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We often say "let F be a field", so I was wondering if we could consider, in ZFC, the set of all fields without some contradictions arising (so that we wouldn't have to use the global axiom of choice in the class of all fields to choose one)

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From the (supposed) set of all fields, or even of all two-element fields, we can produce the "set" of all sets. – André Nicolas Jan 10 at 7:23
In addition to what @AndréNicolas said, you can think about sizes. If there was a set of all fields, you could also take the union of all fields (as sets), and the cardinality of this union would be an upper bound on the cardinalities of all fields. But there are fields of arbitrarily large cardinalities, a contradiction. – Dan Shved Jan 10 at 7:27
@AndréNicolas Thanks, that answers my question (since with two elements, it's easy to make operations that would make it a field, imitating Z/2Z. And then the axiom of union would make the set of all sets) – Mageek Jan 10 at 7:30
up vote 19 down vote accepted

Exercise: every set is an element of (the underlying set of) some field. (HINT: Given a set $x$, let $y$ be any set other than $x$; now give $\{x, y\}$ a field structure.)

If you believe this, then if $F$ were the set of all fields, $\bigcup F$ would be the set of all sets.

Note that there's nothing special about fields here; the same argument works for

  • groups,

  • topological spaces,

  • rings,

  • semigroups with exactly 18 elements,

  • etc.

However, you could ask: "What about up to isomorphism?" That is, is there a set $S$ of fields such that every field is isomorphic to an element of $S$?

Here, the argument above doesn't work, but the answer is still "no" - for $\kappa$ a cardinal, let $F_\kappa$ be a field gotten by adjoining $\kappa$-many transcendentals to $\mathbb{Q}$. This shows that $S$ has to contain sets of arbitrarily large cardinality, which in turn (exercise) can be used to build the set of all ordinals, violating the Burali-Forti paradox.

Again, this same argument can be applied to practically any kind of structure.

However, note that the set of fields with a fixed underlying set does exist (powerset + separation), and therefore so do sets containing (up to isomorphism) every field of size $<\kappa$ for any fixed $\kappa$.

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(just a comment, related to the last paragraph): If we restrict to ordered fields, there are also the Surreal numbers Field (and the isomorphic maximal hyperreals Field) which contain all possible ordered fields as subfields. (Field instead of field because they are classes, not sets). See: Surreals – ypercubeᵀᴹ Jan 10 at 7:59

Check out the Lowenheim-Skolem theorem. It states that, since the first order theory of fields is countable and has an infinite model (i.e. An infinite set exists that satisfies the theory), it must have an infinite model of every cardinality. This implies that the "set" of all fields is a proper class, since there exists at least one field of every cardinality and the "set" of all cardinal numbers is a proper class.

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