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    Splitting fields and normal extensions

    An algebraic extension \(L/K\) is normal if every irreducible polynomial in \(K[x]\) with a root in \(L\) splits into linear factors over \(L\). A finite extension \(L/K\) is normal if and only if \(L\) is the splitting field over \(K\) of some polynomial in \(K[x]\).

    Normal field extension

    Let \(L/k\) be an algebraic field extension. The extension \(L/k\) is normal if every irreducible polynomial \(f\in k[x]\) that has a root in \(L\) splits into linear factors in \(L[x]\).

    Equivalently, for every \(\alpha\in L\), all roots of the minimal polynomial of \(\alpha\) over \(k\) in an algebraic closure of \(L\), the conjugates of \(\alpha\) over \(k\), lie in \(L\). Thus an irreducible \(f\in k[x]\) either splits in \(L[x]\) or has no root in \(L\).

    Normal field extension

    An algebraic field extension \(L/K\) is normal if every irreducible polynomial \(p \in K[x]\) either has no roots in \(L\) or splits into linear factors in \(L[x]\).

    What does it mean for a field extension $L/K$ to be normal?
    What does an irreducible $p(x) \in K[x]$ do in a normal extension $L/K$?

    Splitting field of a polynomial

    Let \(k\) be a field and \(f\in k[x]\) a nonzero polynomial. A splitting field of \(f\) over \(k\) is a field extension \(F/k\) such that \(f\) splits into linear factors in \(F[x]\) and \(f\) does not split into linear factors in \(E[x]\) for any proper subfield \(E\subsetneq F\) containing \(k\).

    Equivalently, \(f\) splits in \(F[x]\) as \(f=c\prod_{i=1}^n(x-\alpha_i)\) and \(F=k(\alpha_1,\ldots,\alpha_n)\): a subfield of \(F\) containing \(k\) over which \(f\) splits contains every root \(\alpha_i\), and \(k(\alpha_1,\ldots,\alpha_n)\) is such a subfield.

    What is the splitting field of a polynomial $f \in k[x]$?

    Characterization of normal algebraic extensions

    An algebraic extension is normal exactly when every base-field embedding preserves the extension.Suggest a slogan

    Let \(k\) be a field, let \(L/k\) be an algebraic extension, and let \(\bar{k}\) be an algebraic closure of \(k\) containing \(L\). Then \(L/k\) is normal if and only if every \(k\)-embedding \(\sigma\colon L\to \bar{k}\) satisfies \(\sigma(L) = L\). In that case every \(k\)-embedding \(\sigma\colon L\to\bar k\) restricts to a \(k\)-automorphism of \(L\):

    Suppose \(L/k\) is normal and let \(\sigma\colon L\to\bar k\) be a \(k\)-embedding. For \(\alpha\in L\) with minimal polynomial \(m_\alpha\in k[x]\), \(\sigma(\alpha)\) is a root of \(m_\alpha\), and all roots of \(m_\alpha\) lie in \(L\), so \(\sigma(L)\subseteq L\). Moreover \(\sigma\) injects the finite set of roots of \(m_\alpha\) in \(L\) into itself, hence permutes it, so \(\alpha\) lies in \(\sigma(L)\); thus \(\sigma(L)=L\).

    Conversely, suppose every \(k\)-embedding \(L\to\bar k\) has image \(L\). Let \(f\in k[x]\) be irreducible with a root \(\alpha\in L\), and let \(\beta\in\bar k\) be any root of \(f\). There is a \(k\)-embedding \(k(\alpha)\to\bar k\) with \(\alpha\mapsto\beta\), and since \(L/k(\alpha)\) is algebraic and \(\bar k\) is algebraically closed, it extends to a \(k\)-embedding \(\sigma\colon L\to\bar k\). Then \(\beta=\sigma(\alpha)\in\sigma(L)=L\), so \(f\) splits in \(L\) and \(L/k\) is normal.

    Normal closure of a field extension

    Let \(K/k\) be an algebraic field extension. A normal closure of \(K/k\) is a field extension \(N/K\) such that \(N/k\) is normal and no proper subfield of \(N\) containing \(K\) is normal over \(k\).

    A normal closure exists: inside an algebraic closure \(\bar K\) of \(K\), take the subfield generated over \(k\) by all roots in \(\bar K\) of the minimal polynomials over \(k\) of the elements of \(K\). It is unique up to an isomorphism that restricts to the identity on \(K\).

    Characterization of finite normal extensions as splitting fields

    Finite normal extensions are exactly splitting fields of nonzero polynomials.Suggest a slogan

    Let \(k\) be a field and \(L/k\) a field extension. Then \(L/k\) is finite and normal if and only if \(L\) is the splitting field over \(k\) of some nonzero polynomial \(f\in k[x]\).

    Fix an algebraic closure \(\bar k\) containing \(L\).

    Suppose \(L/k\) is finite and normal, and write \(L=k(\beta_1,\ldots,\beta_r)\). Let \(f\) be the product of the minimal polynomials of the \(\beta_i\) over \(k\). By normality each factor splits in \(L\), and \(L\) is generated over \(k\) by roots of \(f\), so \(L\) is the splitting field of \(f\).

    Conversely, let \(L=k(\alpha_1,\ldots,\alpha_n)\), where \(\alpha_1,\ldots,\alpha_n\) are the roots of \(f\) in \(L\) and \(f\) splits in \(L\). Then \(L/k\) is finite. A \(k\)-embedding \(\sigma\colon L\to\bar k\) sends each \(\alpha_i\) to a root of \(f\), and \(f\) splits in \(L\), so \(\sigma\) permutes \(\theset{\alpha_1,\ldots,\alpha_n}\) and \(\sigma(L)=L\). By PR-OZYUC, \(L/k\) is normal.

    For a finite extension \(L = K(\alpha_1,\ldots,\alpha_r)\) inside an algebraic closure \(\overline K\), the normal closure of \(L/K\) is the splitting field over \(K\) of the product of the minimal polynomials of \(\alpha_1,\ldots,\alpha_r\); it is the smallest normal extension of \(K\) in \(\overline K\) containing \(L\). For example, the normal closure of \(\QQ(2^{1/4})/\QQ\) is \(\QQ(2^{1/4}, i)\), the splitting field of \(x^4-2\).

    (Normality is not transitive)

    In the tower \(\QQ\subseteq\QQ(\sqrt 2)\subseteq\QQ(2^{1/4})\), both steps have degree \(2\) and are therefore normal, and \(\QQ(2^{1/4})/\QQ\) is not normal: \(x^4-2\) is irreducible over \(\QQ\) and has the root \(2^{1/4}\) in \(\QQ(2^{1/4})\subseteq\RR\), but not its roots \(\pm i2^{1/4}\).

    For a finite Galois extension \(L/K\) and an intermediate field \(K\subseteq E\subseteq L\), \(E/K\) is normal if and only if \(\Gal(L/E)\) is a normal subgroup of \(\Gal(L/K)\).

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