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| In [[mathematics]], the '''fundamental theorem of [[Galois theory]]''' is a result that describes the structure of certain types of [[field extension]]s.
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| In its most basic form, the theorem asserts that given a field extension ''E''/''F'' which is [[finite extension|finite]] and [[Galois extension|Galois]], there is a one-to-one [[correspondence (mathematics)|correspondence]] between its [[intermediate field]]s and [[subgroup]]s of its [[Galois group]]. ([[Intermediate field]]s are fields ''K'' satisfying ''F'' ⊆ ''K'' ⊆ ''E''; they are also called ''subextensions'' of ''E''/''F''.)
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| ==Proof==
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| The proof of the fundamental theorem is not trivial. The crux in the usual treatment is a rather delicate result of [[Emil Artin]] which allows one to control the dimension of the intermediate field fixed by a given group of automorphisms. The automorphisms of a Galois extension ''K''/''F'' are linearly independent as functions over the field ''K''. The proof of this fact follows from a more general notion, namely, the linear independence of [[Character (mathematics)|characters]].
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| There is also a fairly simple proof using the [[primitive element theorem]]. This proof seems to be ignored by most modern treatments, possibly because it requires a separate (but easier) proof in the case of finite fields.<ref>See {{cite book |last=Marcus |first=Daniel |title=Number Fields |others=Appendix 2 |location=New York |publisher=Springer-Verlag |year=1977 |isbn=0-387-90279-1 }}</ref>
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| In terms of its abstract structure, there is a [[Galois connection]]; most of its properties are fairly formal, but the actual isomorphism of the [[poset]]s requires some work.
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| ==Explicit description of the correspondence==
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| For finite extensions, the correspondence can be described explicitly as follows.
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| * For any subgroup ''H'' of Gal(''E''/''F''), the corresponding field, usually denoted ''E<sup>H</sup>'', is the set of those elements of ''E'' which are fixed by every automorphism in ''H''.
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| * For any intermediate field ''K'' of ''E''/''F'', the corresponding subgroup is just Aut(''E''/''K''), that is, the set of those automorphisms in Gal(''E''/''F'') which fix every element of ''K''.
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| For example, the topmost field ''E'' corresponds to the trivial subgroup of Gal(''E''/''F''), and the base field ''F'' corresponds to the whole group Gal(''E''/''F'').
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| ==Properties of the correspondence==
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| The correspondence has the following useful properties.
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| * It is ''inclusion-reversing''. The inclusion of subgroups ''H<sub>1</sub> ⊆ H<sub>2</sub>'' holds if and only if the inclusion of fields ''E<sup>H<sub>1</sub></sup>'' ⊇ ''E<sup>H<sub>2</sub></sup>'' holds.
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| * Degrees of extensions are related to orders of groups, in a manner consistent with the inclusion-reversing property. Specifically, if ''H'' is a subgroup of Gal(''E''/''F''), then |''H''| = [''E'':''E<sup>H</sup>''] and |Gal(''E''/''F'')/''H''| = [''E<sup>H</sup>'':''F''].
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| * The field ''E<sup>H</sup>'' is a [[normal extension]] of ''F'' (or, equivalently, Galois extension, since any subextension of a separable extension is separable) if and only if ''H'' is a [[normal subgroup]] of Gal(''E''/''F''). In this case, the restriction of the elements of Gal(''E''/''F'') to ''E<sup>H</sup>'' induces an [[group isomorphism|isomorphism]] between Gal(''E<sup>H</sup>''/''F'') and the [[quotient group]] Gal(''E''/''F'')/''H''.
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| == Example ==
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| [[File:Lattice diagram of Q adjoin the positive square roots of 2 and 3, its subfields, and Galois groups.svg|thumb|600px|[[Lattice of subgroups]] and subfields]]
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| Consider the field ''K'' = '''Q'''(√2, √3) = '''Q'''(√2)(√3). Since ''K'' is first determined by adjoining √2, then √3, each element of ''K'' can be written as:
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| :<math>(a+b\sqrt{2})+(c+d\sqrt{2})\sqrt{3},</math>
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| where ''a'', ''b'', ''c'', ''d'' are rational numbers. Its Galois group ''G'' = Gal(''K''/'''Q''') can be determined by examining the automorphisms of ''K'' which fix ''a''. Each such automorphism must send √2 to either √2 or −√2, and must send √3 to either √3 or −√3 since the permutations in a Galois group can only permute the roots of an irreducible polynomial. Suppose that ''f'' exchanges √2 and −√2, so
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| :<math>f\left((a+b\sqrt{2})+(c+d\sqrt{2})\sqrt{3}\right)=(a-b\sqrt{2})+(c-d\sqrt{2})\sqrt{3}=a-b\sqrt{2}+c\sqrt{3}-d\sqrt{6},</math>
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| and ''g'' exchanges √3 and −√3, so
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| :<math>g\left((a+b\sqrt{2})+(c+d\sqrt{2})\sqrt{3}\right)=(a+b\sqrt{2})-(c+d\sqrt{2})\sqrt{3}=a+b\sqrt{2}-c\sqrt{3}-d\sqrt{6}.</math>
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| These are clearly automorphisms of ''K''. There is also the identity automorphism ''e'' which does not change anything, and the composition of ''f'' and ''g'' which changes the signs on ''both'' radicals:
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| :<math>(fg)\left((a+b\sqrt{2})+(c+d\sqrt{2})\sqrt{3}\right)=(a-b\sqrt{2})-(c-d\sqrt{2})\sqrt{3}=a-b\sqrt{2}-c\sqrt{3}+d\sqrt{6}.</math>
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| Therefore
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| :<math>G = \left\{1, f, g, fg\right\},</math>
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| and ''G'' is isomorphic to the [[Klein four-group]]. It has five subgroups, each of which correspond via the theorem to a subfield of ''K''.
