Lecture. Algebraic closure [l5]

  • Apr 13, 2026

Example 1. examples of splitting fields [1505]

    1.1. $x^2 - 2$ [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 $\mathbb{Q}(\sqrt{ b^{2} - 4c })$.

      1.2. $x^3 - 2$ [1505B]

        The splitting field of $x^{3} - 2 \in \mathbb{Q}[x]$ is $\mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 })$.

        Exegesis 1.2.1. 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 1.2.2. 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 1.2.3. subfields of $\mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 })$ [1505BC]

              We have the diagram

              Actually, these are all the subfields of $\mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 })$. We will show it later.

              1.3. $x^{n} - 1$ [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 $[\mathbb{Q}(\zeta_{n}) : \mathbb{Q}] = \lvert (\mathbb{Z} / n\mathbb{Z})^{\times} \rvert$.

                1.4. $x^p - 2$ (TODO) [1505D]

                  Theorem 2. [1506]

                    1. All splitting fields of $f(x) \in F[x]$ are isomorphic.
                    2. More generally, let $\varphi : F_{1} \xrightarrow{\cong} F_{2}$ be an isomorphism of fields. Given $f_{1}(x) = a_{n}x^{n} + \cdots + a_{1}x + a_{0} \in F_{1}[x]$, let $f_{2}(x) = \varphi(a_{n})x^{n} + \cdots + \varphi(a_{1})x + \varphi(a_{0}) \in F_{2}[x]$. Suppose that $E_{1}$ is a splitting field of $f_{1}(x)$ over $F_{1}$ and $E_{2}$ is a splitting field of $f_{2}(x)$ over $F_{2}$. Then the isomorphism $\varphi$ extends to an isomorphism $\sigma : E_{1} \xrightarrow{\cong} E_{2}$, i.e., there exists $\sigma$ such that the diagram

                      The induction argument requires the more general statement.

                      The strategy $F \subseteq F(\alpha) \subseteq K$ with directly handling simple extension $F(\alpha) / F$ and apply induction to $E / F(\alpha)$ will be repeatedly used in several proofs.

                    Proof. sketch [1506A]

                      1. Proposition 1503 show that all splitting fields of $f(x)$ contained in a common field extension $K$ of $F$ are the same.
                      2. We prove by induction on $\deg f_{1}(x) = n$.

                        1. Case $n = 1$: in this case, $E_{1} = F_{1}$ and $E_{2} = F_{2}$. The isomorphism $\sigma$ is simply $\varphi$.

                        2. Case $n$: assume the theorem holds for all $\deg f_{1}(x) < n$.

                          Let $\alpha$ be a root of $f_{1}(x)$ over $E_{1}$, we first find its corresponding root $\beta$ of $f_{2}(x)$ over $E_{2}$ such that $\sigma(\alpha) = \beta$. After that, we establish $F_{1}(\alpha) \cong F_{2}(\beta)$ and write $f_{1}(x) = (x - \alpha)g_{1}(x)$ and $f_{2}(x) = (x - \beta)g_{2}(x)$ with $\deg g_{1}(x) < n$ and $\deg g_{2}(x) < n$, and then use assumption on $g_{1}, g_{2}$ over $F_{1}(\alpha), F_{2}(\beta)$ to get $F_{1}(\alpha)(\alpha_{1}, \dots, \alpha_{n-1}) \cong F_{2}(\beta)(\beta_{1}, \dots, \beta_{n-1})$ where $\alpha_{1}, \dots, \alpha_{n-1}$ are the roots of $g_{1}(x)$ and $\beta_{1}, \dots, \beta_{n-1}$ are the roots of $g_{2}(x)$. And finally claim that $E_{1} = F_{1}(\alpha, \alpha_{1}, \dots, \alpha_{n-1}) = F_{1}(\alpha, \dots, \alpha_{n-1})$ and the same for $E_{2}$.

                      Proposition 3. an upper bound for the degree of the splitting field [1507]

                        The splitting field of a polynomial of degree $n$ over a base field is of degree at most $n!$.

                        Proof. sketch [1507A]

                          Let $F$ be the base field and $f(x) \in F[x]$ be the polynomial with $\deg f(x) = n$. Let $\alpha_{1}, \dots, \alpha_{n}$ be the roots of $f(x)$ and $E = F(\alpha_{1}, \dots, \alpha_{n})$ be the splitting field of $f(x)$.

                          Then $[F(\alpha_{1}) : F]$ has degree at most $n$ since $1, \alpha_{1}, \dots, \alpha_{1}^{n}$ is not linearly independent ($f(\alpha_{1}) = 0$). Moreover, by Homework 2, $[F(\alpha_{1}, \alpha_{2}) : F(\alpha_{1})] < [F(\alpha_{1}) : F] \leq n \implies [F(\alpha_{1}, \alpha_{2}): F(\alpha_{1})] \leq n-1$. If we keep doing so, we will finally arrive at $$ \begin{align*} [E : F] &= [F(\alpha_{1}, \dots, \alpha_{n}) : F(\alpha_{1}, \dots, \alpha_{n-1})] \cdots [F(\alpha_{1}, \alpha_{2}) : F(\alpha_{1})][F(\alpha_{1}) : F] \\ &\leq 1 \cdots \cdot (n-1)n \\ &= n! \end{align*} $$

                          Algebraic closure

                          Proposition 4. equivalent definitions of algebraically closed fields [1601]

                            For a field $\Omega$, the following statements are equivalent:

                            1. Every non-constant polynomial in $\Omega[x]$ splits completely in $\Omega$.
                            2. Every non-constant polynomial in $\Omega[x]$ has a root in $\Omega$.
                            3. The irreducible polynomials in $\Omega[x]$ are those of degree $1$.
                            4. If $K / \Omega$ is an algebraic extension, then $K = \Omega$.
                            5. If $K / \Omega$ is a finite extension, then $K = \Omega$.

                            Proof. sketch [1601A]

                              1 $\iff$ 2 $\iff$ 3 is obvious.

                              3 $\implies$ 4: Take $\alpha \in K$ and consider $m_{\alpha, \Omega}(x)$. It must be $m_{\alpha, \Omega}(x) = x - \alpha \in \Omega[x]$ since it is irreducible and monic, hence $\alpha \in \Omega$. So we conclude that $K = \Omega$.

                              4 $\implies$ 5: Obvious because of Proposition 14010.

                              5 $\implies$ 1: Take any $f(x) \in \Omega[x]$ and let $K$ be its splitting field. Then by Proposition 1507, $[K : \Omega]$ is at most $n!$ where $n = \deg f(x)$, hence it is finite. So $K = \Omega$. Then we can conclude that every non-constant polynomial in $\Omega[x]$ splits completely in $\Omega$.

                              Definition 5. algebraically closed field [1602]

                                A field $\Omega$ is called algebraically closed if it satisfies the equivalent conditions in Proposition 1601.