2
$\begingroup$

Does there exist a set containing all rings ?

Possible idea :I think such set is not possible.If S is a set containing all rings i think we can again define a structure on S to make it Ring and that is a contradiction because S cannot contain itself.

$\endgroup$
4
  • $\begingroup$ The collection of all rings would form a proper class. For example, imagine imposing a ring structure on each ordinal set. $\endgroup$ – user88524 Sep 13 '14 at 20:01
  • $\begingroup$ If we do not identify isomorphic rings, there is not even a set of all $2$-element rings. $\endgroup$ – André Nicolas Sep 13 '14 at 20:05
  • $\begingroup$ If $S$ is a set containing all rings, then a ring structure on $S$ is not the same thing as $S$ --- it is, at least, an ordered triple consisting of $S$, an addition rule and a multiplication rule. So the existence of this ring structure does not imply that $S$ contains itself; it implies only that $S$ contains a triple whose first component is $S$. So this argument doesn't quite work. There are, however, perfectly good arguments available (see answer by @William). $\endgroup$ – WillO Sep 13 '14 at 21:10
  • $\begingroup$ Related: math.stackexchange.com/questions/1606455 $\endgroup$ – Watson Aug 22 '16 at 13:15
2
$\begingroup$

Let $\mathcal{R}$ denote the class of all rings.

No, in $\mathsf{ZFC}$ set theory, the class of all ring is not a set for a very simple reason: $\mathcal{R}$ is essentially too big to be a set. The details are below.

Let $\mathbb{Z}$ denote the familiar ring of integers. For any set $x$, define $\mathbb{Z}_x = \{(n, x) : n \in \mathbb{Z}\}$. Defining the ring operation in the natural way, $\mathbb{Z}_x$ is a ring isomorphic to $\mathbb{Z}$.

Then there is an (proper class) injection of $V$ (the class of all sets) into $\mathcal{R}$. It is well-known that $V$ is not a set. Hence $\mathcal{R}$ is not a set.

$\endgroup$
1
  • $\begingroup$ In the same way as William suggested here, we can show that there is no set of all trivial rings. For a slightly more meaningful (imho) interpretation, see my answer. $\endgroup$ – tomasz Sep 13 '14 at 20:22
3
$\begingroup$

It is not possible, even if by "containing all rings" you mean "containing all rings, up to isomorphism".

You can prove it in much the same way you prove that there are infinitely many primes:

  1. Take any set of rings $X$.
  2. Consider the ring $2^{\bigcup X}$ (functions with pointwise addition and multiplication modulo $2$).
  3. This ring is larger than any element of $X$ (by Cantor's theorem), and therefore not isomorphic to any of them, so $X$ does not contain representatives for all isomorphism classes of rings.

The thing is, there are arbitrarily large rings. If a class contains arbitrarily large objects, it can't be a set (not even up to isomorphism, if the isomorphisms are bijections), at least not in ZF set theory, the same argument applies to groups, fields, algebraically closed fields, Banach spaces etc.

On the other hand, it makes sense (and is sometimes useful, i.e. for constructing universal objects) to consider the class of all objects of given size, up to isomorphism. For example, there is certainly a set of all rings of cardinality smaller than $2^{2^{2^{2^{\aleph_0}}}}$, up to isomorphism. This set will likely contain all rings you would ever care to think about (or something isomorphic to them).

$\endgroup$
1
$\begingroup$

Imagine there would be a set $R$ of all rings. Then there would be also the set of all underlying sets (just map each ring to the underlying set). Therefore also the union $U$ of those sets would exist. Since every underlying set of a ring in $R$ would be a subset of that union, it would have to at most at large as that union.

But we can define a ring on the powerset of U by taking the symmetric difference as addition and the intersection as multiplication. Since the powerset of $U$ is larger than $U$, it cannot be a subset of $U$, and therefore not the underlying set of a ring in $R$, in contradiction to the assumption that $R$ contains all rings.

$\endgroup$
0

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.