11
$\begingroup$

What is an example of a ring that has nothing to do with numbers? For example, for groups, we have dihedral groups, quaternions, etc.

I'm missing the analogue of the characteristic of a ring, when the ring is not related to numbers (complex, real, integers,...).

Part of why I want this example is because I want to see if the characteristic of a ring must be an element of the ring itself.

Improve this question
$\endgroup$
  • 2
    $\begingroup$ Does your definition of a ring have a (multiplicative) identity I? If so, then the characteristic is always a number or something 'number-equivalent' because it's I+I+I+... n times. $\endgroup$ – Steven Stadnicki Dec 2 '20 at 19:51
  • 6
    $\begingroup$ The characteristic is always an integer, so if your ring is not related to numbers, the characteristic can't be an element, obviously. $\endgroup$ – user436658 Dec 2 '20 at 19:53
  • $\begingroup$ What about matrices or polynomials? $\endgroup$ – J. W. Tanner Dec 2 '20 at 19:56
  • 3
    $\begingroup$ @StevenStadnicki And in any ring with identity and characteristic $n$ we'd have $n\cdot 1=0$, so the characteristic shows up as $0$ every time through that lens. :( $\endgroup$ – rschwieb Dec 2 '20 at 20:05
  • $\begingroup$ Related, I think :-) $\endgroup$ – Jyrki Lahtonen Dec 2 '20 at 20:28
34
$\begingroup$

Let $S$ be any set. We can turn its powerset $$P(S)=\{A\mid A\subseteq S\}$$ into a ring by defining the sum $$ A+B:=(A\setminus B)\cup (B\setminus A) $$ and the product $$ A\cdot B:=A\cap B. $$ The empty set is the zero element of this ring, and the full set $S$ plays the role of the multiplicative neutral element. Leaving it to you to verify all the ring axioms.

This ring is isomorphic to the ring of functions $f:S\to\Bbb{Z}_2$ with addition and product defined pointwise. I chose to write it as above, because then no numbers are present (as requested in the title).

Clearly the ring $(P(S),+,\cdot,\emptyset,S)$ has characteristic two.

Improve this answer
$\endgroup$
  • 7
    $\begingroup$ Implicitly we have the "numbers" $\Bbb Z/2\Bbb Z$ in it, but that would be too strict, I know. So the example is very good, I think. $\endgroup$ – Dietrich Burde Dec 2 '20 at 20:06
17
$\begingroup$

Part of why I want this example, is because I want to see if the characteristic of a ring must be an element of the ring itself.

It is not really sensible to think of the characteristic of a ring as being an element of the ring, even when it literally is.

Characteristic is a sort of "dimension" on the ring, having a value in the natural numbers.

It's kind of like asking if the dimension of a vector space can be an element in the vector space or not.

Now, coincidentally the characteristic of $\mathbb Z$ is in $\mathbb Z$ if you like $\mathbb N$ being a subset of $\mathbb Z$, but the observation is not really useful.

The closest you can come to putting the characteristic "in" the ring is if the ring has identity so that you can see the homomorphic image of $\mathbb Z$ within your ring. But even then, the characteristic always maps to the $0$ element of your ring, so nothing is achieved.

An example, though?

I dunno, can you think of an abelian group $G$ that doesn't involve numbers? if so, the endomorphism ring $End(G_\mathbb Z)$ is a ring that apparently does not involve numbers.

Improve this answer
$\endgroup$
  • $\begingroup$ There's also the fact that from one point of view, there isn't so much "the ring $\mathbb Z$", but rather the equivalence class of rings of which $\mathbb Z$ is a representation. $\endgroup$ – Acccumulation Dec 3 '20 at 5:31
  • $\begingroup$ @Acccumulation Equivalence classes of pairs of natural numbers? Probably still qualifies as number-related. $\endgroup$ – rschwieb Dec 3 '20 at 13:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.