Timeline for Do groups of units change base nicely, assuming the fields are algebraically closed?
Current License: CC BY-SA 4.0
12 events
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| Jan 5 at 11:25 | history | edited | Will Sawin | CC BY-SA 4.0 |
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| Jan 5 at 3:00 | vote | accept | Neil Epstein | ||
| Jan 5 at 2:28 | comment | added | Neil Epstein | Also, reduced rings don't embed in fields unless they don't have nonzero zero divisors. But I'll go ahead and accept the answer now. Thank you! | |
| Jan 5 at 1:53 | comment | added | Will Sawin | @NeilEpstein I just mean the induced map $id \otimes f \colon \mathcal O(X) \otimes_K S \to \mathcal O(X)$. Now edited to clarify a bit. | |
| Jan 5 at 1:52 | history | edited | Will Sawin | CC BY-SA 4.0 |
added 13 characters in body
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| Jan 4 at 22:30 | comment | added | Neil Epstein | Never mind. I see now how to adapt this to a rigorous proof. | |
| Jan 4 at 22:21 | comment | added | Neil Epstein | ‘t make sense. But I don’t see such a map. | |
| Jan 4 at 22:20 | comment | added | Neil Epstein | Another question: It seems we should need a map from the ring you constructed (let’s call it $S$) to $\mathcal{O}(X)$, rather than to $K$. Otherwise the expression $f(\sum_i a_i b_i)$ doesn | |
| Jan 4 at 20:48 | comment | added | Will Sawin | @NeilEpstein Since $a_0=1$, we can replace $f(b_0)$ by $f(b_0)- \sum_i a_i f(b_i)$ to get a linear relation, which is nontrivial since $f(b_j)\neq 0$. | |
| Jan 4 at 20:42 | comment | added | Neil Epstein | I’m almost with you, except for one step. We have $\sum_i a_i f(b_i) \in K$ and $f(b_j) \neq 0$. How does that contradict linear independence of the $a_i$ over $K$? | |
| Jan 4 at 18:17 | history | edited | GNiklasch | CC BY-SA 4.0 |
summation index grouping
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| Jan 4 at 18:00 | history | answered | Will Sawin | CC BY-SA 4.0 |