Normal extension

For other uses, see Normal closure.

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 in 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.

Definition

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]

• Every embedding of L in ${\displaystyle {\overline {K}}}$ over K induces an automorphism of L.
• L is the splitting field of a family of polynomials in ${\displaystyle K[X]}$.
• Every irreducible polynomial of ${\displaystyle K[X]}$ that has a root in L splits into linear factors in L.

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 (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

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

• The minimal polynomial over K of every element in L splits in L;
• There is a set ${\displaystyle S\subseteq K[x]}$ of polynomials that each splits over L, such that if ${\displaystyle K\subseteq F\subsetneq L}$ are fields, then S has a polynomial that does not split in F;
• All homomorphisms ${\displaystyle L\to {\bar {K}}}$ that fix all elements of K have the same image;
• The group of automorphisms, ${\displaystyle {\text{Aut}}(L/K),}$ of L that fix all elements of K, acts transitively on the set of homomorphisms ${\displaystyle L\to {\bar {K}}}$ that fix all elements of K.

Examples and counterexamples

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 ${\displaystyle \mathbb {Q} ({\sqrt[{3}]{2}}).}$

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 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.

See also

• Galois extension
• Normal basis

Citations

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. Lang 2002, p. 238, Theorem 3.4.

References

• Lang, Serge (2002), Algebra, Graduate Texts in Mathematics, vol. 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