# If the characteristic of a field is two, doesn't the quadratic form associated with a bilinear form exist?

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

• See also this in MO. – Jyrki Lahtonen May 22 at 21:21
• OP, I think you should accept the other answer. – Jonathan Hebert May 23 at 17:41

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.

• OP, please accept this answer, not mine. I did not even answer the question you asked, I answered the reverse. – Jonathan Hebert May 23 at 0:12
• @JonathanHebert You may have to comment under the OP's question instead of under this answer to be certain of grabbing their attention. – Brahadeesh May 23 at 4:58

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 $$0$$.