Timeline for Is there a ring of integers except for Z, such that every extension of it is ramified?
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6 events
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| Jun 2, 2010 at 15:04 | comment | added | Torsten Ekedahl | Well it is even worse I thought about the distinction for $K$ where I noted that it doesn't matter as $K$ is totally complex but forget that it matters for $\mathbb Q[\sqrt{19\cdot151}$. What is much worse I however is that when I asked Magma (and also Kant to doublecheck) about the class number I did for the field obtained by adjoining one root of the polynomial and not all of them. Hence I got class number (which is probably OK) but for the wrong field... Back to the drawing board. | |
| Jun 2, 2010 at 11:07 | comment | added | Franz Lemmermeyer | You only can expect a positive answer if unramified means unramified everywhere. If you allow ramification at infinity, you get a larger tower by adjoining a square root of -19 and then continuing with the Hilbert class field of ${\mathbb Q}(\sqrt{-151})$. BTW, how do you prove that K has class number 1 if Odlyko's bounds are too weak? | |
| Jun 2, 2010 at 5:17 | comment | added | Torsten Ekedahl | Here is a specialisation of the question at the end. The polynomial (an example "dear to Artin" according to Lang, if I remember the comment in his book correctly) $x^5-x+1$ gives an unramified $A_5$-extension $K$ of $\mathbb Q[\sqrt{19\cdot151}]$. Is this the maximal unramified extension of $\mathbb Q[\sqrt{19\cdot151}]$? The class group of $K$ is trivial so a further solvable extension is out. On the other hand Odlyzko's bounds are not strong enough to give anything it seems. | |
| Jun 2, 2010 at 4:06 | history | edited | Cam McLeman | CC BY-SA 2.5 |
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| May 31, 2010 at 3:17 | history | edited | Cam McLeman | CC BY-SA 2.5 |
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| May 31, 2010 at 3:09 | history | answered | Cam McLeman | CC BY-SA 2.5 |