Timeline for A finitely generated $\mathbb{Z}$-algebra that is a field has to be finite
Current License: CC BY-SA 4.0
6 events
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| Apr 9, 2023 at 20:43 | history | edited | LSpice | CC BY-SA 4.0 |
Mild tidying, while this is on the front page
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| Mar 7, 2011 at 1:38 | history | edited | Guillermo Mantilla | CC BY-SA 2.5 |
added 25 characters in body
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| Mar 7, 2011 at 1:21 | comment | added | Guillermo Mantilla | @aglearner: I've added an explanation to what you are wondering. The point is that one version of the Nullstellensatz, which I learned by the name algebraic Nullstellensatz, is the following: A finitely generated extensions of fields $F/K$ is algebraic. | |
| Mar 7, 2011 at 1:16 | history | edited | Guillermo Mantilla | CC BY-SA 2.5 |
added 367 characters in body
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| Mar 7, 2011 at 0:12 | comment | added | aglearner | Unfortunately, I can not understand when you write " By the Nullstellensatz we have that each $\alpha_i$ is algebraic over $\mathbb Q$". Could you please explain this point? Are you using Nullstelensatz over $\mathbb Q$ here? To which ring are you applying it? | |
| Mar 6, 2011 at 3:23 | history | answered | Guillermo Mantilla | CC BY-SA 2.5 |