Timeline for Quick proof of the fact that the ring of integers of $\mathbb Q(\zeta_n)$ is $\mathbb Z[\zeta_n]$?
Current License: CC BY-SA 2.5
7 events
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| May 27, 2010 at 15:27 | comment | added | ACL | @Andrea Ferreti: The discriminant of $\mathbf Z[\lambda]$ is the discriminant of the ring of integers $R$ times the $d$th power of the index of $\mathbf Z[\lambda]$ in $R$, where $d$ is the degree of the extension. So if the discriminant of $\mathbf Z[\lambda]$ has no $d$th powers, one must have $\mathbf Z[\lambda]=R$. | |
| Mar 6, 2010 at 23:06 | comment | added | KConrad | Jim, that involves saying what it means for a prime to ramify in an order, which seems to make things more complicated in terms of a basic explanation. | |
| Mar 6, 2010 at 17:23 | comment | added | stankewicz | @Ferretti: If I were to state it as a theorem, it would be this: Let $K$ be an algebraic number field whose discriminant is a power of $p$ a prime number which is totally ramified in $K$. If $\mathbf{Z}[\lambda]$ is an order of $K$ in which $p$ totally ramifies then $\mathbf{Z}[\lambda]$ is the ring of integers. | |
| Mar 6, 2010 at 17:07 | comment | added | Andrea Ferretti | @stankewicz: Ok, but apart from what we call it, is that true? Namely: if $\lambda$ is an algebraic integer in a number field with norm $p$ and if the discriminant is a power of $p$, then the ring of intgers is $\mathbb{Z}[\lambda]$. | |
| Mar 6, 2010 at 16:58 | comment | added | stankewicz | I don't know that it's a significant enough phrase to call a theorem, but it's a good way to find the ring of integers when you have a totally ramified prime in $\mathbf{Z}[\lambda]$. | |
| Mar 6, 2010 at 16:49 | comment | added | Andrea Ferretti | I fear this is the "computational" proof I already know. By the way, is your second sentence an actual theorem? This may help streamlining the proof. | |
| Mar 6, 2010 at 16:11 | history | answered | stankewicz | CC BY-SA 2.5 |