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..., a '''splitting field''' for a polynomial ''f'' over a field ''F'' is a [[field extension]] ''E''/''F'' with the properties that ''f'' splits completely over ''E'',

...quadratic field''' is a [[Field theory (mathematics)|field]] which is an [[field extension|extension]] of its [[prime field]] of degree two.

...''n''-th root of unity, then the ''n''-th cyclotomic field ''F'' is the [[field extension]] <math>\mathbf{Q}(\zeta)</math>.

{{r|Field extension}}

{{r|Field extension}}

{{r|Field extension}}

...The degree of the minimal polynomial of α is equal to the degree of the [[field extension]] '''Q'''(α)/'''Q'''. ...tisfies. The polynonmial ring ''F''[α] is then a field, and is the simple field extension ''F''(α). This field is a finite dimension al vector space over ''F'', on

Key concepts are [[Field extension|field extensions]] and [[Group theory|groups]], which should be thoroughly ...a subfield and also all the roots of ''f''. This field is known as the [[field extension|extension]] of ''K'' by the roots of ''f'', or the ''[[splitting field]]''

{{r|Field extension}}

*[[Field extension]]

{{r|Field extension}}

In Δ is not a square in ''F'' then the [[field extension]] <math>F(\sqrt\Delta)</math> is [[quadratic field|quadratic]] over ''F'':

Let us take the case that ''G'' is the [[Galois group]] of a [[field extension]] ''L''/''K''. A factor system in H²(''G'',''L''^*) g

An ''algebraic number field'' ''K'' is a finite degree [[field extension]] of the [[field (mathematics)|field]] '''Q''' of [[rational number]]s. Th

...e one. Since <math>\scriptstyle\mathbb{C} = \mathbb{R}[i]</math>, any such field extension also extends <math>\scriptstyle\mathbb{R}</math>. Now, any <math>\scriptsty

* [[Algebraically independent set]]s in a [[field extension]];

Let ''K'' be an [[algebraic number field]], a finite [[field extension|extension]] of '''Q''', and ''E'' an elliptic curve defined over ''K''. Th

...analysis, we could next show that <math>\mathbb{C}</math> has no finite [[field extension|extension]] and must therefore be [[algebraic closure|algebraically closed]

...analysis, we could next show that <math>\mathbb{C}</math> has no finite [[field extension|extension]] and must therefore be [[algebraic closure|algebraically closed]