A algebraic field extension L/k induces of homomorphism between the Wittrings. We get
$\phi: W(k) -> W(L)$. If every anisotropic isometry class of $W(k)$ stays anisotropic, the kernel of $\phi$ consists only of the class of the hyperbolic plane $\mathbb{H}$. Lets have a look on the $u$-invariants of the fields. One might think that for a anisotropic extension they coincide. This is wrong,as the example of in the comments shows. Here comes my question.
Is it possible that the kernel of $\phi$ is not trivial but $u(L)$ $=$ $u(k)$ holds, though?
We spare the case $u=\infty$.
In order for this to happen, there need to be several Pfisterforms $p,q$ of highest possible dimension in $W(k)$, such that the sets of symbols {$\alpha_1$...} , {$\beta_1$..} in $H^1(k,\mu_2)$ ,whose cup product equal $p$ and $q$ is disjoint. Adding the squareroot of $\alpha_1$ to $k$, for example does not influence the $u$-invariant then, because the other Pfisterform stays anisotropic, although the kernel of $\phi$ will contain the class of $\alpha$. So far my plan. (Note that i associate $X^2 + \alpha Y^2$ with $\alpha$)
But this would only work if i find a adequate field, whose Wittring has such Pfisterforms. Is there any example?
Im sorry,this isnt the highest level of research,but i dont get responces in undergraduate boards,on such stuff.