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I have a question: if the characteristic of a field is two, doesn't the quadratic form associated with a bilinear form exist?

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    $\begingroup$ See also this in MO. $\endgroup$ – Jyrki Lahtonen May 22 at 21:21
  • $\begingroup$ OP, I think you should accept the other answer. $\endgroup$ – Jonathan Hebert May 23 at 17:41
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If $b:V\times V\to K$ is a symmetric bilinear form, you can always associate a quadratic form $q_b:V\to K$ by setting $q_b(x)=b(x,x)$ for all $x\in V$.

If $q:V\to K$ is a quadratic form, by very definition the map $b_q:V\times V\to K, (x,y)\mapsto q(x+y)-q(x)-q(y)$ is symmetric and bilinear.

However, there is no 1-1 correspondence anymore between quadratic forms and symmetric bilinear forms when $char(K)=2$.

To see this, just observe that in this case, $b_q$ is alternating; $b_q(x,x)=q(2x)-2q(x)=2q(x)=0$.

On the other hand, if $b$ is alternating, $q_b$ is the zero quadratic form.

Thus we have two very different theories...or even three:

  • the theory of non alternating symmetric nondegenerate bilinear forms: such forms can be diagonalized ($V$ has an orthogonal basis wrt to $b$): see my answer here :Existence of orthogonal base for finite Galois extension over characteristic 2

  • the theory of alternating symmetric nondegenerate bilinear forms: any such form is hyperbolic ($V$ has a symplectic basis wrt to $b$)

  • the theory of non degenerate quadratic forms: any such form is an orthogonal sum of quadratic planes $K^2\to K, (x_1,x_2)\mapsto ax_1^2+x_1x_2+bx_2^2$. Note that a $q$-orthogonal basis never exists in this case.

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  • $\begingroup$ OP, please accept this answer, not mine. I did not even answer the question you asked, I answered the reverse. $\endgroup$ – Jonathan Hebert May 23 at 0:12
  • $\begingroup$ @JonathanHebert You may have to comment under the OP's question instead of under this answer to be certain of grabbing their attention. $\endgroup$ – Brahadeesh May 23 at 4:58
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If you give me a quadratic form $q(x)$, then it is associated to the bilinear form:

$b(x,y) = \frac{1}{2} (q(x+y) - q(x) - q(y))$

The immediate problem for a field of charactistic $2$ is that I do not know what to make of $\frac{1}{2}$. $2$ does not have a multiplicative inverse, because it is the additive identity. It is $0$.

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