|WikiProject Mathematics||(Rated C-class, High-importance)|
I think I remember from math class that it's unknown whether every finite-degree extension is a Galois extension. If that's the case, it should be mentioned here. That always seemed like an amazingly basic open question, if it is one.
- Q(³√2)/Q has finite degree, is normal, but it is not separable, so it is not a Galois extension. --ReiVaX 17:50, 15 July 2005 (UTC)
- Umm, it's the other way around. Every algebraic extension of Q is separable; Q(³√2)/Q is not normal. -- EJ 20:27, 9 January 2006 (UTC)
The question you half-remembered from math class is this: is every finite group the Galois group of some Galois extension of Q? The answer is indeed unknown. This is the inverse Galois problem. AxelBoldt 07:15, 23 March 2006 (UTC)
merge with algebraic extension
- I agree that the information on this page and on algebraic extension could be redistributed in a better way. I would also like to see an example of a galois extension which is not abelian, if a simple example exists. Owen Jones 10:42, 14 January 2006 (UTC)
- Just found this: http://www.emba.uvm.edu/~sands/papers/stpopsub.pdf. See p. 7 where the author gives an example of an non-abelian galois extension. (I didn't check, if he is right ;-)) mathaxiom 23:38, 13 March 2006 (UTC)
Just a sidenote, perhaps soemone can clean up the symbols in this page? It's kind of distracting to read...
Restructering of the article
I am currently restructering the article in order to collect the important definitions, which were scattered in the article, in the Definition section. MathMartin 13:32, 17 April 2006 (UTC)
- Generally looks good, but why did you remove () multiplicitivity of degrees? Dmharvey 18:49, 17 April 2006 (UTC)
- Actually, I wonder if it wouldn't be better to move degree theorem to something like degree of a field extension, and to flesh out that article a bit, including a definition of degree (with the infinite case as well), and perhaps a proof of the multiplicitivity result, and some examples. Also, I don't remember "degree theorem" being a standard term for this result. What do you think? Dmharvey 20:07, 17 April 2006 (UTC)
- Sounds good, I like the new title. But I cannot help you fleshing out the article as I only know about the finite case. MathMartin 21:36, 17 April 2006 (UTC)
- OK, well that was probably more than I intended to write in one go :-) But now we have some overlap, which is not necessarily a bad thing, but perhaps we could focus the examples a bit better. Dmharvey 01:20, 18 April 2006 (UTC)
Is a field extension a special tensor product?
Of course not every tensor product is a field extension, but what about the converse? The multiplication of dimensions makes me suspect this, but it's been awhile since I "studied" this stuff.(but i did ask a teacher if it was a tensor product and he said yes, on the other hand, he didn't spend any time pondering his yes. Also, a different teacher denied that it was a tensor product.) If I'm right, some words on that and a link to tensor product would be valuable.Rich 09:03, 27 November 2006 (UTC)
Error? [K(t):K] = N_0 ?
It is stated that the degree of K(t) over K, for any field K, has countable degree (N_0). I am certainly no expert, but this seems to be false. Any transcendental extension (i.e. K(t)) must have degree at last the cardinality of K, since all the inverses of (1 - x t), for x in K, are linearly independent over K. What do people think? —The preceding unsigned comment was added by 18.104.22.168 (talk) 00:19, 14 December 2006 (UTC).
- The statement was wrong, so I've removed it from the article. --Zundark 08:44, 14 December 2006 (UTC)
[R : Q]
Right now the article says that it's . However, considered as a vector space, a field extension of degree over a field is isomorphic to , which has cardinality , which seems to imply that the degree is actually , as , while . Is this correct? —Preceding unsigned comment added by Schizobullet (talk • contribs) 02:26, 21 January 2009 (UTC)
- Close. The vector space is |K|^(n), the direct sum of n copies of K, not the direct product. In this case Q^(aleph0) has cardinality aleph0, the cardinality of all finite sequences (of unbounded length) of rational numbers. The direct product Q^aleph0 has cardinality c, the cardinality of all countable sequence of rational numbers. So [R:Q] really is c. I think Hamel basis is a relevant search term, though I forget if wikipedia has much on it. JackSchmidt (talk) 04:02, 21 January 2009 (UTC)
I think that having the definition of a subfield contained in this article on field extensions is a little odd. Moreover, the search for subfield (mathematics) directs to this field extension page. The definition of subfield should be in the field article and then, given proper links, knowledge of subfields should be take a read. 22.214.171.124 (talk) 14:24, 29 May 2009 (UTC)
Bases containing 1
Given a finite extension L/K, can a basis of L over K always be chosen to contain 1? I think so, because one can simply choose any element x from the basis and multiply all of the elements from the basis by x-1. GeoffreyT2000 (talk) 21:39, 24 May 2015 (UTC)