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This is a question I've had for a while and really don't know how to go about finding an answer:

Does there exist a pair of binary operations, $\boxplus$ and $\boxtimes$, other than the usual $+$ and $\times$, such that $(\mathbb{Q}, \boxplus, \boxtimes)$ forms a ring?

I realize that there's probably some "axiom of choice" proof that constructs unintelligible binary operations or even a construction using some other countably infinite ring and a bijection to the rationals. So more importantly I ask:

Does there exist such a $\boxplus$ and $\boxtimes$ such that $a \boxplus b$ and $a \boxtimes b$ can be computed from (closed?) formulas that only involve $+$ and $\times$ (or other related properties of $a$ and $b$ such as prime factors, divisors, gcd, partitions, etc.)?

My apologies if there's some sort of easy example out there that I'm missing. (Though in that case I'll push further and ask if it can be generalized to a larger class of examples.)


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What about taking some permutation of $\mathbb Q$ which fixes $0$ or $1$ (if you want this new ring to have the same $0$ and $1$ as the original ring $\mathbb Q$; if not, then you don't have to require even this) and conjugating the operations $+$ and $\times$ by this permutation? –  darij grinberg Aug 31 '11 at 14:50
Read "and" for "or" in my comment above. –  darij grinberg Aug 31 '11 at 14:50
Maybe you want "closed formulas" to mean something like "the operations are rational functions"? –  darij grinberg Aug 31 '11 at 14:51
I wasn't thinking rational functions because I feel like other properties arising from multiplication and addition should be allowed, such as prime factors, gcd, partitions, etc. I'll edit the original post to reflect this. –  Aeryk Aug 31 '11 at 14:55
If you are only slightly generous with the notion of formula, you will pick up all countably infinite computable rings, since the formulas will allow you to specify any computable function. For example, Julia Robinson proved that the integers are definable inside the rational field in a primitive manner, and then any computable function y=f(x) is specified by there being an integer solution to a certain diophantine equation involving x and y and other integer variables. So the answer to the question will depend greatly on the details of what kind of definition you allow. –  Joel David Hamkins Aug 31 '11 at 15:50
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6 Answers

up vote 11 down vote accepted

Given any bijection $f : \mathbb{Q} \to R$ where $(R,\oplus,\otimes)$ is some (necessarily countable) ring, you'll be able to get a new ring structure $(\mathbb{Q},\boxplus,\boxtimes)$ isomorphic to $(R,\oplus,\otimes)$, by setting:

$a \boxplus b = f^{-1}(f(a)\oplus f(b))$
$a \boxtimes b = f^{-1}(f(a)\otimes f(b))$

The nicer $f$ is, the nicer the expressions for $\boxplus$ and $\boxtimes$ will be. Perhaps the simplest examples are if $p \in \mathbb{Q}^\times,\ q\in \mathbb{Q}$, $R = \mathbb{Q}$, then $f(x) = px+q$ will work. This generalizes Neil's and Pace's answers.

The "converse" is trivially true, in that if $f : (\mathbb{Q},\boxplus,\boxtimes) \to (R,\oplus,\otimes)$ is an isomorphism from some ring structure on $\mathbb{Q}$ to a ring $R$, then $f$ is a bijection $\mathbb{Q} \to R$ and


So in some sense, the above method for getting a ring structure on $\mathbb{Q}$ is the only way to do it. The question (more or less) boils down to, "for which rings $(R,\oplus,\otimes)$ is there a 'nice' bijection $\mathbb{Q} \to R$?" It depends, of course, on what you think "nice" means.

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Thanks for the answer. I will now have to think more carefully about what "nice" means. –  Aeryk Aug 31 '11 at 21:56
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The simplest example is just $x\boxplus y=x+y$ and $x\boxtimes y=-xy$. This gives a ring structure with $-1$ as the multiplicative identity. The map $x\mapsto -x$ gives an isomorphism $(\mathbb{Q},\boxplus,\boxtimes)\simeq(\mathbb{Q},+,\times)$.

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Does this generalize? If this is the simplest example, is there a next simplest example? –  Aeryk Aug 31 '11 at 18:48
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$x\oplus y=\mathrm{min}(x,y)$, $x \otimes y=x+y$ gives a semiring. The direct sum of a countable collection of finite rings gives a ring. Or lots of other things. There is certainly no classification of the countable rings.

