As a layman, I suspect my question is ill-formed and I will do my best to explain what it is I mean.

I find the notion of algebraic structures absolutely fascinating. I understand that there exist structures that require certain properties of addition and multiplication to hold. As an example, a ring requires an abelian group with a second binary operation that is associative and distributes over the operation in the abelian group. A field is an example of another similar structure.

The Question

Given some set $A$ Is there a name for a structure $S$ that requires certain properties of $(A, +)$ and $(A, \cdot)$ to hold that also hold given $(A, \cdot)$ is the additive structure and $(A, +)$ the multiplicative structure? I am trying to say that it matters not which structure we consider as additive and multiplicative as the properties of the $S$ hold regardless. In this sense $(A, +)$ and $(A, \cdot)$ are dual to eachother. Do these objects exist and are they interesting to study?

My gut feeling says that I am failing to write my intent accurately and apologize in advance.

  • $\begingroup$ Are you referring to, for example, the fact that the real numbers $\mathbb R$ with respect to addition is isomorphic as groups to the positive real numbers $\mathbb R^+$ with respect to multiplication, the isomorphism being $x\mapsto e^x$. $\endgroup$ Apr 29, 2015 at 14:38
  • 2
    $\begingroup$ Are you asking: does there exist a set $A$ and two binary operations $\oslash$ and $\odot$ so that $(A,\oslash,\odot)$ and $(A, \odot, \oslash)$ produce the same algebraic structure? $\endgroup$
    – Sloan
    Apr 29, 2015 at 14:40
  • $\begingroup$ This is what I am asking Sloan. I think i could find some trivial examples however I wonder if there are interesting examples. $\endgroup$
    – nSheahan
    Apr 29, 2015 at 14:41
  • 4
    $\begingroup$ In particular, boolean algebras maybe? $(B,\vee,\wedge)\cong(B,\wedge,\vee)$ $\endgroup$ Apr 29, 2015 at 14:52

2 Answers 2


As I understand it, you're looking for a set $A$ with two operations $\cdot$ and $+$ such that $(R, +, \cdot)$ is a ring and $(R, \cdot, +)$ is a ring.

Unfortunately, this can't really happen. It's not hard to show that in any ring, $0 \cdot r = 0$ for all $r \in R$. So $1 + r = 1$ for all $r \in R$ also, since $(R, \cdot, +)$ is a ring. But this implies $r = 0$ for all $r \in R$, so the only example is the trivial ring.

  • $\begingroup$ Thank you for your answer. This is interesting as it relates to rings but I am wondering what happens if we remove the requirements that the structure is a ring. Are there any interesting examples of structures that have $(A, +, \cdot) = (A, \cdot, +)$ in general, not necessarily a ring? $\endgroup$
    – nSheahan
    Apr 29, 2015 at 14:47

One example that fits this sort of "swap the operations" requirement is a lattice, though these two operations don't really behave like addition and multiplication. For these structures the operations are called join ($\vee$) and meet ($\wedge$) and are required to obey the following so-called absorption laws:

$a \vee (a \wedge b) = a$

$a \wedge (a \vee b) = a$

As you can see this is very much not a ring. Due to the symmetry of the definition, you'll note that if $(L,\vee,\wedge)$ is a lattice then $(L,\wedge,\vee)$ is also a lattice, though not necessarily the same one. Lattices essentially have one foot in abstract algebra and the other in order theory, and this duality is precisely the duality of the corresponding posets. If a lattice is isomorphic to its dual it is called self-dual though this property is not shared by all lattices. Also the word "lattice" is used elsewhere in algebra to refer to something completely different, so you need to be carful which one is being discussed when reading.


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