My belief is that no result in algebraic geometry that does
not explicitly engage the universe concept will fully
require the use of universes. Indeed, I shall advance an
argument that no such results actually need anything beyond
ZFC, and indeed, that they need much less than this. (But please note, I answer as a set theorist rather than an algebraic geometer.)
My reason is that there are several hierarchies of weakened
universe concepts, which appear to be sufficient to carry
out all the arguments that I have heard using universes,
but which are set-theoretically strictly weaker hypotheses.
A universe, as you know, is known in set theory as
$H_\kappa$, the set of all
hereditarily-size-less-than-$\kappa$ sets, for some
inaccessible cardinal $\kappa$. Every such universe also
has the form $V_\kappa$, in the cumulative Levy hierarchy,
because $H_\kappa=V_\kappa$ for any beth-fixed point, which
includes all inaccessible cardinals. Thus, the Universe
Axiom, asserting that every set is in a universe, is
equivalent to the large cardinal assertion that there are
unboundedly many inaccessible cardinals.
This hypothesis is relatively low in the large cardinal
hierarchy, and so from this perspective, it is relatively
mild to just go ahead and use universes. In consistency
strength, for example, it is strictly weaker than the
existence of a single Mahlo cardinal, which is considered
to be very weak large-cardinal theoretically, and much
stronger hypotheses are routinely considered in set theory.
Nevertheless, these hypotheses do definitely exceed ZFC in
strength, unless ZFC is already inconsistent, and so your
question is a good one. It follows the pattern in set
theory of inquiring the exact large cardinal strength of a
given hypothesis.
The weaker universe concepts that I propose to use in
replacement of universes are the following, where I take
the liberty of introducing some new terminology.
- A weak universe is some $V_\alpha$ which models ZFC.
The Weak Universe axiom is the assertion that every set is
in a weak universe.
This axiom is strictly weaker than the Universe Axiom,
since in fact, every universe is already a model of it.
Namely, if $\kappa$ is inaccessible, then there are
unboundedly many $\alpha\lt\kappa$ with $V_\alpha$ elementary in $V_\kappa$, by the Lowenheim-Skolem theorem, and so
$V_\kappa$ satisfies the Weak Universe Axiom by itself.
From what I have seen, it appears that most of the applications of
universes in algebraic geometry could be carried out with
weak universes, if one is somewhat more careful about how
one treats universes. The difference is that when using
weak universes, one must pay attention to whether a given
construction is definable inside the universe or not, in
order to know whether the top level $\kappa$ of the weak
universe, which may now be singular (and this is the
difference), is reached.
- Let us say that a very weak universe is simply a
transitive set model of ZFC. (In set theory, one would want just to call these universes, but here that word is taken to have the meaning above; so we could call them set-theoretic universes.) The Very Weak Universe Axiom (or Set-theoretic Universe Axiom)
is the assertion that every set is an element of a very
weak universe.
The difference between a very weak universe and a weak
universe is that a very weak universe $M$ may be wrong
about power sets, even though it satisfies its own version
of the Powerset Axiom. Set theorists are very attentive to
such very weak universes, and pay attention in a
set-theoretic construction to which model of set theory it
is undertaken. If the algebraic geometers were to give
similar attention to this point, thereby turning themselves
into set theorists, I believe that all of their arguments
using universes could be essentially replaced with very
weak universes. Another important point is that while
universes are always linearly ordered by inclusion, this is
no longer true for very weak universes.
Now, even the Very Weak Universe Axiom transcends ZFC in
consistency strength, because it clearly implies Con(ZFC).
So let me now describe how one might provide an even
greater reduction in the strength of the hypothesis, and
capture a use of universes within ZFC itself.
The key is to realize that algebraic geometry does not
really use the full force of ZFC. (Please take this with
some skepticism, given my comparatively little exposure to
algebraic geometry.) It seems to me unlikely, for example,
that one needs the full Replacement Axiom in order to carry
out the principal goal constructions of algebraic geometry.
Let me suppose that these arguments can be carried out in
some finite fragment $ZFC_0$ of $ZFC$, for example, $ZFC$
restricted to formulas of complexity $\Sigma_N$ for some
definite number $N$, such as $100$ or so. In this case, let
me define that a good-enough-universe is $V_\kappa$,
provided that this satisfies $ZFC_0$. All such good-enough
universes have $V_\kappa=H_\kappa$, just as with universes,
since these will be beth-fixed points. The Good-enough
Universe Axiom is the assertion that every set is a member
of a good-enough universe.
