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Are there non-isomorphic number fields (say of the same degree and signature) that have the same discriminant and regulator? I'm guessing the answer is no - why?

And focusing on fields of small degree (n=3 and n=4), what about a less restrictive question: can we find two such fields that have the same regulator (no discriminant restrictions)?

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3 Answers

up vote 23 down vote accepted

Yes, see e.g. the paper "Arithmetically equivalent number fields of small degree" (Google for it) by Bosma and de Smit.

In brief: two number fields $K$ and $K'$ are said to be arithmetically equivalent if they have the same Dedekind zeta function. A famous group-theoretic construction of Perlis (Journal of Number Theory, 1977) gives many nontrivial (i.e., non-isomorphic) pairs of arithmetically equivalent number fields. Remarkably, this construction works equally well to construct isospectral, non-isometric Riemannian manifolds, as was later shown by Sunada.

Arithmetically equivalent number fields necessarily share many of the simplest invariants, for instance they have equal discriminants.

As the aformentioned paper explains, for arithmetically equivalent $K$ and $K'$, comparing zeta functions gives

$h(K)r(K) = h(K')r(K')$,

where $h$ is the class number and $r$ is the regulator. Therefore, to get an affirmative answer to your question you want a nontrivial pair of arithmetically equivalent number fields $K$ and $K'$ with $h(K) = h(K')$. The paper by Bosma and de Smit gives such examples.

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Thanks! Just what I was looking for. Skimming through their paper, the examples are for n>=8, right? If so, what can be said about n=3 and n=4 for my simpler question, about fields with equal regulators? –  danseetea May 5 '10 at 12:25
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You're welcome. I don't know anything about fields of small degree with equal regulators. If that's important to you, you might consider an edit to make that particular part of the question more prominent, or perhaps even a new question. –  Pete L. Clark May 5 '10 at 12:29
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It's impossible in small degree. Perlis showed arithmetically equivalent number fields of degree up to 6 are isomorphic and he gave examples in degree 7 and 8. Rzedowski-Calderon and Villa-Salvador showed there are no examples in degree 9 or 10. A pair of arithmetically equivalent number fields in degree 8 is Q(3^(1/8)) and Q(48^(1/8)). These fields have class number 1, discriminant -2^24*3^7, and equal regulator. –  KConrad May 5 '10 at 17:41
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Pete- It seems pretty bizarre to ascribe the construction to Sunada, when in his paper (jstor.org/stable/1971195?seq=17) he cites Cassels and Froehlich for it (what Sunada did was the extension to Riemannian geometry). This construction was definitely well known 10 years before Sunada's paper (see Perlis's 1977 paper ams.org/mathscinet-getitem?mr=447188), and I believe it's much older than that. –  Ben Webster Nov 9 '11 at 20:56
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@Ben: My attribution to Sunada was not so much bizarre as...incorrect. Despite being a number theorist, I learned about the construction first in the case of Riemannian manifolds, and never went back to check whether my personal history was in accord with the true chronology. –  Pete L. Clark Nov 9 '11 at 22:26
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Building on G.Myerson's answer and KConrad's explanation, it's not hard to construct pairs $K,K'$ of quartic fields that have both the same discriminant and the same regulator. [Edited to add examples where $K$ and $K'$ do not have the same roots of unity.]

Namely, start with a real quadratic field $F\phantom.$ with fundamental unit $\epsilon$, and let $K,K'$ be totally imaginary quadratic extensions of $F$, not isomorphic with $F\phantom.((-\epsilon)^{1/2})$ and with no roots of unity other than $\pm 1$, whose relative discriminants $d_{K/F}$ and $d_{K'/F}$ have the same norm in ${\bf Q}$. Then $K$ and $K'$ have the same discriminant over ${\bf Q}$, and each has the same unit group $\pm \epsilon^{\bf Z}$ as $F$, so they have the same regulator.

For an explicit example, take $F = {\bf Q}(r)$ with $r=\sqrt{2}$, and let $K = F\phantom.(\sqrt{ab})$ and $K'=F\phantom.(\sqrt{a'b})$ where $a,a' = 7 \pm 2r$ have norm $41$ and $b=5+2r$ has norm $17$. Then $K$ and $K'$ are generated by roots of $x^4+54x^2+697$ and $x^4+86x^2+697$, and are not isomorphic (e.g. the rational prime $7$ splits completely in $K'$ but not in $K$) but both have discriminant $44608 = 2^6 17 \cdot 41$ and unit group $\pm \epsilon^{\bf Z}$ where $\epsilon=1+r$.

