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.

  • $\begingroup$ 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. $\endgroup$ – Jyrki Lahtonen Feb 4 at 21:25
  • $\begingroup$ 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. $\endgroup$ – Jyrki Lahtonen Feb 4 at 21:32
  • $\begingroup$ I need to consult my bible of Finite Fields (in my office) for more, so it will have to wait. $\endgroup$ – 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.)


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.

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