Consider, for instance, the ring $(\mathbb{Z}, +, \cdot)$. We can create a new ring $(\mathbb{Z}, \oplus, \odot)$ by defining $$ a \oplus b = a + b - 1 \quad \text{and} \quad a \odot b = a + b - ab. $$ There are also other operations that we could use as 'addition' and 'multiplication' (for example we could define $a \oplus b = a + b - 1$ and $a \odot b = ab - (a + b) + 2$, etc.)

My question is: is there a finite number of different operations $\oplus$ and $\odot$ that we could define on $\mathbb{Z}$ and retain the ring structure, or are there infinitely many such operations? what about on $\mathbb{R}$? what about on $\mathbb{Z} / n \mathbb{Z}$?

  • $\begingroup$ You're asking how many ring isomorphism classes there are for sets with countable cardinality. Is that what you wanted to ask? Or did you have in mind something more specific? $\endgroup$
    – Jack M
    Sep 26, 2013 at 20:24
  • $\begingroup$ @JackM I mean, for some ring $(R, +, \cdot)$, how many different operations $\oplus$ and $\odot$ replacing $+$ and $\cdot$ (keeping $R$ fixed) such that $(R, \oplus, \odot)$ is still a ring. $\endgroup$
    – tylerc0816
    Sep 26, 2013 at 20:28
  • $\begingroup$ @tylerco816 That still just depends on the cardinality of R. $\endgroup$
    – Jack M
    Sep 26, 2013 at 20:30

2 Answers 2


On $\Bbb {Z/nZ}$ there must be finitely many as there are only finitely many operations on a finite set.

Your examples are derived by considering replacing each element $x$ with $x-1$ and seeing what happens to the operations. The constant is arbitrary, so you have an infinite number of choices on $\Bbb {Z, R}$

  • $\begingroup$ Can there exist an infinite set $S$ such that there are only finitely many operations $+$ and $\cdot$ such that $(S, +, \cdot)$ is a ring? $\endgroup$
    – tylerc0816
    Sep 26, 2013 at 20:36
  • 1
    $\begingroup$ No. Given any infinite ring $S$, there are an infinite number of permutations of the elements. For each permutation $p(x)$, I can define $a \oplus b =p^{-1}( p(a)+p(b)) \quad \text{and} \quad a \odot b = p^{-1}( p(a) \cdot p(b))$ $\endgroup$ Sep 26, 2013 at 20:39
  • $\begingroup$ @tylerc0816 Of course, everything changes if you fix one of the operations. For example, if you took an abelian group $(G,+)$ and asked how many operations $\cdot:G\times G\to G$ make $(G,+,\cdot)$ into a ring, the answer can be none! This is true even if you ask $G$ to be infinite (e.g $(\mathbb{Q}/\mathbb{Z},+)$). $\endgroup$ Sep 26, 2013 at 23:16

Given a ring, one way to put a new ring structure on it is by relabelling the elements. For example, we might decide to call $1$ $2$, call $2$ $3$, call $-2$ $-1$ and so on. All that is important is that this map is a bijection - a one-to-one mapping. Once we have relabelled the elements, we can then apply the usual addition and multiplication operations to the new labels. Mathematically, if we have a bijection $\phi$ from the ring to itself, then the operations given by conjugating the original operations with $\phi$; i.e.:

\begin{align} a\oplus b&=\phi^{-1}(\phi(a)+\phi(b))\\ a\otimes b&=\phi^{-1}(\phi(a)\phi(b))\\ \end{align}

Since there are uncountably many permutations of the elements of $\mathbb Z$, we can put uncountably many different ring structures on it. However, these ring structures aren't really different, and in mathematics we say that they are isomorphic; they have the same structure up to rearranging and relabelling the elements. Specifically, $\phi$ is, pretty much by definition, an isomorphism from the old ring to the new one.

Taking this example a bit further, we see that even if we retain the multiplication operation on $\mathbb Z$, there are infinitely many choices for the addition operation.

