Timeline for To prove the Nullstellensatz, how can the general case of an arbitrary algebraically closed field be reduced to the easily-proved case of an uncountable algebraically closed field?
Current License: CC BY-SA 2.5
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| Feb 25, 2010 at 10:51 | vote | accept | user2734 | ||
| Feb 18, 2010 at 7:53 | comment | added | user2734 | Wow! this looks like what I am looking for. It will take me some time to process this proof, though. | |
| Feb 18, 2010 at 6:52 | history | edited | BCnrd | CC BY-SA 2.5 |
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| Feb 18, 2010 at 6:44 | history | edited | BCnrd | CC BY-SA 2.5 |
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| Feb 18, 2010 at 6:27 | comment | added | BCnrd | The previous post had an integrality argument that didn't apply when k'/k is not algebraic. The ironic thing is that my immediate reaction upon seeing the question was "Oh, it's just the old spread out and specialize business", and while typing that I thought I found an even "slicker" argument (the original post) which I realized was not right about 2 seconds after I posted it. So I went back to my original idea, which is correct. Better to follow one's instincts and not try to be too slick. :) | |
| Feb 18, 2010 at 6:22 | comment | added | Pete L. Clark | @Brian: when you edit a post significantly, it is nice to give some indication of what you have changed. Was there something wrong with your previous argument? | |
| Feb 18, 2010 at 6:15 | history | edited | BCnrd | CC BY-SA 2.5 |
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| Feb 18, 2010 at 6:05 | comment | added | Pete L. Clark | +1: This works nicely. | |
| Feb 18, 2010 at 5:52 | history | answered | BCnrd | CC BY-SA 2.5 |