# Can there be different ring structures on the same set?

Say we have the set of real numbers. Can we construct a different ring than the usual ring of real numbers?

I am trying to wrap my head around the idea of rings and I couldn't find two other operations that will hold distributive property other than usual addition and multiplication.

If we can find another ring with the set of real numbers, is there a set with only one ring structure? If we can't find another ring with the set of real numbers, is there a set with different ring structures?

• Definitely. Consider that $\Bbb Z$ and $\Bbb Q$, for example, can be considered the same set since they are in bijection. Their usual ring structures are clearly not-equivalent since $\Bbb Q$ is a field, for example. Jun 1, 2021 at 23:53
• Any set with two elements has precisely one ring structure: $\mathbb{Z}/2\mathbb{Z}$
– E G
Jun 1, 2021 at 23:57

Here is a more explicit example if my comment about $$\Bbb Z$$ and $$\Bbb Q$$ was unconvincing.

Let $$S=\{a,b,c,d\}$$. Let $$R_1$$ be the ring structure on $$S$$ inherited from $$\Bbb Z/(2)\times\Bbb Z/(2)$$ under the bijection $$a\mapsto (0,0)$$, $$b\mapsto (1,0)$$, $$c\mapsto (0,1)$$, and $$d\mapsto (1,1)$$. For example, $$ab=a$$, $$b+d=c$$, and $$cd=c$$.

Now let $$R_2$$ be the ring on $$S$$ inherited from $$\Bbb Z/(4)$$ with the bijection $$a\mapsto 0$$, $$b\mapsto 1$$, $$c\mapsto 2$$, and $$d\mapsto 3$$.

You can verify that $$R_1$$ and $$R_2$$ are not isomorphic since, for example, $$R_1$$ has no element of (additive) order $$4$$.

The other examples given are nice, but to give an example where the set is $$\mathbb{R}$$ as you were originally wondering about you take $$\mathbb{R}$$ with the normal multiplication but where we define the multiplication operation to always evaluate to $$0$$. Under this definition we have a different ring structure and it's certainly not isomorphic since this structure is not a field. This isn't a particularly nice ring since it doesn't have unity.

• I don’t think that will be a unital ring since $(2)(2)=(1+1)(1+1)=(1)(1)+\cdots+(1)(1)=4$ Jun 2, 2021 at 0:09
• @ElliotG correct, but OP didn't ask for that Jun 2, 2021 at 0:18

Given any bijection $$\,f:\mathbb{R}\to\mathbb{R},\,$$ define new addition and multiplication operations by $$\,x\oplus y:=f^{-1}(f(x)+f(y))\,$$ and $$\,x\otimes y:=f^{-1}(f(x)\cdot f(y)).\,$$ These two operations form $$\,\mathbb{R}\,$$ into a ring isomorphic the the usual ring of reals. As a concrete example, try $$\,f(x):=x+1.\,$$ This example is given in an answer to MSE question 1911294 "Operations on the vector set $$\mathbb{R}$$ that will provide a vector space".

• Being isomorphic it is not a "different ring structure". This is known as transport of structure Jun 2, 2021 at 10:33
• @BillDubuque That depends on exactly what the definition of "different ring structure" is. The question was not precise on that point. I did mention that the new ring is isomorphic to the original ring. Finding a non trivial non isomorphic ring on $\mathbb{R}$ is a much more difficult question. Jun 2, 2021 at 10:46