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Given any algebraic object $X$, say group, ring, integral domain, etc., and a special subset $I$ of $X$ namely normal subgroup, ideal etc., it is always possible to put a structure on $X/I$ induced from $X$.

Now, I will forget $I$ and the existing structure of $X$ also. I will consider $X$ as a set only.

If $X$ is finite set, then we can always give group structure or ring structure on it. The actual problem will come if $X$ is infinite set, and I don't know whether we can always make $X$ into a group or ring with some binary operations.

Question: Given any infinite set $S$, is it always possible to put (1) structure of group on $S$? (2) structure of ring on $S$? (3) structure of field on $S$?

Only thing I know that given any infinite set $S$, the power set of $S$ can be made into an algebra (hence group and ring structure is coming here; but where this is exactly? It is on the power set of $S$, not necessarily on $S$. Thus, in question, I stress on only set $S$ given in our hand, and try to make it a group or ring or field. Is this always possible?

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1  
Yes, yes, yes. See the Löwenheim–Skolem theorem. – bof 3 hours ago
2  
The first two paragraphs don't seem to have anything to do with the rest of the question. The first paragraph is also a bit misleading: in general, quotient objects are described by more complicated things than special subsets. For example, quotients of monoids cannot be described in terms of "normal submonoids." – Qiaochu Yuan 3 hours ago
    
First two paragraphs: I am familiar with putting structures on some new objects from structures on old objects; in algebra, I didn't see any other examples than "structure on quotient". My point was to put structure on an object, without help of other objects. – p Groups 2 hours ago
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To answer all three questions it suffices to construct a field of any given infinite cardinality, since a field is a ring which is additively a group. To that effect, let $A$ be an infinite set. Then the field $\mathbb Q(A)$ of rational functions over $\mathbb Q$ using the elements of $A$ as indeterminates is a field with the same cardinality as $A$.

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This is much clear, how a structure on $A$ can be given. – p Groups 3 hours ago
3  
@p Groups: if I understand your question correctly, you need to pick a bijection between $A$ and $\mathbb{Q}(A)$, then transport the structure from the latter to the former. It's worth pointing out that the axiom of choice is required to do this: in fact the axiom of choice is equivalent to the claim that every nonempty set has a group structure (mathoverflow.net/questions/12973/…). – Qiaochu Yuan 2 hours ago
up vote 4 down vote

Yes, many results of this form follow from the Löwenheim–Skolem theorem, which asserts that if a first-order theory (a particular way to write down axioms something should satisfy; this includes groups, rings, fields, and more) has an infinite model then it has a model of every infinite cardinality.

The Löwenheim–Skolem theorem has much weirder consequences than this; for example, it implies that there is a countable model of ZF set theory (which is supposed to be a first-order theory of all sets), as well as an uncountable model of (first-order) Peano arithmetic (which is supposed to be a first-order theory of the natural numbers).

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Good answer, but to be strictly accurate I think you have to say something about the theory being countable. For instance, if you formalize the notion of a real vector space as an Abelian group augmented with uncountably many unary operations (one for scalar multiplication by each real number), there is no model of cardinality $\aleph_0.$ – bof 2 hours ago

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