# Fine moduli space and modular curves

I am a bit confused about the definition of fine moduli spaces. As far as I understand, the difference between fine moduli space and coarse moduli space is the existence of universal family. For fine moduli space, suppose that $U\to M$ is the universal family, then given any family $X\to B$, there exists a unique morphism $B\to M$ such that we have a $B$-isomorphism $\alpha: X\to B\times_M U$. Is $\alpha$ require to be unique as well?

On one hand, I feel that the requirement of $\alpha$ to be unique is kind of unnatural. When we consider $M$ as a representable functor from the category of schemes to that of sets, it is natural to send a base $B$ to the isomorphic class of families. On the other hand, here is a proof from nlab that shows the $j$-line is not a fine moduli space. I don't see where the contradictions arises if $\alpha$ is not required to be unique.

In general, the modular curve $X_0(N)$ is not a fine moduli space for any $N$. I wonder if the fact that $-1$ is an automorphism of any family has anything to do with this result.

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In the examples I know of, e.g., the various moduli problems for elliptic curves with level structure, you're always looking at isomorphism classes of families, so the isomorphism $\alpha$ is not important. You just ask for an equality of isomorphism classes $[X]$ and $[B\times_MU]$. – Keenan Kidwell Apr 23 '12 at 16:54

I am working in the category of schemes over some fixed algebraically closed field $k$. Let $M$ be a putative moduli space for isomorphism classes of elliptic curves. Suppose there is a family $X \to S$ of elliptic curves such that all of the fibers $X_s$ are isomorphic to the same curve $E/k$ for closed points $s \in S$, but such that $X$ is not itself isomorphic to $S \times E$. It's easy to show that such a family exists (e.g. take $y^2 = x^3 + t$ over the affine $t$-line minus the origin).
There is then a map $f: S \to M$ such that $X$ is the pullback of the putative universal family. Traditionally, one concludes here that every closed point of $S$ must map to the same point of $M$ (the one corresponding to $E$) and so $S$ maps to a single point. This is a contradiction since the trivial family $S \times E$ maps to the same point.
However, the argument that $S$ must map to a point uses the uniqueness of the map Spec $k \to X$ corresponding to the elliptic curve $E$ over Spec $k$, which is the requirement you wish to drop.
If we drop the uniqueness requirement, we DO get a "moduli space" - here's an example of one - just take the disjoint union of all bases of families of elliptic curves, together with the obvious universal family (OK, there are set theoretic issues here, but they can be worked around). The point is that such a thing is horribly not unique, since why not just take two copies of any particular family? For a given functor $F$, once you've stopped asking that $F(X) = \text{Hom}(X, Y)$ for some $Y$, but only demanded that there be a surjection of functors $\text{Hom}(\text{_}, Y) \to F$, you've given up on Yoneda's lemma, which is what gives the uniqueness of $Y$. Note that you don't even get uniqueness up to isomorphism.