Proving No Solution in Extension Field

[Szichedelic]

Homework Statement

Let $\beta=\omega\sqrt[3]{2}$, where $\omega=e^{2\pi i/3}$, and let $K=\mathbb{Q}(\beta)$. Prove that the equation $x^{2}_{1}+\cdots+x^{2}_{k}=-1$ has no solution with $x_{i}$ in $K$.

Homework Equations

The irreducible polynomial f is the monic polynomial of lowest degree in F[x} that has an element $\alpha$ of K as a root.

The Attempt at a Solution

My first idea is to move the one over and then view f(x)+1 as a polynomial in F[x]. Then, I want to show that there is no element in K which is a root to this polynomial... I believe this would show that there is no solution to the above equation with $x_{i}$ in $K$. However, I'm not sure how to show this for an arbitrary element of K and if this is even the right approach. I was also wondering if it had to do with the fact that $\beta$ contains a cube root, which would mean we would need at least a degree 3 polynomial with coefficients in the rationals... We only have a degree 2.

 

[mighty2000]

Your approach is on the right track. To prove that there is no solution to the equation with x_i in K, you can use the fact that the irreducible polynomial f of \beta over \mathbb{Q} has degree 3. This means that any element in K can be written as a polynomial in \beta of degree at most 2. So, if we consider the polynomial f(x)+1, we can write it as a polynomial in \beta of degree at most 2. Now, if we assume that there exists an element in K that is a root of f(x)+1, then this would mean that we can write this element as a polynomial in \beta of degree at most 2. But, since f(x)+1 has degree 3, this is a contradiction. Therefore, there is no element in K that is a root of f(x)+1, and hence there is no solution to the equation with x_i in K.