Now, my claim is first, that this Good-enough Universe
Axiom is sufficient to carry out most or even all of the
applications of universes in algebraic geometry, provided
that one is sufficiently attentive to the set-theoretic
issues, and second, that this axiom is simply a theorem of
ZFC. Indeed, one can get more, that the various good-enough
universes agree with each other on truth.
Theorem. There is a definable closed unbounded class
of cardinals $\kappa$ such that every $V_\kappa$ is a
good-enough universe and furthermore, whenever
$\kappa\lt\lambda$ in $C$, then $\Sigma_N$-truth in
$V_\kappa$ agrees with $\Sigma_N$-truth in $V_\lambda$, and
moreover, agrees with $\Sigma_N$ truth in the full universe
$V$.
This theorem is exactly an instance of the Levy Reflection
Theorem.
OK, so if I am right, then the algebraic geometers can
carry out their universe arguments by paying a lot of
careful attention to the set-theoretic complexity of their
constructions, and using good-enough universes instead of
universes.
But should they do this? For most purposes, I don't think
so. The main purpose of universes is as a simplifying
device of convenience to stratify the full universe by
levels, which can be fruitfully compared by local notions
of large and small. This makes for a very convenient
theory, having numerous local concepts of large and small.
I can imagine, however, a case where one has used the
Universe theory to prove an elementary result, such as
Fermat's Last Theorem, and one wants to know what are the
optimal hypotheses for the proof. The question would be
whether the extra universe hypotheses are required or not.
The thrust of my answer here is that such a question will
be answered by replacing the universe concepts that are
used in the proof with any of the various weakened universe
concepts that I have mentioned above, and thereby realizing
the theorem as a theorem of ZFC or much less.
Here at least is the usual justification for moving from AC for sets to what is normally called the global axiom of choice, which asserts that there is a class well-ordering of the (first-order) universe.
Theorem.
The global axiom of choice, when added to the ZFC or
GB+AC axioms of set theory, leads to no new theorems about sets.
That is, the first-order assertions about sets that are provable
in GBC are precisely the same as the theorems of ZFC.
Furthermore, every model of ZFC can be extended (by forcing) to a
model of GBC, in which the global axiom of choice is true, while
adding no new sets (only classes).
In particular, the global axiom of choice is safe in the
sense that it will not cause inconsistency, unless the underlying
system without the global axiom of choice was already inconsistent.
Proof. Suppose that $M$ is any model of ZFC. Consider the class
partial order $\mathbb{P}$ consisting of all well-orderings in $M$
of any set in $M$, ordered by end-extension. As a forcing notion,
this partial order is $\kappa$-closed for every $\kappa$ in $M$,
since the union of a chain of (end-extending) well-orderings is
still a well-order. If $G\subset\mathbb{P}$ is $M$-generic for
this partial order, then $G$ is, in effect, a well-ordering of all
the sets in $M$. Furthermore, one can prove by the usual forcing
technology that the structure $\langle M,{\in},G\rangle$ satisfies
$\text{ZFC}(G)$, that is, where the predicate $G$ is allowed to
appear in the replacement and other axiom schemes.
Essentially, what we've done is add a global well-ordering of the
universe generically. And since the forcing was closed, no new
sets were added, and so $M[G]$ has the same first-order part as
$M$.
It follows now that GBC is conservative over ZFC for first-order
assertions, since any first-order statement $\sigma$ that is true
in all GBC models will be true in $M[G]$ and therefore also in
$M$, and so $\sigma$ is true in all ZFC models as well. QED
Best Answer
Universes provide a convenient framework for stacking many levels of largeness. But a class in NBG cannot contain any other class. The most basic application which shows that universes are more convenient is that the category of functors $[\mathcal{C},\mathcal{D}]$ between two $U$-categories $\mathcal{C},\mathcal{D}$ is again a $V$-category if $U \in V$ are two nested universes. You cannot even write down this construction in NBG (unless you use inaccessible cardinals, which then is more or less equivalent to the universe approach).