Is $-1$ sum of squares in the field $\mathbb{Q(\beta)}$ where $\beta = 2^{1/3}e^{2\pi i /3}$

Question

Prove that $−1$ is not a sum of squares in the field $\mathbb{Q(\beta)}$ where $\beta = 2^{1/3}e^{2\pi i /3}$

My attempt : In fact $\Bbb Q(\beta)$ and $\Bbb Q(2^{1/3})$ are naturally isomorphic. So I need to show $−1$ is not a sum of squares in the field $\Bbb Q(2^{1/3})$. But I don't know if field isomorphism preserves the ordering or not. If this happen then I can tell some thing that as $\Bbb Q(2^{1/3})$ ordered but $-1$ is not positive. Or may be there is something tricky

Actually I can't find any way to do this after this point. Any help/hint in this regards would be highly appreciated. Thanks in advance!

Answer 1

$−1$ is not a sum of squares in the field $\mathbb{Q}(2^{1/3})$, since $\mathbb{Q}(2^{1/3}) \subset \mathbb{R}$.

That's exactly the point.

But a field isomorphism maps a sum of squares to a sum of squares, and maps $-1$ to $-1$, hence if two fields are isomorphic, and if $-1$ is a not a sum of square in one of the fields, then it's also not a sum of squares in the other.

Answer 2

Here's my attempt at a proof:

So the field extension has degree 3, so let's try to write everything in the basis $\{1,\beta,\beta^2\}$.

$$(p+q\beta+r\beta^2)^2=(p^2+4qr)+(2r^2+2pq)\beta+(q^2+2pr)\beta^2$$

where we use the fact that $\beta^3=2$. So now consider the sums of squares:

$$\sum (p+q\beta+r\beta^2)^2=-1$$

which implies

$$\sum p^2+4qr=-1\implies -4q\cdot r>p\cdot p$$ $$\sum 2r^2+2pq=0\implies -p\cdot q=r\cdot r$$ $$\sum q^2+2pr=0\implies -2p\cdot r=q\cdot q$$

I need to derive a contradiction from this somehow..

Answer 3

Each element in $\;\Bbb Q(\beta)\;$ can be written as $\;a+b\beta+c\beta^2\;,\;\;a,b,c\in\Bbb Q\;$ , since $\;\dim_{\Bbb Q}\Bbb Q(\beta)=3\;$ . Now, we can define

$$\phi: \Bbb Q(\beta)\to\Bbb Q(\sqrt[3]2)\,,\;\text{by defining}\;\phi\beta:=\sqrt[3]2$$

and extending linearly:

$$\phi(a+b\beta+c\beta^2):=a+b\sqrt[3]2+c\sqrt[3]4$$

Check the above is well defined indeed (trivial, from linear algebra) and that it indeed is a fields isomorphism (almost trivial since both $\;\beta,\,\sqrt[3]2\;$ are roots of the same (irreducible) polynomial $\;x^3-2\in\Bbb Q[x]\,$) .

Suppose now that there are $\;x_1,...x_n\in\Bbb Q(\beta)\;$ s.t. $\;\sum\limits_{k=1}^n x_i^2=-1\;$ , then

$$-1=\phi(-1)=\sum_{k=1}^n\left(\phi x_i\right)^2$$

yet this is impossible since the rightmost sum happens in $\;\Bbb Q(\sqrt[3]2)\;$ , which is a real field .