Witt index of the sum of 24 squares

Question

Consider the quadratic forms $q(x)=x_1^2+\dots +x_8^2$ in 8 variables and $p(x)=x_1^2+\dots+x_{24}^2$ in 24 variables over the field of rational numbers $\mathbb{Q}$. Let $E/\mathbb{Q}$ be a field extension. Is it true that if $q_E$ is anisotropic over $E$, then the Witt inddex of $p_E$ over $E$ is less or equal 8?

Answer 1

The answer is YES. The quadratic form $q$ from the question is a Pfister form, and $p=q\oplus q\oplus q$. Therefore, the affirmative answer follows from the next lemma.

Lemma. Let $q$ be an anisotropic Pfister quadratic form of dimension $m=2^n$ over a field $k$ of characteristic different from $2$. Let $a,b,c\in k^\times$. Consider the quadratic form $p:=aq\oplus bq\oplus cq$ of dimension $3m$. Then the Witt index $i(p)\le m$.

Proof. We may assume that $a=1$. Consider the quadratic form $s:=q\oplus bq$ of dimension $2m$, it is again a Pfister form, and by Pfister's theorem it is either anisotropic or (completely) split.

If $s$ is split, then $p=s\oplus cq$ is a direct sum of the split form $s$ and the anisotropic form $cq$, and by Witt's decomposition theorem $i(p)=i(s)=m$.

If $s$ is anisotropic, then $p\oplus (-cq)=s\oplus cq\oplus(-cq)$ is a direct sum of the anisotropic form $s$ and the split form $cq\oplus(-cq)$ of dimension $2m$ and of Witt index $m$. By Witt's decomposition theorem $i(p\oplus(-cq))=i(cq\oplus(-cq))=m$, and therefore, $i(p)\le m$.

Answer 2

The answer seems to be affirmative.

It is well known that if the level of a number field is finite then it belongs to $\{1,2,4\}$.

Now in your situation, as $q_E$ is anisotropic, $E$ should be formally real. So $p_E$ is anisotropic as well, so its Witt index is $0$.