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 over L. 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 $L/K$ be an algebraic extension (i.e., L is an algebraic extension of K), such that $L\subseteq \bar{K}$ (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:

• Every embedding of L in $\bar{K}$ over K induces an automorphism of L.
• L is the splitting field of a family of polynomials in $K[X]$.
• Every irreducible polynomial of $K[X]$ 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.
• 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.

Equivalent conditions for normality

Let $L/K$ 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 $S\subseteq K[x]$ of polynomials that each splits over L, such that if $K\subseteq F⊊L$ are fields, then S has a polynomial that does not split in F;
• All homomorphisms $L\to \overline{K}$ that fix all elements of K have the same image;
• The group of automorphisms, $\text{Aut}(L/K),$ of L that fix all elements of K, acts transitively on the set of homomorphisms $L\to \overline{K}$ that fix all elements 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 $\bar{\mathbb{Q}}$ of algebraic numbers is the algebraic closure of $\mathbb{Q},$ and thus it contains $\mathbb{Q}(\sqrt[3]{2}).$ Let $\omega$ be a primitive cubic root of unity. Then since, $$\mathbb{Q}(\sqrt[3]{2})=\{a+b\sqrt[3]{2}+c\sqrt[3]{4}\in \bar{\mathbb{Q}}|a,b,c\in \mathbb{Q}\}$$ the map $$\{\begin{array}{c}\sigma :\mathbb{Q}(\sqrt[3]{2})⟶\bar{\mathbb{Q}}\\ a+b\sqrt[3]{2}+c\sqrt[3]{4}⟼a+b\omega \sqrt[3]{2}+c{\omega }^2\sqrt[3]{4}\end{array}$$ is an embedding of $\mathbb{Q}(\sqrt[3]{2})$ in $\bar{\mathbb{Q}}$ whose restriction to $\mathbb{Q}$ is the identity. However, $\sigma$ 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 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