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Let $\;$$K = \Bbb Q$[$\sqrt{-5}$] $\;$and let$\;$ $L = \Bbb Q$[$\sqrt{-6}$]$\,$. While it is clear that, as fields, K and L are distinct, if each arithmetic operation is considered separately, they seem to share a number of common features. For example, on the additive side, K and L are isomorphic since each is just a two dimensional vector space over the rationals. On the multiplicative side, their respective rings of integers share the same trivial unit group (w = 2) and the same nontrivial class group (h = 2). Can one say more?

Question: Are the multiplicative groups of K and L isomorphic?

Thanks.

user2052
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  • Well they're not isomorphic as fields. And their number rings have different irreducibles. – Gregory Grant Dec 22 '15 at 19:30
  • Purely group-theoretically $K^/{\pm1}$ and $L^/{\pm1}$ are torsion-free abelian groups of countable rank without divisible elements. Does this make them free $\Bbb Z$-modules of the same rank? – Lubin Dec 22 '15 at 22:13
  • @Lubin, the direct product of countably many copies of $\mathbf{Z}$ is torsion-free of countable rank without divisible elements, but is not free. – Epargyreus Dec 26 '15 at 20:48
  • And, @Epargyreus, the direct product of countably many copies of $\Bbb Z$ is not of countable rank. – Lubin Dec 26 '15 at 21:42
  • @Lubin, yes, my mistake. Your question probably does have a positive answer, although in this case I think my answer below is possibly more transparent. – Epargyreus Dec 26 '15 at 23:07

3 Answers3

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No, the quadratic number fields are not isomorphic, see here for several proofs concerning possible isomorphism between $\mathbb{Q}(\sqrt{d})$ and $\mathbb{Q}(\sqrt{d'})$. Moreover their discriminant and the Minkowski bound of their rings of integers is different. For $\mathcal{O}_K=\mathbb{Z}[\sqrt{-5}]$ it is $\frac{2}{\pi}\sqrt{20}$, and for $\mathcal{O}_K=\mathbb{Z}[\sqrt{-6}]$ it is $\frac{2}{\pi}\sqrt{24}$. Still, the class number is $2$ in both cases.

Community
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Dietrich Burde
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The similarities between the two cases are probably best expressed in binary quadratic forms with integer coefficients and variable values. The statements about primes are in Cox, Primes of the Form $x^2 + n y^2.$ The shorthand "good, bad, medium" was taught to me for use in writing programs, by Irving Kaplansky. Turns out there is a second edition (2013) of COX.

The form $x^2 + 5 y^2$ integrally represents $5$ and all primes $p \equiv 1,9 \pmod {20}.$ Call these the "good" primes. The form $2x^2 + 2xy + 3 y^2$ integrally represents $2$ and all primes $p \equiv 3,7 \pmod {20}.$ Call these the "medium" primes. The "bad" primes are all $q \equiv 11,13,17,19 \pmod {20}.$

A positive integer $n$ is integrally represented by $x^2 + 5 y^2$ if and only if, in factoring $n,$ the exponent of each bad prime is even, and the sum of all the exponents of medium primes is even.

The form $x^2 + 6 y^2$ integrally represents all primes $p \equiv 1,7 \pmod {24}.$ Call these the "good" primes. The form $2x^2 + 3 y^2$ integrally represents $2,3$ and all primes $p \equiv 5,11 \pmod {24}.$ Call these the "medium" primes. The "bad" primes are all $q \equiv 13,17,19,23 \pmod {20}.$

A positive integer $n$ is integrally represented by $x^2 + 6 y^2$ if and only if, in factoring $n,$ the exponent of each bad prime is even, and the sum of all the exponents of medium primes is even.

There is a resemblance at the level of represented numbers. Each genus has only one class, and each discriminant has just two genera.

Will Jagy
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The question is "are the multiplicative groups $K^{\times}$ and $L^{\times}$ isomorphic? The answer is yes, but this is fairly weak result. Let $\mu_K$ be the subgroup of roots of unity of $K^{\times}$.

Lemma Let $K$ be a number field. Then $K^{\times}$ is isomorphic to the direct sum $\Gamma$ of $\mu_K$ and countably many copies of $\mathbf{Z}$.

Proof This is the idea. Suppose that the ring of integers $\mathcal{O}_K$ had class number one. Then every element of $K^{\times}$ could be factored uniquely into a product of a unit and powers (possibly negative) of irreducible elements. Choosing a basis for the unit group (which is isomorphic to the direct sum of $\mu_K$ and a finitely many copies of $\mathbf{Z}$) and choosing an irreducible element for each prime of $\mathcal{O}_K$ gives the desired isomorphism of $K^{\times}$ to $$\Gamma = \mu_K \oplus \bigoplus_{\mathbf{N}}\mathbf{Z}.$$

In general, the ring $\mathcal{O}_K$ will not have unique factorization, but that is easily fixed -- the ring $\mathcal{O}_K[1/S]$ will have unique factorization for some (many) choices of integer $S$, and the unique group $\mathcal{O}^{\times}_K[1/S]$ will be finitely generated, and so the same argument applies.

Epargyreus
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