Normal extension

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

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

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

Let L be an extension of a field K. Then:

For example, ${\displaystyle \mathbb {Q} ({\sqrt {2}})}$ is a normal extension of ${\displaystyle \mathbb {Q} }$, since it is a splitting field of x² − 2. On the other hand, ${\displaystyle \mathbb {Q} ({\sqrt[{3}]{2}})}$ is not a normal extension of ${\displaystyle \mathbb {Q} }$ since the irreducible polynomial x³ − 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 ${\displaystyle \mathbb {Q} }$, i.e., it contains ${\displaystyle \mathbb {Q} ({\sqrt[{3}]{2}})}$. Since, ${\displaystyle \mathbb {Q} ({\sqrt[{3}]{2}})=\{a+b{\sqrt[{3}]{2}}+c{\sqrt[{3}]{4}}\in {\overline {\mathbb {Q} }}\,|\,a,b,c\in \mathbb {Q} \}}$ and, if ω is a primitive cubic root of unity, then the map

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