Determine splitting field $K$ over $\mathbb{Q}$ of the polynomial $x^3 - 2$

Determine the splitting field $K$ over $\mathbb{Q}$ of the polynomial $x^3 - 2$

Also determine the basis over $\mathbb{Q}$ and its degree. Can I do this using only first principles?

Solution 1:

The roots of $x^3-2=0$ are $x_1=\sqrt[3]{2},$ $x_2= \sqrt[3]{2}\omega$, $x_3=\sqrt[3]{2}\omega^2$, where $\omega=\frac{-1+i\sqrt{3}}{2}$. It follows that the splitting field $K$ is $$K=\mathbb{Q}(x_1,x_2,x_3)=\mathbb{Q}(\sqrt[3]{2},i\sqrt{3})$$

Since $1, \sqrt[3]{2}, \sqrt[3]{2}^2$ form a basis for $\mathbb{Q}(\sqrt[3]{2})$ over $\mathbb{Q}$ and $1, i\sqrt{3}$ form a basis for $\mathbb{Q}(\sqrt[3]{2}, i\sqrt{3})$ over $\mathbb{Q}(\sqrt[3]{2})$, a basis for $\mathbb{Q}(\sqrt[3]{2},i\sqrt{3})$ over $\mathbb{Q}$ must be $1, \sqrt[3]{2}, \sqrt[3]{2}^2,i\sqrt{3}, \sqrt[3]{2}i\sqrt{3},\sqrt[3]{2}^2i\sqrt{3} $.

Edit: I have used the following facts, taken from Robert Ashs excellent book Abstract Algebra: The Basic Graduate Year:

3.1.7

Let $E/F$ be a field extension and suppose $\alpha\in E$ is algebraic over $F$. Then $$1, \alpha, \alpha^2, ..., \alpha^{n-1}$$ form a basis of $F(\alpha)$ over $F$, where $n=$deg Irr$(\alpha, F)$.

3.1.8

If $E/K/F$ are field extensions such that $\{\alpha_i\}_{i\in I}$ is a basis for $E$ over $K$ and $\{\beta_j\}_{j\in J}$ is a basis for $K$ over $F$, then $\{\alpha_i\beta_j\}_{i\in I, j\in J}$ form a basis for $E$ over $F$.