For example, given an integer $n$, define $\phi(n)$ as follows: prime factorize $n$, and then swap the exponents on $2$ and $3$. So $4=2^2$ becomes $3^2=9$ and $60=2^2\times3^1\times5^1$ becomes $2^1\times3^2\times5^1=90$. Then let the addition operator $\oplus$ be defined by:

$$ a\oplus b := \phi^{-1}(\phi(a)+\phi(b)) $$

[Note: in this case, $\phi^{-1}=\phi$, but that won't always be the case. Now let's check that our new ring $(\mathbb Z,\oplus,\times)$ still satisfies the ring axioms.

It's fairly clear that $\mathbb Z$ is still an abelian group under the new operation (the conjugation with $\phi$ can be thought of as a 'relabelling' of the members of $\mathbb Z$, as above), so we just need to check the distributive law. Before we do that, we should state one important property of the function $\phi$: it is multiplicative; i.e., it satisfies $\phi(ab)=\phi(a)\phi(b)$. This is fairly easy to show, if you think about it. Note also that $\phi^{-1}$ is multiplicative (for example, since $\phi^{-1}=\phi$). Then:

\begin{align} a(b\oplus c)&=a\phi^{-1}(\phi(b)+\phi(c))\\ &=\phi^{-1}(\phi(a))\phi^{-1}(\phi(b)+\phi(c))\\ &=\phi^{-1}(\phi(a)(\phi(b)+\phi(c)))\\ &=\phi^{-1}(\phi(ab)+\phi(ac)\\ &=ab\oplus ac \end{align}

Now the above proof works equally well if we replace the integers with any ring, and the function $\phi$ with any multiplicative function on the ring. We interchanged the roles of $2$ and $3$, but you can permute the prime numbers in any way you like and still have this work. So there are again uncountably many possibilities.

Unfortunately for us, the new ring is still isomorphic to the old one. Once again, $\phi$ provides us with an isomorphism: $phi(a+b)=\phi(a)\oplus\phi(b)$ almost by definition, and $\phi$ is multiplicative as before. In fact, this is really a special case of the example in the first paragraph: $\phi$ is a permutation of $\mathbb Z$ that is specially chosen to preserve the multiplication operation.

If we want to come up with an example of a ring structure on $\mathbb Z$, though, we still can. First of all, notice that the following defines a ring structure on the set of pairs of numbers $(a,b)$:

\begin{align} (a,b)+(c,d)&=(a+b,c+d)\\ (a,b)(c,d)&=(ab,cd) \end{align}

This ring is certainly not isomorphic to $\mathbb Z$: indeed, in $\mathbb Z$, any two non-zero numbers multiply to give a non-zero number, while here, we have $(1,0)(0,1)=(0,0)$. Now it is well known that we can put this set into one-to-one correspondence with $\mathbb Z$ arrange the pairs of numbers $(a,b)$ in increasing order of $|a|+|b|$, then by $|a|$, then by $a$, then by $b$:

$$ (0,0),(0,-1),(0,1),(-1,0),(1,0),(0,-2),(0,2),(-1,-1),(-1,1),(1,-1),(1,1),(-2,0),(2,0),\dots $$

(even if you don't quite follow my explanation of the ordering, I hope you appreciate that there's some way to write these pairs as a sequence). Then go along the sequence labelling each pair by $0,-1,1,-2,2,-3,3,\dots$. This in turn gives a relabelling on $\mathbb Z$, which can be used to define a new ring structure on $\mathbb Z$ which is not isomorphic to the original ring. This trick works for any ring with countably infinitely many elements (i.e., one that can be put into bijection with $\mathbb Z$), and in general, the number of non-isomorphic ring structures that can be put on a ring is equal to the number of rings with the same number of elements (or same cardinality) - just take a map from one ring to the other, and conjugate the addition and multiplication operations with it as in the first paragraph.


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