normal extension

In abstract algebra, a normal extension is an algebraic field extension ''L''/''K'' for which every irreducible polynomial over ''K'' which has a root in ''L'', splits into linear factors in ''L''. These are one of the conditions for algebraic extensions to be a Galois extension. Bourbaki calls such an extension a quasi-Galois extension.

Definition

Let ''$L/K$'' be an algebraic extension (i.e. ''L'' is an algebraic extension of ''K''), such that $L\subseteq \overline$ (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 $\overline$ induces an automorphism of ''L''.
* ''L'' is the splitting field of a family of polynomials in K\left \right/math>.
* Every irreducible polynomial of K\left \right/math> which 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 /math> of polynomials that simultaneously split over ''L'', such that if $K\subseteq F\subsetneq L$ are fields, then ''S'' has a polynomial that does not split in ''F'';
* All homomorphisms $L \to \bar$ have the same image;
* The group of automorphisms, $\text(L/K),$ of ''L'' which fixes elements of ''K'', acts transitively on the set of homomorphisms $L \to \bar.$

Examples and counterexamples

For example, $\Q(\sqrt)$ is a normal extension of $\Q,$ since it is a splitting field of $x^2-2.$ On the other hand, \Q(\sqrt is not a normal extension of $\Q$ since the irreducible polynomial $x^3-2$ has one root in it (namely, \sqrt /math>), but not all of them (it does not have the non-real cubic roots of 2). Recall that the field $\overline$ of algebraic numbers is the algebraic closure of $\Q,$ that is, it contains \Q(\sqrt . Since,
\Q (\sqrt =\left. \left \
and, if $\omega$ is a primitive cubic root of unity, then the map
\begin \sigma:\Q (\sqrt \longrightarrow\overline\\ a+b\sqrt c\sqrt longmapsto a+b\omega\sqrt c\omega^2\sqrt end
is an embedding of \Q(\sqrt in $\overline$ whose restriction to $\Q$ is the identity. However, $\sigma$ is not an automorphism of \Q (\sqrt .

For any prime $p,$ the extension \Q (\sqrt \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 \Q (\sqrt \zeta_3) is the normal closure (see below) of \Q (\sqrt .

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, that is, 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.

See also

* Galois extension
* Normal basis

Citations

References

*
* {{citation , last = Jacobson , first = Nathan , author-link = Nathan Jacobson , title = Basic Algebra II, edition = 2nd , year = 1989 , publisher = W. H. Freeman , isbn = 0-7167-1933-9 , mr = 1009787

Field extensions