The same technique generates arbitrarily large packets of quartic fields with the same discriminant and regulator. More generally, for any totally real field $F\phantom.$ of degree $d>1$ there are arbitrarily large packets of totally imaginary quadratic extensions $K$ of $F\phantom.$ with the same discriminant over ${\bf Q}$ and the same unit group as $F\phantom.$: by the Dirichlet unit theorem each $K$ has the same unit rank as $F$, so — as long as $K$ has no new roots of unity and is not generated by the square root of a unit of $F\phantom.$ — all the units of $K$ are contained in $F$.

[The requirement that $K$ have no roots of unity other than $\pm 1$ is used for this conclusion $O_K^* = O_F^*\phantom.$, but is not needed for equality of regulators. EDIT Indeed it may happen that in such a pair of quartic fields $K$ had more roots of unity than $K'$: if $\epsilon \equiv 1 \bmod 4$ then ($\epsilon$ is totally positive and) $K=F\phantom.(\sqrt{-3})$ has sixth roots of unity while $K'=F\phantom.(\sqrt{-3\epsilon})$ does not. The regulators are still the same unless $K = F\phantom.((-\epsilon)^{1/2})$, that is, unless $3\epsilon$ is a square in $F$, in which case the regulator of $K'$ is twice that of $K$. For example we can take $F = {\bf Q}(\sqrt{203})$, which has $\epsilon = 57 + 4 \sqrt{203}$, but not $F = {\bf Q}(\sqrt{39})$ because then $\epsilon = 25 + 4 \sqrt{39} = (6+\sqrt{39})^2/3$ so $K$ contains the square roots of $-\epsilon$. TIDE]

Degree $4$ is likely minimal here: in degree $2$ (and $1$), number fields are uniquely determined by their discriminant; and as for degree $3$, while there can be arbitrarily large packets of cubic number fields of the same discriminant, it seems most unlikely (though hard to disprove in the totally real case) that any two would have the same regulator.

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Looking at page 607 in Eduardo Friedman, Analytic formulas for the regulator of a number field, Inventiones 98 (1989) 599-622, I notice that there are two quartic fields, one of discriminant 125 and the other of discriminant 225, both having regulator .96242. Of course, that's rounded to 5 decimals; I don't know whether the fields really have the same regulator.

Friedman's numbers come from page 311 of Pohst, Weiler, and Zassenhaus, On effective computation of fundamental units, II, Math Comp 38 (1982) 293-329. Maybe there's enough information in that paper to decide whether the regulators are equal or merely agree to a few decimals.

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Great, I'll check it out. Thanks! –  danseetea May 6 '10 at 12:01
    
Yes, I found the fundamental units in Pohst, Weiler, and Zassenhaus. The regulators are indeed equal. That's interesting. –  danseetea May 6 '10 at 13:17
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It's no surprise that those quartic fields have equal regulator. One is the 5th cyclotomic field (the one with discriminant 125) and the other is Q(sqrt(5),sqrt(-3)). Both are quartic CM fields with the same maximal real subfield Q(sqrt(5)). By Prop 4.16 in Washington's Cyclotomic Fields, the regulator of a quartic CM field and the regulator of its maximal real subfield have a ratio that is 1 or 2 (see Theorem 4.12 for the definition of Q that is used in Prop 4.16), and in this case the ratio for both is 2. The fund. unit (1+sqrt(5))/2 in Q(sqrt(5)) is a fund. unit for both quartic fields. –  KConrad May 23 '10 at 19:58
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In fact the regulator .96242... for the two quartic fields is just twice the regulator of Q(sqrt(5)): 2*log((1+sqrt(5))/2) = .9624236... –  KConrad May 23 '10 at 19:59
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Letting u = (1+sqrt(5))/2, the unit group of Q(zeta_5) is mu_{10} x <u> and the unit group of Q(sqrt(-3),sqrt(5)) is mu_{6} x <u>. Here mu_n is the group of n-th roots of unity. –  KConrad May 23 '10 at 20:01
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