $$b$$ on $$(Fe_1)^\perp$$ is nondegenerate and non alternating, by assumptions on $$e_2$$, and we may conclude by induction (pick a $$b$$-orthogonal basis of $$(Fe_1)^\perp$$ and add $$e_1$$ to it).

Proof of the claim.

Since $$b$$ is non alternating, pick $$e_1\in E$$ such that $$\lambda= b(e_1,e_1)\neq 0$$. Hence the restriction of $$b$$ to $$Fe_1$$ is non degenerate, so $$E=Fe_1\oplus (Fe_1)^\perp$$.

If $$b$$ is non alternating on $$(Fe_1)^\perp$$, pick any $$e_2\in (Fe_1)^\perp$$ such that $$b(e_2,e_2)\neq 0$$.

If $$b$$ is alternating on $$(Fe_1)^\perp$$, pick a nonzero $$e_2\in (Fe_1)^\perp$$. Since the restriction of $$b$$ to $$(Fe_1)^\perp$$ is non degenerate, there exists $$e_3\in (Fe_1)^\perp$$ such that $$b(e_2,e_3)=1$$. Notice that we have $$b(e_3,e_3)=0$$. Set $$e'_1=e_1+\lambda e_2+\lambda e_3$$ and $$e'_2=e_1+e_2$$. Then $$b(e'_1,e'_1)=\lambda\neq 0$$ and $$b(e'_1, e'_2)=0$$. Now $$b(e'_2,e'_2)=\lambda\neq 0$$, and we are done.