# Anisotropy over bigger fields

• Dec 4th 2011, 07:03 PM
redsoxfan325
Anisotropy over bigger fields
My goal for the problem is to classify all equivalence classes of anisotropic quadratic forms over $\mathbb{F}_3((t))$ and then use that to determine 1) which of these forms stay anistropic and 2) which become isometric to each other over:

1. $\mathbb{F}_9((t))$
2. $\mathbb{F}_3((\sqrt{t}))$

The anisotropic equivalence classes for dimensions 1 through 4 (which is sufficient since $u(\mathbb{F}_3((t))=4$) are:

$\langle1\rangle, \langle2\rangle, \langle t\rangle, \langle2t\rangle$
$\langle1,1\rangle, \langle t,t\rangle, \langle1,t\rangle, \langle2,t\rangle$
$\langle1,1,t\rangle,\langle1,1,2t\rangle,\langle1, t,t\rangle,\langle2,t,t\rangle$
$\langle1,1,t,t\rangle$

where the notation $\langle a,b\rangle$, for example, means the quadratic form $q(x)=ax^2+by^2$.

1. Over $\mathbb{F}_9((t))$, $2$ becomes a square, so we have

$\langle1\rangle\cong\langle2\rangle, \langle t\rangle\cong\langle2t\rangle$
$\langle1,1\rangle, \langle t,t\rangle, \langle1,t\rangle\cong\langle2,t\rangle$
$\langle1,1,t\rangle\cong\langle1,1,2t\rangle, \langle 1,t,t\rangle\cong\langle2,t,t\rangle$
$\langle1,1,t,t\rangle$

Nothing becomes isotropic.

2. Over $\mathbb{F}_3((\sqrt{t}))$, $t$ becomes a square, so all that remains is

$\langle1\rangle\cong\langle t\rangle,\langle2\rangle\cong\langle2t\rangle$
$\langle1,1\rangle\cong\langle t,t\rangle\cong\langle1,t\rangle$

Everything else becomes isotropic.

Thanks.