Talk:Algebraically closed field

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~~Can link if and only if: ...ors]]. It can be shown that a field is algebraically closed if and only if it has no proper [[algebraic extension]], and this is somet...,~~ Done Paul August ☎ 16:37, Jun 19, 2005 (UTC)

Inward links

Additionally, there are some other articles which may be able to linked to this one (also known as "backlinks"):

~~In Divisor, can backlink algebraically closed: ...visor]]s. The concepts agree on nonsingular varieties over algebraically closed fields. Any Weil divisor is a locally finite [[linear comb...~~ done Paul August ☎ 16:38, Jun 19, 2005 (UTC)
In Projective line, can backlink algebraically closed: ...lar]] curve of [[genus (mathematics)|genus]] 0. If ''K'' is algebraically closed, it is the unique such curve over ''K'', up to isomorphism.... done Paul August ☎ 16:38, Jun 19, 2005 (UTC)
In Almost complex manifold, can backlink algebraically closed: ... of ''V'', :<math>V^C=V\otimes C</math> because ''C'' is algebraically closed, ''J'' is guaranteed to have eigenvalues which satisfy &lam... done Paul August ☎ 16:38, Jun 19, 2005 (UTC)
In Cubic surface, can backlink algebraically closed field: ...e of a smooth cubic surface. A smooth cubic surface over an algebraically closed field famously contains 27 lines. The arrangement of these lines ... done Paul August ☎ 16:38, Jun 19, 2005 (UTC)

Algebraic numbers are algebraically closed?

I don't see how the algebraic numbers are algebraically closed. The definition I know of the algebraic numbers is that they are the set of real solutions to polynomials over the integers. Therefore 1 is an algebraic number. However $x^{2}+1=0$ has no solution in the algebraic numbers.

Is my definition of algebraic number wrong?

—Preceding unsigned comment added by Luqui (talk • contribs) 08:59, 14 December 2005

Definitively: Algebraic numbers are all the complex(!) solutions of polynomials with rational(!) coefficients. —Preceding unsigned comment added by 193.6.218.9 (talk) 11:46, 7 November 2007 (UTC)[reply]

I did a bit of googling, and apparently some definitions of an algebraic nunber require it to be real, and some do not. I do not know which is more standard. - Jwwalker 23:30, 31 December 2005 (UTC)[reply]

this is probabely mixing two different things:

definition: let L be a ring, K be a subset of L, then l in L is algebraic over K when: there is a polynomial p with coefficients in K s.t. p(l)=0. in some cases K is required to be a subring or something not just a subset.
definition: let K be a field then K is algebraically closed when all polynomials with deg>=1 have at least one root (as defined in the article).

probabely the real algebraic numbers are: the Real numbers that are algebraic over Z.

the general are the algebraic closure of Z, which is C. --itaj 00:49, 14 November 2006 (UTC)[reply]

The algebraic closure of Z is not C. The algebraic numbers are the complex numbers that are algebraic over Q. These form a countable subfield of C. The real algebraic numbers are the algebraic numbers that lie in R. --Zundark 10:00, 14 November 2006 (UTC)[reply]

ahh, yes ofcourse, i'm sorry. the algebraic closure of R is C, and the closure of Z is a subfield of C. --itaj 16:08, 24 November 2006 (UTC)[reply]

I changed "algebraic number" to "(complex) algebraic number". Hopefully this does not create more confusion than it solves, but I certainly had the same reaction of "how can the (real) algebraic numbers be algebraicly closed?". The one book I looked up (Prime Obsession: Bernhard Riemann and the Greatest Unsolved Problem in Mathematics) does define algebraic numbers as complex, so if this whole concept of "real algebraic number" was just a misconception of mine, I'm OK with a revert. Kingdon 19:54, 22 March 2007 (UTC)[reply]

What field is this article in?

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)[reply]

I've changed it to algebra. --Zundark 16:56, 13 May 2007 (UTC)[reply]

Some suggestions for improvement

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)Plclark[reply]

Algebraically closed rings

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

"Relative algebraic closure" listed at Redirects for discussion

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. 1234qwer 1234qwer 4 19:19, 4 May 2022 (UTC)[reply]

The finite field of one element [0] is algebraically closed

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)[reply]

No, because there isn't a field with only one element: the definition of a field requires

0\neq 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)[reply]