Example. examples of splitting fields [1505]
Example. examples of splitting fields [1505]
1. $x^2 - 2$ [1505A]
1. $x^2 - 2$ [1505A]
The splitting field for $x^{2} - 2 \in \mathbb{Q}[x]$ is $\mathbb{Q}(\sqrt{ 2 })$. $\pm \sqrt{ 2 } \in \mathbb{Q}(\sqrt{ 2 })$. In general, the splitting of $x^{2} + bx + c \in \mathbb{Q}[x]$ is $\mathbb{Q}(\sqrt{ b^{2} - 4c })$.
2. $x^3 - 2$ [1505B]
2. $x^3 - 2$ [1505B]
The splitting field of $x^{3} - 2 \in \mathbb{Q}[x]$ is $\mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 })$.
Exegesis 2.1. why $\mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 })$ is the splitting field [1505BA]
Exegesis 2.1. why $\mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 })$ is the splitting field [1505BA]
Suppose $K$ is a splitting field of $x^{3} - 2$ in $\mathbb{Q}[x]$. The roots of $x^{3} - 2$ are $$\theta_{1} = \sqrt[3]{ 2 }, \quad \theta_{2} = \sqrt[3]{ 2 } e^{ 2 \pi i / 3 } = \sqrt[3]{ 2 } \frac{-1 + \sqrt{ -3 }}{2}, \quad \theta_{3} = \sqrt[3]{ 2 } e^{ 4 \pi i / 3 } = -\sqrt[3]{ 2 } \frac{-1 + \sqrt{ -3 }}{2}.$$ They all belong to $\mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 })$, hence $K \subseteq \mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 })$. On the other hand, if $K$ contains all the roots, then $\theta_{1} \in K$ and $\theta_{1}^{2}\theta_{2} = -1 + \sqrt{ -3 } \in K \implies \sqrt{ -3 } \in K$, and $\mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 }) \subseteq K$. Therefore, $K = \mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 })$.
Exegesis 2.2. degree of $\mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 })$ over $\mathbb{Q}$ [1505BB]
Exegesis 2.2. degree of $\mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 })$ over $\mathbb{Q}$ [1505BB]
We have $$ [\mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 }) : \mathbb{Q}] = [\mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 }) : \mathbb{Q}(\sqrt[3]{ 2 })] [\mathbb{Q}(\sqrt[3]{ 2 }) : \mathbb{Q}] = 2 \times 3 = 6. $$ $[\mathbb{Q}(\sqrt[3]{ 2 }) : \mathbb{Q}] = 3$ is because $m_{\sqrt[3]{ 2 }, \mathbb{Q}}(x) = x^{3} - 2$, which has degree $3$, and then by Proposition 1407.
Similarly, $x^{2} + 3$ is monic and irreducible in $\mathbb{Q}[x]$, so $m_{\sqrt{ -3 }, \mathbb{Q}}(x) = x^{2} + 3$. Moreover, $\mathbb{Q} \subseteq \mathbb{Q}(\sqrt[3]{ 2 })$, hence we must have $m_{\sqrt{ -3 }, \mathbb{Q}(\sqrt[3]{ 2 })}(x) \mid m_{\sqrt{ -3 }, \mathbb{Q}}(x)$, which means either $\deg m_{\sqrt{ -3 }, \mathbb{Q}(\sqrt[3]{ 2 })}(x) = 1$ or $2$. The former case implies $x - \sqrt{ -3 } \in \mathbb{Q}\left( \sqrt[3]{ 2 } \right)[x]$, which means $\sqrt{ -3 } \in \mathbb{Q}\left( \sqrt[3]{ 2 } \right)$. But $\sqrt{ -3 }$ is a pure imaginary number, it cannot exist in $\mathbb{Q}\left( \sqrt[3]{ 2 } \right)$. Hence $\deg m_{\sqrt{ -3 }, \mathbb{Q}\left( \sqrt[3]{ 2 } \right)}(x) = 2$, which means $[\mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 }) : \mathbb{Q}(\sqrt[3]{ 2 })] = 2$.
It is easy to see that $\mathbb{Q}(\sqrt{ -3 }), \mathbb{Q}(\theta_{1}), \mathbb{Q}(\theta_{2})$ and $\mathbb{Q}(\theta_{3})$ are all proper subfields of $\mathbb{Q}(\sqrt{ 2 }, \sqrt{ -3 })$ and $$ [\mathbb{Q}(\theta_{j}) : \mathbb{Q}] = 3 \quad (j = 1, 2, 3), \qquad [\mathbb{Q}(\sqrt{ -3 }) : \mathbb{Q}] = 2. $$
Exegesis 2.3. subfields of $\mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 })$ [1505BC]
Exegesis 2.3. subfields of $\mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 })$ [1505BC]
We have the diagram
3. $x^{n} - 1$ [1505C]
3. $x^{n} - 1$ [1505C]
The splitting field of $x^{n} - 1 \in \mathbb{Q}[x]$ is $\mathbb{Q}(\zeta_{n})$.
The polynomial splits completely in $\mathbb{C}$. Let $\zeta_{n} = e^{ 2\pi i / n } = \cos \left( \frac{2\pi}{n} \right) + i \sin \left( \frac{2\pi}{n} \right) \in \mathbb{C}$. Then $\zeta_{n}$ is a root of $x^{n} - 1$ and $\zeta_{n}^{k}, 0 \leq k \leq n -1$ are distinct roots of $x^{n} - 1$. It follows that $$ (x - \zeta_{n}^{0})(x - \zeta_{n}^{1}) \cdots (x - \zeta_{n}^{n-1}) \mid x^{n} - 1 $$ in $\mathbb{C}[x]$. By comparing degrees, we know that $$ x^{n} - 1 = (x - \zeta_{n}^{0})(x - \zeta_{n}^{1}) \cdots (x - \zeta_{n}^{n-1}). $$ Therefore, a splitting field of $x^{n} - 1$ is $\mathbb{Q}(\zeta_{n}, \dots, \zeta_{n}^{n-1}) = \mathbb{Q}(\zeta_{n}) = \mathbb{Q}[\zeta_{n}]$.
If $k$ is coprime to $n$, then $\mathbb{Q}(\zeta_{n}) = \mathbb{Q}(\zeta_{n}^{k})$.
When $n = p$ is prime, we have shown that $\Phi_{p}(x) = \frac{x^{p} - 1}{x - 1} = x^{p-1} + \cdots + x + 1$ is irreducible in $\mathbb{Q}[x]$ in MATH 111B. And $\Phi_{p}(\zeta_{p}) = 0$ and $\Phi_{p}(x)$ is monic, so $m_{\zeta_{p}, \mathbb{Q}}(x) = \Phi_{p}(x)$. Therefore, $[\mathbb{Q}(\zeta_{p}) : \mathbb{Q}] = \deg \Phi_{p}(x) = p-1$.
Later, we will show that $[\mathbb{Q}(\zeta_{n}) : \mathbb{Q}] = \lvert (\mathbb{Z} / n\mathbb{Z})^{\times} \rvert$.