Does the set of all fields exist ?

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.

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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 $\kappa$.