Timeline for Does a "composite field" always exist?
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Feb 7, 2012 at 20:15 | vote | accept | Pace Nielsen | ||
| Feb 7, 2012 at 11:03 | comment | added | Martin Brandenburg | (This also reminds me of the concise tensor product construction of the algebraic closure of a field: In $k' := \bigotimes_{0 \neq f \in k[x]} k[x]/(f)$ every polynomial as a root, thus the colimit of $k \subseteq k' \subseteq k'' \subseteq k''' \subseteq ...$ is an algebraic closure of $k$. One can show $k''=k'$, but this is not trivial.) | |
| Feb 7, 2012 at 11:00 | comment | added | Martin Brandenburg | Years ago, I first learnt the solution in David's answer and could not really find anything enlightening or memorable about it. Later the argument using tensor products was used in a text on algebraic geometry (namely the lemma that $|X \times_S Y| \to |X| \times_{|S|} |Y|$ is surjective) and of course now everything was clear as crystal. As a side remark, both proofs use variants of the axiom of choice (even twice). | |
| Feb 6, 2012 at 21:01 | history | edited | Angelo | CC BY-SA 3.0 |
added 2 characters in body
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| Feb 6, 2012 at 20:34 | history | answered | Angelo | CC BY-SA 3.0 |