Timeline for On algebraic field extensions
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 18, 2009 at 11:29 | vote | accept | Nick Gill | ||
| Nov 17, 2009 at 1:18 | answer | added | Greg Kuperberg | timeline score: 3 | |
| Nov 12, 2009 at 18:40 | comment | added | Mariano Suárez-Álvarez | As you note, both questions are trivial when A is finite, therefore they are trivial in general because K(A) is the union of of fields K(B) with B a finite subset of A. | |
| Nov 12, 2009 at 16:48 | comment | added | Nick Gill | Sorry I meant that "question (2) is answered in the affirmative" by the final observation. And you don't dot t's, you cross them. :-) | |
| Nov 12, 2009 at 16:37 | comment | added | Nick Gill | The lemma that Greg gave certainly answers (1). The definition of K(A) is that it is the smallest subfield of L that contains both K and A. Everyone I've asked here thought both questions were trivial, but couldn't quite dot the i's and t's (at least on the first parse). However after some thought I believe that question (1) is answered in the affirmative by recognising that K(A) is equal to the set of rational expressions with indeterminates in A, and coefficients in K. The point being that these rational expressions contain a FINITE set of elements of A. | |
| Nov 12, 2009 at 16:07 | comment | added | Greg Kuperberg | I think what is missing in (1) is not the definition, but the lemma that in any field extension M of K, the sum and product of two elements algebraic over K are both algebraic over K. Reciprocals are also algebraic. Hence the algebraic subset of M is a subfield of M, and per (1) it can only be M itself. I agree that (2) is by definition. | |
| Nov 12, 2009 at 14:59 | comment | added | David E Speyer | This is skirting kind of close to what could be homework for a Galois theory course, but could also be honest confusion. Both of these come down to remembering definitions. For (1), the definition of "algebraic" is that every element of beta of M obeys a polynomial with coefficients in K. For (2), what it the definition of K(A)? | |
| Nov 12, 2009 at 14:51 | history | asked | Nick Gill | CC BY-SA 2.5 |