Talk:Algebraically closed field

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[edit] Outward links

• 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)

[edit] 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)

[edit] 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 x² + 1 = 0 has no solution in the algebraic numbers.

Is my definition of algebraic number wrong?

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)

this is probabely mixing two different things:

1. 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.
2. 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)

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)

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)

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)