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Normal extension - Wikipedia

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In abstract algebra, a normal extension is an algebraic field extension L/K for which every irreducible polynomial over K that has a root in L splits into linear factors over L.[1][2] This is one of the conditions for an algebraic extension to be a Galois extension. Bourbaki calls such an extension a quasi-Galois extension. For finite extensions, a normal extension is identical to a splitting field.

Let {\displaystyle L/K} be an algebraic extension (i.e., L is an algebraic extension of K), such that {\displaystyle L\subseteq {\overline {K}}} (i.e., L is contained in an algebraic closure of K). Then the following conditions, any of which can be regarded as a definition of normal extension, are equivalent:[3]

Let L be an extension of a field K. Then:

  • If L is a normal extension of K and if E is an intermediate extension (that is, L ⊇ E ⊇ K), then L is a normal extension of E.[4]
  • If E and F are normal extensions of K contained in L, then the compositum EF and E ∩ F are also normal extensions of K.[4]

Equivalent conditions for normality

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Let {\displaystyle L/K} be algebraic. The field L is a normal extension if and only if any of the equivalent conditions below hold.

Examples and counterexamples

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For example, {\displaystyle \mathbb {Q} ({\sqrt {2}})} is a normal extension of {\displaystyle \mathbb {Q} ,} since it is a splitting field of {\displaystyle x^{2}-2.} On the other hand, {\displaystyle \mathbb {Q} ({\sqrt[{3}]{2}})} is not a normal extension of {\displaystyle \mathbb {Q} } since the irreducible polynomial {\displaystyle x^{3}-2} has one root in it (namely, {\displaystyle {\sqrt[{3}]{2}}}), but not all of them (it does not have the non-real cubic roots of 2). Recall that the field {\displaystyle {\overline {\mathbb {Q} }}} of algebraic numbers is the algebraic closure of {\displaystyle \mathbb {Q} ,} and thus it contains {\displaystyle \mathbb {Q} ({\sqrt[{3}]{2}}).} Let {\displaystyle \omega } be a primitive cubic root of unity. Then since, {\displaystyle \mathbb {Q} ({\sqrt[{3}]{2}})=\left.\left\{a+b{\sqrt[{3}]{2}}+c{\sqrt[{3}]{4}}\in {\overline {\mathbb {Q} }}\,\,\right|\,\,a,b,c\in \mathbb {Q} \right\}} the map {\displaystyle {\begin{cases}\sigma :\mathbb {Q} ({\sqrt[{3}]{2}})\longrightarrow {\overline {\mathbb {Q} }}\\a+b{\sqrt[{3}]{2}}+c{\sqrt[{3}]{4}}\longmapsto a+b\omega {\sqrt[{3}]{2}}+c\omega ^{2}{\sqrt[{3}]{4}}\end{cases}}} is an embedding of {\displaystyle \mathbb {Q} ({\sqrt[{3}]{2}})} in {\displaystyle {\overline {\mathbb {Q} }}} whose restriction to {\displaystyle \mathbb {Q} } is the identity. However, {\displaystyle \sigma } is not an automorphism of {\displaystyle \mathbb {Q} ({\sqrt[{3}]{2}}).}

For any prime {\displaystyle p,} the extension {\displaystyle \mathbb {Q} ({\sqrt[{p}]{2}},\zeta _{p})} is normal of degree {\displaystyle p(p-1).} It is a splitting field of {\displaystyle x^{p}-2.} Here {\displaystyle \zeta _{p}} denotes any {\displaystyle p}th primitive root of unity. The field {\displaystyle \mathbb {Q} ({\sqrt[{3}]{2}},\zeta _{3})} is the normal closure (see below) of {\displaystyle \mathbb {Q} ({\sqrt[{3}]{2}}).}

If K is a field and L is an algebraic extension of K, then there is some algebraic extension M of L such that M is a normal extension of K. Furthermore, up to isomorphism there is only one such extension that is minimal, that is, the only subfield of M that contains L and that is a normal extension of K is M itself. This extension is called the normal closure of the extension L of K.

If L is a finite extension of K, then its normal closure is also a finite extension.

  1. ^ Lang 2002, p. 237, Theorem 3.3, NOR 3.
  2. ^ Jacobson 1989, p. 489, Section 8.7.
  3. ^ Lang 2002, p. 237, Theorem 3.3.
  4. ^ a b Lang 2002, p. 238, Theorem 3.4.