Timeline for Is every Noetherian Commutative Ring a quotient of a Noetherian Domain?
Current License: CC BY-SA 3.0
5 events
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| Apr 4, 2014 at 7:45 | history | edited | Jérôme Poineau | CC BY-SA 3.0 |
In the last line, changed $\bar{\mathbb Q} \times \mathbb C$ into $\mathbb Q \times \bar{\mathbb Q}$.
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| Apr 4, 2014 at 7:43 | comment | added | Jérôme Poineau | @SiddharthVenkatesh: You are right, this should work with $\bar{\mathbb Q}$ instead of $\mathbb Q$ (although the ring we get is not explicitly in Harbater's paper). Actually, I have changed the wrong factor and should have written $\mathbb Q \times \bar{\mathbb Q}$ instead of $\bar{\mathbb Q} \times \mathbb C$. I will edit. | |
| Apr 4, 2014 at 2:06 | comment | added | Siddharth Venkatesh | That's a cool answer. Why would this not work with $Q$ replaced with $\overline{Q}$? Does the Noetherian property not hold for it? | |
| Apr 3, 2014 at 23:20 | comment | added | Joël | Very nice answer. | |
| Apr 3, 2014 at 21:16 | history | answered | Jérôme Poineau | CC BY-SA 3.0 |