Lecture. Normal extensions and Galois extensions [l11]

  • May 6, 2026

Normal extensions and Galois extensions

Definition 1. normal extension [2201]

    An algebraic extension $K / F$ is normal if for every $\alpha \in K$, its minimal polynomial $m_{\alpha, F}(x) \in F[x]$ splits completely in $K$.

    Definition 2. splitting field for $A$ [2202]

      Let $F$ be a field and $A$ be a subset of $F[x]$. An algebraic field extension $K$ of $F$ is called a splitting field for $A$ (over $F$) if

      1. every $f(x) \in A$ splits completely in $K[x]$
      2. if $F \subseteq L \subseteq K$ and every $f(x) \in A$ splits completely in $L[x]$, then $L = K$.

      Remark 3. special case that $A$ is finite [2203]

        If $A$ is a finite subset of $F[x]$, say $A = \{ f_{1}(x), \dots, f_{n}(x) \}$. Then a field extension $K$ of $F$ is a splitting field for $A$ over $F$ if and only if $K$ is a splitting field for $$ f(x) = f_{1}(x) f_{2}(x) \cdots f_{n}(x) \in F[x]. $$

        Proposition 4. existence and uniqueness of the splitting field for $A$ [2204]

          Let $F$ be a field and $A$ be a subset of $F[x]$.

          1. A splitting field for $A$ over $F$ exists.
          2. If $K$ and $L$ are both splitting fields for $A$ over $F$, then there exists field isomorphism $\varphi : K \to L$ with $\varphi|_{F} = \operatorname{id}_{F}$.

          Proof. sketch [2204A]

            1. One can check that $F(S)$ is a splitting field for $A$ over $F$, where $S \subseteq \overline{F}$ is the subset consisting of roots of the polynomials in $A$. (cf. Proposition 1503)
            2. It suffices to show that if $K$ is a splitting field for $A$ over $F$, then there exists a field isomorphism $\varphi : K \to F(S)$ with $\varphi |_{F} = \operatorname{id}_{F}$. Apply Proposition 1603 to
              then we obtain a field embedding $\varphi : K \to \overline{F}$. Every polynomial splits completely over $K$ $\implies$ $S \subseteq \varphi(K)$ $\implies$ $F(S) \subseteq \varphi(K)$ $\implies$ $\varphi ^{-1}(F(S)) \subseteq K$. By (2) of Definition 2202, since $K$ is a splitting field for $A$ over $F$, it follows that $\varphi ^{-1}(F(S)) = K$ $\implies$ $F(S) = \varphi(K)$. Moreover, $\varphi$ is injective and $\varphi |_{F} = \operatorname{id}_{F}$. So $\varphi$ is a field isomorphism that maps $K$ to $F(S)$ with $\varphi |_{F} = \operatorname{id}_{F}$.

            Proposition 5. equivalent definitions of normal extensions [2205]

              Let $K / F$ be an algebraic extension. TFAE:

              1. $K / F$ is normal.
              2. Any irreducible polynomial in $F[x]$ that has a root in $K$ splits completely in $K$.
              3. There exists a subset $S \subseteq K$ such that $K = F(S)$ and $m_{\alpha, F}(x)$ splits completely in $K$ for every $\alpha \in S$.
              4. $K$ is the splitting field for a subset of $F[x]$.
              5. Fix an algebraic closure of $F$ with $F \subseteq K \subseteq \overline{F}$. (For example, $\overline{F} = \overline{K}$.) Any $F$-embedding $\varphi : K \to \overline{F}$ satisfies $\varphi(K) = K$.
              6. Fix an algebraic closure of $F$ with $F \subseteq K \subseteq \overline{F}$. Any $F$-embedding $\varphi : K \to \overline{F}$ satisfies $\varphi(K) \subseteq K$.

              Remark 6. the textbook definition for normal extensions [2206]

                The textbook uses (5) of Proposition 2205 as the definition of normal extensions.