Welcome to MathOverflow

Question

20

6

If so, do people expect certain invariants (regulator, # of complex embeddings, etc) to fully 'discriminate' between number fields?

flag	asked Oct 30 2009 at 16:04 Ari Shnidman 516●2●9

D. Savitt · Accepted Answer · 2009-10-30 16:56:51Z UTC

John Jones has computed tables of number fields of low degree with prescribed ramification. Though the tables just list the defining polynomials and the set of ramified primes, and not any other invariants, it's not hard to search them to find, e.g., that the three quartic fields obtained by adjoining a root of x^4 - 6, x^4 - 24, and x^4 - 12*x^2 - 16*x + 12 respectively all have degree 4, class number 1, and discriminant -2^11 3^3. On the other hand these three fields are non-isomorphic (e.g. the regulators distinguish them, the splitting fields distinguish them...).

Rob Harron · Answer 2 · 2012-09-17 19:29:51Z UTC

3

Here's a degree three example with minimal discriminant: the two cubic fields given by $x^3-21x-35$ and $x^3-21x+28$ both have discriminant $3969=63^2$, both are Galois, both have class number 3. The regulator of the first is about 4.20169 and that of the second is about 12.59419. This is the smallest (positive) discriminant of a degree 3 field to have two non-isomorphic fields of that discriminant.

edited Sep 17 at 19:29

answered Jun 25 2010 at 4:51

Rob Harron
3,315●1●8●16

It's easier (well, if you were on the proverbial desert island without a computer) to check they are not isomorphic by looking at prime splitting: 2 is inert in the first and splits in the second. – KConrad Jun 25 2010 at 5:31

1

Indeed. Hopefully desert islands will have free wifi some time soon. – Rob Harron Jun 25 2010 at 7:21

1

There are examples with smaller absolute discriminant: the "pure" cubic fields obtained by adjoining cube roots of 6 and 12. Both have discriminant -972 and class number 1, but they're not isomorphic. Pure cubics give lots of examples like this -- I was able to find 5 fields with common discriminant and class number. – Ari Shnidman Mar 16 2012 at 6:10

@Ari: yeah, sorry, I meant of positive discriminant. – Rob Harron Sep 17 at 19:31

Ben Linowitz · Answer 3 · 2009-10-30 17:01:12Z UTC

11

An example (this is exercise 21 in Chapter 2.2 of Borevich and Shafarevich) is given by the three cubic fields obtained by adjoining a root of x^3-18x-6, x^3-36x-78 and x^3-54x-150. The resulting fields are non-isomorphic cubics with class number one and discriminant 22356.

answered Oct 30 2009 at 17:01

Ben Linowitz
2,512●15●28

3

This is the smallest discriminant shared by three nonisomorphic totally real cubic fields. – KConrad Jun 25 2010 at 5:29

David Speyer · Answer 4 · 2009-10-30 16:29:02Z UTC

This isn't quite your question, but there are nonisomorphic number fields with the same zeta function. They are built in this way: Take a Galois extension K/Q, with Galois group G. Find subgroups H₁ and H₂ so that G/H₁ and G/H₂ are nonisomorphic as sets with G-action but C^{G/H₁} and C^{G/H₂} are isomorphic as vector spaces with a G-action. Then look at the fixed fields of H₁ and H₂.

Because there is a lot of freedom in this construction, I would guess that you can manipulate it to avoid detection by any of the easy invariants you mention.

Are there two non-isomorphic number fields with the same degree, class number and discriminant?

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

4 Answers

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.