Timeline for Is the transcendence degree of a domain over a subfield the same as that of the fraction field of that domain?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| S Mar 26, 2017 at 9:29 | history | suggested | Glorfindel | CC BY-SA 3.0 |
grammar corrections
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| Mar 26, 2017 at 9:08 | review | Suggested edits | |||
| S Mar 26, 2017 at 9:29 | |||||
| Sep 13, 2011 at 8:30 | vote | accept | Georges Elencwajg | ||
| Sep 13, 2011 at 7:53 | answer | added | user2035 | timeline score: 4 | |
| Sep 12, 2011 at 15:57 | comment | added | Georges Elencwajg | a-fortiori, I (and certainly other users too) would be very happy to upvote you if you transformed your comment above into an answer. | |
| Sep 12, 2011 at 15:52 | history | edited | Georges Elencwajg | CC BY-SA 3.0 |
added "Edit"
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| Sep 12, 2011 at 14:55 | comment | added | Georges Elencwajg | You are absolutely right,a-fortiori, thanks a lot! And I can't even plead ignorance: I knew the result you mention! I'll edit my question. | |
| Sep 12, 2011 at 14:02 | history | edited | KConrad | CC BY-SA 3.0 |
Change "alg. indep. over K" to "alg. indep. over k"
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| Sep 12, 2011 at 13:53 | comment | added | user2035 | The usual strengthening of the existence theorem for transcendence bases states: If $K$ is algebraic over $k(A)$ for some subset $A\subset K$, there is a transcendence basis contained in $A$. | |
| Sep 12, 2011 at 13:30 | comment | added | M P | If you are willing to well-order your set $I$, I think that you can first reduce to the extension of a single element of $K$ and then observe that at least one among "numerator" or "denominator" of the next element must be transcendental over the domain you have so far constructed. | |
| Sep 12, 2011 at 13:05 | history | asked | Georges Elencwajg | CC BY-SA 3.0 |