Jump to content
Main menu
Main menu
move to sidebar
hide
Navigation
Main page
Recent changes
Random page
Help about MediaWiki
Special pages
Niidae Wiki
Search
Search
Appearance
Create account
Log in
Personal tools
Create account
Log in
Pages for logged out editors
learn more
Contributions
Talk
Editing
Galois theory
(section)
Page
Discussion
English
Read
Edit
View history
Tools
Tools
move to sidebar
hide
Actions
Read
Edit
View history
General
What links here
Related changes
Page information
Appearance
move to sidebar
hide
Warning:
You are not logged in. Your IP address will be publicly visible if you make any edits. If you
log in
or
create an account
, your edits will be attributed to your username, along with other benefits.
Anti-spam check. Do
not
fill this in!
==Solvable groups and solution by radicals== The notion of a [[solvable group]] in [[group theory]] allows one to determine whether a polynomial is solvable in radicals, depending on whether its Galois group has the property of solvability. In essence, each field extension {{math|''L''/''K''}} corresponds to a [[factor group]] in a [[composition series]] of the Galois group. If a factor group in the composition series is [[cyclic group|cyclic]] of order {{math|''n''}}, and if in the corresponding field extension {{math|''L''/''K''}} the field {{math|''K''}} already contains a [[Root of unity|primitive {{math|''n''}}th root of unity]], then it is a radical extension and the elements of {{math|''L''}} can then be expressed using the {{math|''n''}}th root of some element of {{math|''K''}}. If all the factor groups in its composition series are cyclic, the Galois group is called ''solvable'', and all of the elements of the corresponding field can be found by repeatedly taking roots, products, and sums of elements from the base field (usually {{math|'''Q'''}}). One of the great triumphs of Galois Theory was the proof that for every {{math|''n'' > 4}}, there exist polynomials of degree {{math|''n''}} which are not solvable by radicals (this was proven independently, using a similar method, by [[Niels Henrik Abel]] a few years before, and is the [[Abel–Ruffini theorem]]), and a systematic way for testing whether a specific polynomial is solvable by radicals. The Abel–Ruffini theorem results from the fact that for {{math|''n'' > 4}} the [[symmetric group]] {{math|''S''<sub>''n''</sub>}} contains a [[simple group|simple]], noncyclic, [[normal subgroup]], namely the [[alternating group]] {{math|''A''<sub>''n''</sub>}}. ==={{anchor|Abel-Ruffini theorem}}A non-solvable quintic example=== [[image:Non solvable quintic.svg|right|thumb|For the polynomial {{math|''f''(''x'') {{=}} ''x''<sup>5</sup> − ''x'' − 1}}, the lone real root {{math|''x'' {{=}} 1.1673...}} is algebraic, but not expressible in terms of radicals. The other four roots are [[complex numbers]].]] [[Bartel Leendert van der Waerden|Van der Waerden]]<ref>van der Waerden, Modern Algebra (1949 English edn.), Vol. 1, Section 61, p.191<!-- p. 191 isn't in the preview, but this instance might be a better ref. {{cite book |first=B.L. |last=van der Waerden |first2=Emil |last2=Artin |first3=Emmy |last3=Noether |title=Algebra: Based in Part on Lectures by E. Artin and E. Noether. ... |url=https://books.google.com/books?id=XDN8yR8R1OUC&pg=PP1 |date=2003 |publisher=Springer |isbn=978-0-387-40624-4 |volume=1 |origyear=1970}} --></ref> cites the polynomial {{math|''f''(''x'') {{=}} ''x''<sup>5</sup> − ''x'' − 1}}. By the [[rational root theorem]], this has no rational zeroes. Neither does it have linear factors modulo 2 or 3. The Galois group of {{math|''f''(''x'')}} modulo 2 is cyclic of order 6, because {{math|''f''(''x'')}} modulo 2 factors into polynomials of orders 2 and 3, {{math|(''x''<sup>2</sup> + ''x'' + 1)(''x''<sup>3</sup> + ''x''<sup>2</sup> + 1)}}. {{math|''f''(''x'')}} modulo 3 has no linear or quadratic factor, and hence is irreducible. Thus its modulo 3 Galois group contains an element of order 5. It is known<ref>{{cite book |first=V.V. |last=Prasolov |chapter=5 Galois Theory Theorem 5.4.5(a) |doi=10.1007/978-3-642-03980-5_5 |title=Polynomials |publisher=Springer |series=Algorithms and Computation in Mathematics |volume=11 |year=2004 |isbn=978-3-642-03979-9 |pages=181–218 }}</ref> that a Galois group modulo a prime is isomorphic to a subgroup of the Galois group over the rationals. A permutation group on 5 objects with elements of orders 6 and 5 must be the symmetric group {{math|''S''<sub>5</sub>}}, which is therefore the Galois group of {{math|''f''(''x'')}}. This is one of the simplest examples of a non-solvable quintic polynomial. According to [[Serge Lang]], [[Emil Artin]] was fond of this example.<ref>{{cite book |title=Algebraic Number Theory|volume=110|series=Graduate Texts in Mathematics|first=Serge|last=Lang|author-link=Serge Lang|publisher=Springer|year=1994|isbn=9780387942254|page=121|url=https://books.google.com/books?id=u5eGtA0YalgC&pg=PA}}</ref>
Summary:
Please note that all contributions to Niidae Wiki may be edited, altered, or removed by other contributors. If you do not want your writing to be edited mercilessly, then do not submit it here.
You are also promising us that you wrote this yourself, or copied it from a public domain or similar free resource (see
Encyclopedia:Copyrights
for details).
Do not submit copyrighted work without permission!
Cancel
Editing help
(opens in new window)
Search
Search
Editing
Galois theory
(section)
Add topic