Note that $\alpha= \sqrt[3]{2}e^{2\pi i/3}$ is a root of the polynomial $x^3-2$.
The polynomial $x^3-2$ is irreducible over $\Q$ by Eisenstein’s criterion with prime $p=2$.

The roots of this polynomial are
\[\sqrt[3]{2}, \sqrt[3]{2}e^{2\pi i/3}, \sqrt[3]{2}e^{4\pi i/3}.\]
Then it follows that we have an isomorphism
\begin{align*}
\Q(\alpha)=\Q(\sqrt[3]{2}e^{2\pi i/3}) \cong \Q[x]/(x^3-2)\cong \Q(\sqrt[3]{2}).
\end{align*}
Let us denote this isomorphism by $\phi:\Q(\alpha) \to \Q(\sqrt[3]{2})$, which sends $\alpha$ to $\sqrt[3]{2}$ and fixed $\Q$ elementwise.

Seeking a contradiction, we assume that there exist $x_1, \dots, x_k \in \Q(\alpha)$ such that
\[x_1^2+\cdots +x_k^2=-1.\]

Then we apply the isomorphism $\phi$ and obtain
\begin{align*}
-1&=\phi(-1)=\phi(x_1^2+\cdots +x_k^2)\\
&=\phi(x_1)^2+\cdots +\phi(x_k)^2.
\end{align*}

However, this equality does not hold since $\phi(x_i) \in \Q(\sqrt[3]{2})$ and the field $\Q(\sqrt[3]{2})$ consists of real numbers.

Thus, we have reached a contradiction, hence there is no solution of $x_1^2+\cdots +x_k^2=-1$ in the field $\Q(\alpha)$.

$x^3-\sqrt{2}$ is Irreducible Over the Field $\Q(\sqrt{2})$
Show that the polynomial $x^3-\sqrt{2}$ is irreducible over the field $\Q(\sqrt{2})$.
Hint.
Consider the field extensions $\Q(\sqrt{2})$ and $\Q(\sqrt[6]{2})$.
Proof.
Let $\sqrt[6]{2}$ denote the positive real $6$-th root of of $2$.
Then since $x^6-2$ is […]

Irreducible Polynomial $x^3+9x+6$ and Inverse Element in Field Extension
Prove that the polynomial
\[f(x)=x^3+9x+6\]
is irreducible over the field of rational numbers $\Q$.
Let $\theta$ be a root of $f(x)$.
Then find the inverse of $1+\theta$ in the field $\Q(\theta)$.
Proof.
Note that $f(x)$ is a monic polynomial and the prime […]

Example of an Infinite Algebraic Extension
Find an example of an infinite algebraic extension over the field of rational numbers $\Q$ other than the algebraic closure $\bar{\Q}$ of $\Q$ in $\C$.
Definition (Algebraic Element, Algebraic Extension).
Let $F$ be a field and let $E$ be an extension of […]

Galois Group of the Polynomial $x^p-2$.
Let $p \in \Z$ be a prime number.
Then describe the elements of the Galois group of the polynomial $x^p-2$.
Solution.
The roots of the polynomial $x^p-2$ are
\[ \sqrt[p]{2}\zeta^k, k=0,1, \dots, p-1\]
where $\sqrt[p]{2}$ is a real $p$-th root of $2$ and $\zeta$ […]

Degree of an Irreducible Factor of a Composition of Polynomials
Let $f(x)$ be an irreducible polynomial of degree $n$ over a field $F$. Let $g(x)$ be any polynomial in $F[x]$.
Show that the degree of each irreducible factor of the composite polynomial $f(g(x))$ is divisible by $n$.
Hint.
Use the following fact.
Let $h(x)$ is an […]

Application of Field Extension to Linear Combination
Consider the cubic polynomial $f(x)=x^3-x+1$ in $\Q[x]$.
Let $\alpha$ be any real root of $f(x)$.
Then prove that $\sqrt{2}$ can not be written as a linear combination of $1, \alpha, \alpha^2$ with coefficients in $\Q$.
Proof.
We first prove that the polynomial […]

Galois Extension $\Q(\sqrt{2+\sqrt{2}})$ of Degree 4 with Cyclic Group
Show that $\Q(\sqrt{2+\sqrt{2}})$ is a cyclic quartic field, that is, it is a Galois extension of degree $4$ with cyclic Galois group.
Proof.
Put $\alpha=\sqrt{2+\sqrt{2}}$. Then we have $\alpha^2=2+\sqrt{2}$. Taking square of $\alpha^2-2=\sqrt{2}$, we obtain […]

The Polynomial $x^p-2$ is Irreducible Over the Cyclotomic Field of $p$-th Root of Unity
Prove that the polynomial $x^p-2$ for a prime number $p$ is irreducible over the field $\Q(\zeta_p)$, where $\zeta_p$ is a primitive $p$th root of unity.
Hint.
Consider the field extension $\Q(\sqrt[p]{2}, \zeta)$, where $\zeta$ is a primitive $p$-th root of […]