Normal extension

In abstract algebra, an algebraic field extension L/K is said to be normal if every polynomial that is irreducible over K either has no root in L or splits into linear factors in L. Bourbaki calls such an extension a quasi-Galois extension.

Definition

The algebraic field extension L/K is normal (we also say that L is normal over K) if every irreducible polynomial over K that has at least one root in L splits over L. In other words, if α ∈ L, then all conjugates of α over K (i.e., all roots of the minimal polynomial of α over K) belong to L.

Equivalent properties

The normality of L/K is equivalent to either of the following properties. Let K^a be an algebraic closure of K containing L.

Every embedding σ of L in K^a that restricts to the identity on K, satisfies σ(L) = L (σ is an automorphism of L over K.)
Every irreducible polynomial in K[X] that has one root in L, has all of its roots in L, that is, it decomposes into linear factors in L[X]. (One says that the polynomial splits in L.)

If L is a finite extension of K that is separable (for example, this is automatically satisfied if K is finite or has characteristic zero) then the following property is also equivalent:

There exists an irreducible polynomial whose roots, together with the elements of K, generate L. (One says that L is the splitting field for the polynomial.)

Other properties

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 (i.e., L ⊃ E ⊃ K), then L is a normal extension of E.
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.

Examples and Counterexamples

For example, $\mathbb {Q} ({\sqrt {2}})$ is a normal extension of $\mathbb {Q}$ , since it is a splitting field of $x 2 - 2$ . On the other hand, $\mathbb {Q} ({\sqrt[{3}]{2}})$ is not a normal extension of $\mathbb {Q}$ since the irreducible polynomial $x 3 - 2$ has one root in it (namely, ${\sqrt[{3}]{2}}$ ), but not all of them (it does not have the non-real cubic roots of 2). Recall that the field ${\overline {\mathbb {Q} }}$ of algebraic numbers is the algebraic closure of $\mathbb {Q}$ , i.e., it contains $\mathbb {Q} ({\sqrt[{3}]{2}})$ . Since, $\mathbb {Q} ({\sqrt[{3}]{2}})=\{a+b{\sqrt[{3}]{2}}+c{\sqrt[{3}]{4}}\in {\overline {\mathbb {Q} }}\,|\,a,b,c\in \mathbb {Q} \}$ and, if ω is a primitive cubic root of unity, then the map

{\begin{array}{rccc}\sigma :&\mathbb {Q} ({\sqrt[{3}]{2}})&\longrightarrow &{\overline {\mathbb {Q} }}\\&a+b{\sqrt[{3}]{2}}+c{\sqrt[{3}]{4}}&\mapsto &a+b\omega {\sqrt[{3}]{2}}+c\omega ^{2}{\sqrt[{3}]{4}}\end{array}}

is an embedding of $\mathbb {Q} ({\sqrt[{3}]{2}})$ in ${\overline {\mathbb {Q} }}$ whose restriction to $\mathbb {Q}$ is the identity. However, σ is not an automorphism of $\mathbb {Q} ({\sqrt[{3}]{2}})$ .

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

Normal closure

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 which is minimal, i.e., the only subfield of M which contains L and which 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.

References

Lang, Serge (2002), Algebra, Graduate Texts in Mathematics, 211 (Revised third ed.), New York: Springer-Verlag, ISBN 978-0-387-95385-4, MR 1878556
Jacobson, Nathan (1989), Basic Algebra II (2nd ed.), W. H. Freeman, ISBN 0-7167-1933-9, MR 1009787

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