Timeline for Is the statement that every field has an algebraic closure known to be equivalent to the ultrafilter lemma?
Current License: CC BY-SA 2.5
8 events
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| Nov 20, 2010 at 16:06 | comment | added | Andrés E. Caicedo | @Joel: Thanks! Nice argument, as usual! | |
| Nov 20, 2010 at 12:02 | comment | added | Joel David Hamkins | Andres, I added an answer explaining it. | |
| Nov 20, 2010 at 2:06 | comment | added | Andrés E. Caicedo | @Joel: And uniqueness? | |
| Nov 20, 2010 at 0:17 | comment | added | Joel David Hamkins | Can't you get it fairly easily from the Compactness theorem, since for any field $F$ you can write down the axioms of what it means to be a field extension of $F$ in which every polynomial over $F$ has a root. It is finitely consistent because of the finite extensions of $F$. If there is a model of the full theory, then the collection of all algebraic elements over $F$ will be algebraically closed, no? | |
| Nov 19, 2010 at 22:45 | comment | added | Eivind Dahl | Interesting! No, he just mentions it in a footnote, sorry. | |
| Nov 19, 2010 at 22:44 | history | edited | Eivind Dahl | CC BY-SA 2.5 |
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| Nov 19, 2010 at 22:42 | comment | added | Andrés E. Caicedo | Hi Elvind, Compactness is in fact equivalent to the ultrafilter theorem. Does Aluffi give a reference, other than Banaschewski's paper? | |
| Nov 19, 2010 at 22:39 | history | answered | Eivind Dahl | CC BY-SA 2.5 |