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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.

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  • $\begingroup$ I do not understand your claims at the beginning : if $k=\mathbb{C}$ and $L=\mathbb{C}((t))$, $\phi$ is injective, $u(k)=1$ and $u(L)=2$. $\endgroup$ Mar 5, 2014 at 22:50
  • $\begingroup$ You are right. We should restrict to the case of a algebraic field extension. I will edit this. Besides that, i have to admit that in the case of the extension you suggest,the kernel of $\phi$ is trivial. $\endgroup$ Mar 6, 2014 at 2:12
  • $\begingroup$ I am still worried by your statements. Consider a Galois extension $F$ of $\mathbb{Q}$ with Galois group $G=\mathcal{A}_5$. Choose a subgroup $\mathbb{Z}/2\mathbb{Z}\subset G$ Let $k$ be the quadratic closure of $\mathbb{Q}$ and $L$ be the subextension of $Fk/k$ corresponding to $\mathbb{Z}/2\mathbb{Z}$, so that $L$ is not quadratically closed. Then $\phi$ is injective, $u(k)=1$ but $u(L)\geq 2$. $\endgroup$ Mar 6, 2014 at 9:05
  • $\begingroup$ I have to delete this inaccurate statement of mine. It does not affect my question still. $\endgroup$ Mar 6, 2014 at 12:34

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