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What algebraic properties imply that the rings $\Bbb Z [\sqrt 2]$ and $ \Bbb Z [\sqrt 3]$ are not isomorphic?

Of course, there are elementary reasons, such that $x^2=2$ has not solution in $ \Bbb Z [\sqrt 3]$, as in this question.

I'm interested in properties of rings that can tell $\Bbb Z [\sqrt 2]$ and $ \Bbb Z [\sqrt 3]$ apart.

For instance, although both rings are UFDs, are their lattice of ideals isomorphic? Are their lattice of prime ideals isomorphic? Do the two rings have some non-isomorphic quotients?

What, if anything, can be said in general, for $\Bbb Z [\sqrt d]$, when $\Bbb Z [\sqrt d]$ is the ring of integers of $\Bbb Q (\sqrt d)$? Or even for quadratic rings of integers in general. I know that the discriminant works for this but I'm interested in ring properties.

lhf
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  • An invariant is the Minkowski bound $B_K$, which is different for those two rings. In general it is $\sqrt{|d|}$ or $\frac{1}{2}\sqrt{|d|}$ for the ring of integers in $\Bbb Q(\sqrt{d})$, but it is related to the discriminant again. – Dietrich Burde Dec 18 '23 at 14:29
  • Jack's anwer here uses "Legendre symbols". If these rings are isomorphic, so are its fraction fields. But they are not isomorphic. – Dietrich Burde Dec 18 '23 at 14:46
  • Dietrich Burde already gave reasonable answers, so I wanted to stress that many of these things you mention ARE ring properties, in the strictest sense. The inability to solve $x^2=2$ is precisely why you have no ring homomorphisms between $\Bbb{Z}[\sqrt{2}]$ and $\Bbb{Z}[\sqrt{3}]$: Any such homomorphism would fix $\Bbb{Z}$, and thus the image $\alpha$ of $\sqrt{2}$ must satisfy $\alpha^2=2$, but no such element exists. You can also turn this into a statement about polynomial rings over these rings. – SomeCallMeTim Dec 19 '23 at 17:00
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    These are answers to the question, not just a comment. Comments should only be used to clarify, not answer the question. See How do comments work for more information. Therefore, please post your answer as an answer. This brings extra visibility to the answer und puts the question off the unanswered list. Also notice that comments are not indexed by the full text search, don't have any revision history, can only be edited for 5 minutes and allow only limited markup. – Martin Brandenburg Dec 19 '23 at 20:54
  • @Martin I don't agree that those comments are answers (and likely the authors did not either). In any case they will likely not see your comment if not explicitly pinged. – Bill Dubuque Dec 20 '23 at 00:52
  • @lhf What precisely is your definition of a "ring property" (and why do you exclude the discriminant)? – Bill Dubuque Dec 20 '23 at 00:56
  • @BillDubuque, no precise definition, but something structural (no precise definition for that either). I've edited the question to include examples about ideals and quotients. – lhf Dec 20 '23 at 10:08

1 Answers1

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The first elementary reason you gave covers all domains of the form $\mathbf Z[\sqrt{d}]$, if you phrase things in the right way.

Let $d$ be an integer that is not a square in $\mathbf Z$. (I am allowing $d$ to have a square factor, e.g., $d = 8$ or $12$.) The number $d$ is a square in $\mathbf Z[\sqrt{d}]$: $d = \sqrt{d}^2$. What are all the integers that are not squares in $\mathbf Z$ but are squares in $\mathbf Z[\sqrt{d}]$?

A general element of $\mathbf Z[\sqrt{d}]$ is $a+b\sqrt{d}$, where $a, b \in \mathbf Z$. Since $$ (a+b\sqrt{d})^2 = a^2 + db^2 + 2ab\sqrt{d}, $$ if an integer $m$ is $(a+b\sqrt{d})^2$ then the term $2ab\sqrt{d}$ must be $0$, so $a = 0$ or $b = 0$. Having $a = 0$ makes $m = db^2$ and having $b = 0$ makes $m = a^2$. The integers $db^2$ with $b \not= 0$ are not squares in $\mathbf Z$, while the integers $a^2$ are squares in $\mathbf Z$. We have proved the following result.

Theorem. The set of integers that are not squares in $\mathbf Z$ but are squares in $\mathbf Z[\sqrt{d}]$ is $$ \{db^2 : b \in \mathbf Z, b \not= 0\}. $$

All such integers share the same sign (namely the sign of $d$) and the least such integer in absolute value is $d$. So we have reconstructed $d$ from $\mathbf Z[\sqrt{d}]$ by a ring-theoretic property: $d$ is the least integer in absolute value that is not a square in $\mathbf Z$ but is a square in $\mathbf Z[\sqrt{d}]$. Thus different $d$'s lead to nonisomorphic rings $\mathbf Z[\sqrt{d}]$.

Remark. You wrote at the end of your post that you don't consider the value of the discriminant to be a ring property. You need to appreciate discriminants more: they distinguish orders in all quadratic fields from each other and that is the standard way to prove they are nonisomorphic rings. When you work with rings that are finite-free $\mathbf Z$-modules, their discriminants are important numerical invariants even though they are not defined for all rings.

KCd
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  • Excellent answer, thanks! – lhf Jan 07 '24 at 17:06
  • I’m still looking for something about ideals, though. – lhf Jan 07 '24 at 17:06
  • Well, when $p$ is prime the quotient $\mathbf Z[\sqrt{d}]/(p)$ is isomorphic to $(\mathbf Z/(p))[x]/(x^2-d)$, and when $p \not= 2$ the ring structure is determined by the Legendre symbol $(d|p)$: it's a field when $(d|p) = -1$, a product of two fields when $(d|p) = 1$, and it's the nonreduced ring $(\mathbf Z/(p))[x]/(x^2)$ when $(d|p) = 0$. Prime density calculations imply that knowledge of $(d|p)$ for all but finitely many $p$ determines the squarefree part of $d$, so we know $d$ up to a square factor from knowing when $\mathbf Z[\sqrt{d}]/(p)$ is a field as $p$ runs over the odd primes. – KCd Jan 07 '24 at 21:05
  • When $\mathcal O$ is an order in a quadratic field $K$, its conductor ideal is the biggest ideal of $\mathcal O_K$ that is also an ideal in $\mathcal O$. This ideal is $m\mathcal O$ where $m$ is the index $[\mathcal O_K:\mathcal O]$, so $m$ is the unique generator of that ideal in $\mathbf Z^+$. There is only one order with each index in $\mathcal O_K$, and the conductor ideal of $\mathcal O$ determines the index $[\mathcal O_K:\mathcal O]$, so knowing the conductor ideal of $\mathcal O$ determines $\mathcal O$. – KCd Jan 07 '24 at 21:15
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    Among ideals $\mathfrak a$ in $\mathcal O_K$ and $\mathfrak b$ in $\mathcal O$ that are relatively prime to the conductor of $\mathcal O$, there are inverse bijections by $\mathfrak a \mapsto\mathfrak a \cap\mathcal O$ and $\mathfrak b \mapsto\mathfrak b\mathcal O_K$, and corresponding ideals have isomorphic residue rings: see Thm 3.8 in https://kconrad.math.uconn.edu/blurbs/gradnumthy/conductor.pdf. In particular, there's a bijection between all but finitely many prime ideals in $\mathcal O_K$ and $\mathcal O$ preserving residue rings. So distinguishing those rings via ideals is subtle. – KCd Jan 08 '24 at 19:00