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| * The trivial subgroup (containing only the identity element) corresponds to all of ''K''.
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| * The entire group ''G'' corresponds to the base field '''Q'''.
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| * The two-element subgroup {1, ''f'' } corresponds to the subfield '''Q'''(√3), since ''f'' fixes √3.
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| * The two-element subgroup {1, ''g''} corresponds to the subfield '''Q'''(√2), again since ''g'' fixes √2.
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| * The two-element subgroup {1, ''fg''} corresponds to the subfield '''Q'''(√6), since ''fg'' fixes √6.
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| == Example ==
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| [[File:Lattice diagram of Q adjoin a cube root of 2 and a primitive cube root of 1, its subfields, and Galois groups.svg|thumb|600px|[[Lattice of subgroups]] and subfields]]
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| The following is the simplest case where the Galois group is not abelian.
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| Consider the [[splitting field]] ''K'' of the polynomial ''x''<sup>3</sup>−2 over '''Q'''; that is, ''K'' = '''Q''' (θ, ω),
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| <!-- aargggghhhh can't decide whether to assume that ''K'' is a subfield of '''C''' or whether to do it the pure algebraic way :-) - Dmharvey -->
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| where θ is a cube root of 2, and ω is a cube root of 1 (but not 1 itself). For example, if we imagine ''K'' to be inside the field of complex numbers, we may take θ to be the real cube root of 2, and ω to be
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| :<math>\omega = \frac{-1}2 + i\frac{\sqrt3}2.</math> | |
| It can be shown that the Galois group ''G'' = Gal(''K''/'''Q''') has six elements, and is isomorphic to the group of permutations of three objects. It is generated by (for example) two automorphisms, say ''f'' and ''g'', which are determined by their effect on θ and ω,
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| :<math>f(\theta) = \omega \theta, \quad f(\omega) = \omega,</math>
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| :<math>g(\theta) = \theta, \quad g(\omega) = \omega^2,</math>
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| <!-- hmmm might be nice to explain a little more clearly the effect of these automorphisms - Dmharvey -->
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| and then
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| :<math>G = \left\{ 1, f, f^2, g, gf, gf^2 \right\}.</math>
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| The subgroups of ''G'' and corresponding subfields are as follows:
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| * As usual, the entire group ''G'' corresponds to the base field '''Q''', and the trivial group {1} corresponds to the whole field ''K''.
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| * There is a unique subgroup of order 3, namely {1, ''f'', ''f'' <sup>2</sup>}. The corresponding subfield is '''Q'''(ω), which has degree two over '''Q''' (the minimal polynomial of ω is ''x''<sup>2</sup> + ''x'' + 1), corresponding to the fact that the subgroup has [[Index of a subgroup|index]] two in ''G''. Also, this subgroup is normal, corresponding to the fact that the subfield is normal over '''Q'''.
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| * There are three subgroups of order 2, namely {1, ''g''}, {1, ''gf'' } and {1, ''gf'' <sup>2</sup>}, corresponding respectively to the three subfields '''Q'''(θ), '''Q'''(ωθ), '''Q'''(ω<sup>2</sup>θ). These subfields have degree three over '''Q''', again corresponding to the subgroups having index 3 in '''G'''. Note that the subgroups are ''not'' normal in ''G'', and this corresponds to the fact that the subfields are ''not'' Galois over '''Q'''. For example, '''Q'''(θ) contains only a single root of the polynomial ''x''<sup>3</sup>−2, so it cannot be normal over '''Q'''.
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| <!--
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| would be nice to have a diagram of subgroups and subfields somewhere here!!! - and for the previous example too! - Dmharvey -->
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| ==Applications==
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| The theorem converts the difficult-sounding problem of classifying the intermediate fields of ''E''/''F'' into the more tractable problem of listing the subgroups of a certain [[finite group]].
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| For example, to prove that the [[general quintic equation]] is not [[solvable by radicals]] (see [[Abel–Ruffini theorem]]), one first restates the problem in terms of [[radical extension]]s (extensions of the form ''F''(α) where α is an ''n''-th root of some element of ''F''), and then uses the fundamental theorem to convert this statement into a problem about groups that can then be attacked directly.
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| Theories such as [[Kummer theory]] and [[class field theory]] are predicated on the fundamental theorem.
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| ==Infinite case==
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| There is also a version of the fundamental theorem that applies to infinite [[algebraic extension]]s, which are [[normal extension|normal]] and [[separable extension|separable]]. It involves defining a certain [[topological structure]], the [[Krull topology]], on the Galois group; only subgroups that are also [[closed set]]s are relevant in the correspondence.
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| ==References==
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| <references/> | |
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| {{Fundamental theorems}}
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| {{DEFAULTSORT:Fundamental Theorem Of Galois Theory}}
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| [[Category:Field theory]]
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| [[Category:Group theory]]
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| [[Category:Galois theory]]
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| [[Category:Theorems in algebra]]
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| [[Category:Fundamental theorems|Galois theory]]
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While in the drop of 1999, Tour: Luke Bryan And which in fact had an amazing list of , including City. “It’s almost like you are getting a authorization to go to the next level, affirms all those designers that had been a part of the Concert toursmore than in a bigger measure of performers.” It wrapped as the best excursions in its 10-season background.
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