Talk:Algebraically closed field

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thread_title: Untitled

message: Is this really supposed to be number theory rather than algebra?  I'm relucant to just change it in case I missed something subtle, but at first glance Algebraically closed fields is very much an algebraic topic. Kingdon 16:24, 13 May 2007 (UTC)
replies: text: I've changed it to algebra. --Zundark 16:56, 13 May 2007 (UTC)
thread_title:  What field is this article in? 

message: This is a good start.  Here are some further ideas:
1) Spell out more explicitly what it means for a polynomial to have a root: could 
it hurt?
2) The fact that x^2+1 = 0 has no root in the real numbers shows 
moreover that no subfield of the real numbers is algebraially 
closed.  This would be a good opportunity to link to formally real fields, 
in which -1 is not a sum of squares -- they cannot be algebraically closed either.  
3) Other nonalgebraically closed fields: pseudo-real closed fields, p-adic fields, 
formally p-adic fields, Laurent series fields, pseudo-p-adically closed fields, fields which are finitely generated over their prime subfields or over a field of rational functions in any number of indeterminates over any field, any field which is elementarily equivalent to a non-algebraically closed field, non-perfect fields.
4) Other algebraically closed fields: for any field k, the field of all elements 
which are algebraic over k is algebraically closed.  (With k = Q, this recovers 
the algebraic numbers.)
5) Another characterization of algebraically closed fields: fields which are perfect and have trivial absolute Galois group.  
6) References for the equivalent properties of algebraic closure?
7) Under "other properties", mention the maximal solvable extension of Q as a field 
which satisfies the given conditions.  
8) Define algebraic closure before asserting that every field has one.  (A better 
definition is an algebraic field extension which is algebraically closed, as is given in the link.)   Give examples of algebraic closure of Q and R.  
9) The statement "every field has a unique algebraic closure" is subtly inaccurate.  
There are many algebraic closures of a given field.  What is true is that any 
two of them are isomorphic as F-algebras.  However, there are many isomorphisms, 
one for each element of the absolute Galois group of F, which if F was neither 
algebraically closed nor real closed to begin with, is an uncountably infinite group.
68.154.96.118 11:33, 17 September 2007 (UTC)
thread_title:  Some suggestions for improvement 

message: Is it worth mentioning, or making another article, about more general algebraically closed rings (possibly noncommutative)?  Daviddwd (talk) 13:37, 2 September 2014 (UTC)
thread_title:  Algebraically closed rings 

message: An editor has identified a potential problem with the redirect :Relative algebraic closure and has thus listed it for discussion. This discussion will occur at Wikipedia:Redirects for discussion/Log/2022 May 4#Relative algebraic closure until a consensus is reached, and readers of this page are welcome to contribute to the discussion.  1234qwer1234qwer4 19:19, 4 May 2022 (UTC)
thread_title:  "Relative algebraic closure" listed at [[Wikipedia:Redirects for discussion|Redirects for discussion]] 

message: The field F_1 has the unique property that 1=0.
Trivially, any polynomial in F_1 has a root in F_1. This contradicts the statement that no finite field is algebraically closed.
This breaks the argument, as P(x)+1 = P(x) + 0 = P(X).
Should this be added in the text (for correctness' sake)? CanaDavid1 (talk) 06:06, 31 May 2022 (UTC)
replies: text: No, because there isn't a field with only one element: the definition of a field requires 0\ne 1. (See, however, field with one element - but such a thing isn't actually a field with only one element.) --Zundark (talk) 08:06, 31 May 2022 (UTC)
thread_title:  The finite field of one element [0] is algebraically closed