Does there exist a semi-ring $(R,+,\cdot)$ (like a ring, but there must be no additive inverses and the $0$ is multiplicatively absorbing by axiom) with isomorphic additive and multiplicative structures? This means there should be a monoid-isomorphism


First thing I noticed is that there must be an absorbing element for the addition, lets call it $\omega=\varphi^{-1}(0)$. There follow many more such "strange" elements, e.g. $\varphi^{-1}(\omega)$, etc. Because addition and multiplication are so similar, the multiplication is automatically commutative and has an identity element $1$. But this is essentially as far as I came.

If the answer to above question is "Yes", then it would be very interesting to see if also $\Bbb N$ can be extended to such a semi-ring.

Note that any natural number $n$ can be represented as a sum of two natural numbers in exactly $\lceil (n+1)/2\rceil$ ways. So for any $n\in\Bbb N$ there are at most two natural numbers that can be represented in exactly $n$ different ways as sum of two other natural numbers. But there are infinitely many numbers that can be written as product of two natural numbers in exactly $n$ ways. This means that this extension of $\Bbb N$ must contain many additional elements.

  • $\begingroup$ Consider the tropical semiring, equivalently min-plus algebra. $\endgroup$ – Wuestenfux Jul 14 '17 at 8:02
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    $\begingroup$ @Wuestenfux Ok, I searched for tropical semi-rings, but I am a bit confused. As I read, the addition is idempotent, the multiplication is not. Addition and multiplication seem not to induce isomorphic monoids, right? $\endgroup$ – M. Winter Jul 14 '17 at 8:18
  • $\begingroup$ On extending $\mathbb{N}$, do want just addition preserved or multiplication as well? $\endgroup$ – badjohn Jul 14 '17 at 9:50
  • $\begingroup$ @badjohn Both. $\Bbb N$ should be sub-semiring of $R$. $\endgroup$ – M. Winter Jul 14 '17 at 9:54

Here's an example: $\left(\left[0,1\right],\max,\min\right)$, i.e. the max-min semiring.

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    $\begingroup$ More generally, any distributive lattice $L$ equipped with an isomorphism $\varphi:L^{op} \rightarrow L$ has this property. For example, if $L$ is a Boolean algebra, then we can take $\varphi$ to be negation. $\endgroup$ – goblin Jul 14 '17 at 9:07

By "must be no", do you mean actually prohibited or just not required? If prohibited then there is a problem with 0.

I was just working on some simple examples but Guy beat me to one of them.

The trivial semi-ring with one element would appear to qualify.

As well as Guy's two element example, both operations can be min. In this case 0 is both the additive and the multiplicative identity.

  • $\begingroup$ Thank you! Of course there can be additive inverses. I changed it. However, can you think about an example extending $\Bbb N$? $\endgroup$ – M. Winter Jul 14 '17 at 8:56
  • $\begingroup$ @badjohn I would just like to note that my example is not a two-element example, but rather the whole interval $\left[0,1\right]$ (although you could restrict it to $\left\{0,1\right\}$) $\endgroup$ – Guy Jul 14 '17 at 9:08
  • $\begingroup$ @Guy Sorry, I read your response too quickly and mistook it for one on my notepad. $\endgroup$ – badjohn Jul 14 '17 at 9:28

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