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The answer is yes (unless I made a mistake somewhere).

For example, you can replace addition with the operation $a\oplus b=a+b-1$. This is a commutative, associative binary operation with identity $1$ and the inverse of $a$ is given by $2-a$.

You replace multiplication by $a\odot b= a+b-ab$. This is a commutative, associative binary operation with identity $0$.

All that remains is to show that the distributive laws hold.

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You don’t need to show any laws, just observe that $x\mapsto 1-x$ is an isomorphism of your structure with the usual $(\mathbb Q,+,\cdot)$. –  Emil Jeřábek Aug 31 '11 at 18:48
True. But that seems to me to be just about as difficult (although it does give an idea of where it comes from). –  Pace Nielsen Aug 31 '11 at 18:54
@Emil: I think this gives me some good insight. True or False: Any linear map $x \to ax+b$ will induce corresponding operations under which 1) $\mathbb{Q}$ is a ring and 2) the linear map gives the isomorphism with the standard operations? I would conjecture now that this is true. –  Aeryk Aug 31 '11 at 18:55
Yes, any bijection will do, it does not have to be linear, but I gather you learned this meanwhile from AKG’s answer. –  Emil Jeřábek Sep 1 '11 at 10:42
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Since the question is somewhat fuzzy, I am not 100% sure what would satisfy you as an answer. Based on nothing, I am guessing you want something a little stranger than the examples Neil and Pace have provided, but a little simpler or more concrete than Joel's family of examples.

There is a somewhat cute, if useless, ring structure for $\mathbb{Q_+}$ (the positive rationals) which might still be of interest to you. I haven't thought of whether it can be extended to $\mathbb{Q}$ in a nice way.

The idea is that the fundamental theorem of arithmetic (existence and uniqueness of factorization in $\mathbb{N}$) gives a bijection between $\mathbb{Q}_+$ and sequences of integers with finitely many nonzero elements. Namely each number is mapped to the sequence of powers of primes in its factorization, i.e., $\frac{3}{4} = 2^{-2}\cdot 3^1 \mapsto (-2,1,0,0,\ldots)$.

The set of sequences of integers with finitely many nonzero elements already has a familiar ring structure, namely that of $\mathbb{Z}[x]$. We can use the map above to transfer this ring structure to $\mathbb{Q}_+$, in which case $\boxplus$ is what we usually call multiplication (the cute part) and $\boxtimes$ is defined by the ring axioms and the condition that $p_i \boxtimes p_j = p_{i+j}$ where $p_0=2,p_1=3, p_2 = 5,\ldots $ is the sequence of primes. For example, you can check that $\frac{3}{4}\boxtimes 6 = \frac{5}{12}$. Computationally the downside is that you need to be able to factor into primes to be able to do $\boxtimes$ (as far as I know).

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Nice example. I suppose the 'nicest' example of a different ring structure on $\Q$ would be to take Cantor's bijection $f:\mathbb{Q}\rightarrow \mathbb{Z}$ and then apply Amit's answer, using the usual ring structure on $\mathbb{Z}$. –  Pace Nielsen Aug 31 '11 at 19:30
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A more general formulation may be as follows, and close to what you desire. Given the set Q of rational numbers and a (necessarily finite) set of basic operations of finite arity, consider the set T of all term operations formed through composition from the basic operations. (Some of the operations can have arity 0, so they look like constants or constant functions.) How many pairs (a,m) of terms from T can one form so that the structure < Q, a, m > is a ring?

I do not know where in the general algebra literature this is covered. Search terms that come to mind are cryptomorphism, term equivalent or polynomially equivalent algebras, and interpreting one structure inside another. I hope this helps. If I had to guess, I would guess to the above that there are infinitely many pairs of terms making Q into infinitely many distinct rings, given + , *, 0 and 1.

Gerhard "Ask Me About System Design" Paseman, 2011.08.31

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I haven't checked the details, but I nominate (a term corresponding to) a+a+b+b as a new form of addition. If that works, then there are countably many distinct examples indeed. Gerhard "Ask Me About Quick Guessing" Paseman, 2011.08.31 –  Gerhard Paseman Aug 31 '11 at 15:32
Nope, associativity fails. I guess I had best review my notes on hyperassociativity. Gerhard "Back To The Scratch Pad" Paseman, 2011.08.31 –  Gerhard Paseman Aug 31 '11 at 15:37
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