most recent 30 from http://mathoverflow.net 2013-05-24T20:28:50Z http://mathoverflow.net/feeds/user/10143 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/28776/does-the-ideal-class-of-the-different-of-a-number-field-have-a-canonical-square/42596#42596 Stéphane Vinatier 2010-10-18T08:22:26Z 2010-10-18T08:22:26Z

<p>Let $N/K$ be a finite Galois extension of number fields, with Galois group $G$, then Hilbert's formula for the valuation of the different at a prime ideal $P$ of $N$ states that: $$v_P(D)=\sum_{i\geq 0}(|G_i(P)|-1)$$ where $G_i(P)$ is the $i$-th ramification group (in lower notation) at $P$ in $N/K$, and $|G_i(P)|$ is its order. The sum is finite since $G_i(P)={1}$ for large $i$; it is an even integer if $G$ is of odd order, since this forces its subgroups $G_i(P)$ to be of odd order as well for any $i$, any $P$.</p> <p>It follows that, in any odd degree Galois extension of number fields, there exists a fractional ideal whose square equals the inverse different. This fractional ideal, known as the "square root of the inverse different", has rich properties as a Galois module and as an hermitian module, which were first revealed and studied by Boas Erez, see e.g.</p> <p>MR1128708 (92g:11108) Erez, B. The Galois structure of the square root of the inverse different. Math. Z. 208 (1991), no. 2, 239–255.</p> <p>In conclusion we get a bit more than what was asked in the odd degree Galois extension case (an ideal whose square equals the inverse different, not only with the same class). But this does not say anything in the even degree or non-Galois case. </p>