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] These are one of the conditions for algebraic extensions to be a Galois extension. Bourbaki calls such an extension a quasi-Galois extension.

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.

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]

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.

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.

• Galois extension
• Normal basis