Timeline for Massive cancellations
Current License: CC BY-SA 3.0
12 events
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| Nov 26, 2014 at 18:28 | comment | added | Will Sawin | @Scott aaronson it's $r+c$, where $r$ is the number of real embedding sand $c$ is the number of complex embeddings. You have $r+2c=d$, where $d$ is the degree of the field. If your field is generated by one element, $d$ is the degree of the minimal polynomial and $r$ is the number of real roots. | |
| Nov 26, 2014 at 18:10 | comment | added | Scott Aaronson | I see; thanks. Is there an easy way to compute the number of such Archimedean valuations? | |
| Nov 26, 2014 at 17:48 | comment | added | David E Speyer | @ScottAaronson There are 4 Archimedean valuations on $\mathbb{Q}[\sqrt{2}, \sqrt{3}]$: we can take $w+x\sqrt{2}+y \sqrt{3} + z \sqrt{6}$ to be $|w+x\sqrt{2}+y \sqrt{3} + z \sqrt{6}|$, $|w-x\sqrt{2}+y \sqrt{3} - z \sqrt{6}|$, $|w+x\sqrt{2}-y \sqrt{3} - z \sqrt{6}|$ or $|w-x\sqrt{2}-y \sqrt{3} + z \sqrt{6}|$. In general, you have to look at all the embeddings of $K$ into $\mathbb{R}$ or $\mathbb{C}$, not just the obvious one. | |
| Nov 26, 2014 at 15:55 | comment | added | Scott Aaronson | Timothy: Thanks so much; that's extremely helpful! Just one thing I'm confused about: for a general number field $K$, what are the other Archimedean valuations? So for example, suppose $K=\mathbb{Q}[\sqrt{2},\sqrt{3}]$---then what are the Archimedean valuations besides the usual absolute value? | |
| Nov 26, 2014 at 14:27 | history | edited | Timothy Chow | CC BY-SA 3.0 |
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| Nov 26, 2014 at 14:23 | comment | added | Timothy Chow | Note that for $K=\mathbb Q$, the archimedean valuation is just the usual absolute value, and the non-archimedean valuations are the $p$-adic norms. The product formula says that if, for each prime $p$, you pull out the highest power (possibly negative, for fractions) of $p$ dividing $x$, and multiply these together, then you get $|x|$. So the product formula is, in a sense, "equivalent" to the unique factorization theorem. For number fields you have to use prime ideals to get unique factorization, but it's all totally analogous. | |
| Nov 26, 2014 at 13:59 | comment | added | Timothy Chow | There's one non-archimedean valuation for each prime ideal in the ring of integers $R$ (and hence countably many) plus at most $d$ archimedean valuations where $d$ is the degree of the field. The principal ideal $xR$ generated by an element $x$ factors (uniquely) into a finite product of prime ideals and a non-archimedean valuation is $\ne1$ only for primes appearing in this factorization. Thus $|x|_v\ne1$ for only finitely many $v$ and the product is well-defined. | |
| Nov 26, 2014 at 5:18 | comment | added | Julian Rosen | There's relevant info in Cassels' Local Fields (10.1-2) and Neukirch's Algebraic Number Theory (III.1). You might also try Section 7 of Milne's algebraic number theory course notes (have a look at p. 97). A number field has countably many places (= valuations up to equivalence). For fixed $x$, $|x|_v=1$ for all but finitely many $v$, so the product is finite. | |
| Nov 26, 2014 at 3:53 | comment | added | Scott Aaronson | Thanks so much, Julian! I'm trying to unpack this proof---can anyone suggest a good reference for reading about why the product formula holds for valuations, and why there are only finitely many valuations $v$ for which $|x|_v$ can exceed $1$? Also, for an algebraic number field, how many valuations are there in total---countably many? How do we even see that $\prod_v |x|_v$ is well-defined? (Sorry for asking such basic questions.) | |
| Nov 25, 2014 at 6:12 | history | edited | Julian Rosen | CC BY-SA 3.0 |
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| Nov 25, 2014 at 5:49 | history | edited | Julian Rosen | CC BY-SA 3.0 |
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| Nov 25, 2014 at 5:31 | history | answered | Julian Rosen | CC BY-SA 3.0 |