## Flat Modules and Morphisms

Especially after yesterday, today’s going to feel very algebraic. That’s because we’re gearing up and pulling out the tools we’ll need to discuss one of the major success stories of algebraic geometry – moduli space theory. The part that matters is the notion of a family of objects. Yes, objects, we’re not only going to care about varieties, but also families of schemes, families of sheaves, families of divisors, and, eventually, we may even mention families of functors.

Now, we want the families to be continuous in some sense. The sense we want isn’t, however, actual continuity. For one thing, in the Zariski topology, continuity is a funny thing. For another, even though we’re working primarily over $\mathbb{C}$, we still want to be algebraic about this, because it’s hard to tell what it means for a family of sheaves to be continuous. The notion we need is flatness.

To define it, we first jump all the way to some category theory. I apologize for mentioning functors in the Blowups post, and now I will define them. If $\mathcal{C},\mathcal{D}$ are two categories, a functor is a map $F:\mathcal{C}\to \mathcal{D}$ taking objects to objects and morphisms $f:C\to D$ to morphisms $F(f):F(C)\to F(D)$ such that $F(1_C)=1_{F(C)}$, $F(f\circ g)=F(f)\circ F(g)$.

So now, we call a functor faithful if the maps $\hom(M,N)\to\hom(F(M),F(N))$ are all injective. And now I use a sleight of hand and ignore the phrase “abelian category” and specialize to categories of modules. That is, for any ring, we define $Mod(R)$ to be the category of modules over that ring. For categories of modules, a functor is faithful if and only if the only map $u:M\to N$ taken to the zero map $0:F(M)\to F(N)$ is the zero map itself.

Now, we say that a functor on module categories is exact if it takes short exact sequences to short exact sequences. We’re now ready to give a very unhelpful for conception but great for theory definition of what it means for a module to be flat. We let $M$ be an $A$-module. Then we can define a functor by $M\otimes_A -$, that is, on objects, it tensors with $M$ and on morphisms it tensors with the identity morphism of $M$. If this functor is exact, then we say that $M$ is flat. If it is faithful and exact, then $M$ is faithfully flat.

Now, a couple of quick consequences. The faithfulness requirement, once we assume flat, is just saying that $M\otimes_A N=0$ if and only if $N=0$ in the first place. So next, let $A\to B$ be a ring homomorphism and $M,N\in Mod(A)$ and $P\in Mod(B)$. Then we have that $M,N$ are flat (or faithfully flat) implies that $M\otimes_A N$ is too (compositions of exact and faithful functors are exact and faithful), if $M$ is flat over $A$ then $M\otimes_A B$ is flat over $B$, and if $B$ is flat over $A$ and $P$ is flat over $B$, then $P$ is flat over $A$.

So there’s a pile of algebra. Time to use this to define a class of morphism. Let $f:X\to Y$ be a morphism of locally ringed spaces, and let $\mathscr{F}$ be an $\mathscr{O}_X$-module. We say that $\mathscr{F}$ is flat over $Y$ at $x\in X$ if $\mathscr{F}_x$ is flat as an $\mathscr{O}_{f(x)}$-module. It’s flat over $y\in Y$ if it is flat at each preimage of $y$, and it is flat over $Y$ if it is flat over every $y\in Y$. We say that it is faithfully flat over $Y$ if $\mathscr{F}$ is flat over $Y$ and $\mathscr{F}\otimes k(y)\neq 0$ for every $y\in Y$, where $k(y)$ is the structure sheaf of the point $y$ pushed forward along its inclusion. Equivalently, this tensor product is just $\mathscr{F}$ as a sheaf on $X$ tensored with the residue field $\mathscr{O}_{y,Y}/\mathfrak{m}_y$ at $y$. Finally, we call a morphism $f:X\to Y$ of schemes flat if $\mathscr{O}_X$ is flat over over $Y$ and $f$ is surjective. (and faithfully flat if it is faithfully flat)

Now, a few quick statements about flat maps:

1. Open immersions are flat (an open immersion is an inclusion of an open subscheme)
2. Composition of flat maps are flat
3. Products of flat morphisms are flat

Now, we take a moment and look at fiber products. These can be used to do one really nice thing: define the fibers of a map. This should be unsurprising considering the name. Let $f:X\to Y$ be a morphism and let $i:\mathrm{Spec} k(y)\to Y$ be the inclusion of a point. Then the fiber over $y$ will just be $X_y=X\times_Y \mathrm{Spec}k(y)$ via these maps. As we care about the geometric situation, we’re only going to look at the “nice” points, ie, those with structure sheaf $\mathbb{C}$, and so the $k(y)$ are all the same. This emphasizes that this is a fibered product of the morphisms, and that the morphisms matter greatly.

Now, I’m not going to prove the following more geometric statements, but here are a few nice things about flat morphisms:

• If $f:X\to Y$ is a flat morphism of varieties, then $\dim X_y=\dim X-\dim Y$. (note, we are requiring varieties to be irreducible, so the dimension doesn’t depend on the point we choose to look at)
• Let $Y$ be a curve, and let $X$ be a subscheme of $\mathbb{P}^n\times (Y\setminus P)$ for some point $P\in Y$. We have the natural map $X\to Y\setminus P$. Assume this is flat. Then there exists a unique subscheme $\bar{X}\subset \mathbb{P}^n\times Y$ which is flat over $Y$ and restricts to $X$. That is, if we have a flat family of projective schemes over a punctured curve, we can take the limit and there’s a uniquely determined scheme over the missing point that completes the family.
• Let $T$ be an integral noetherian scheme, say, a variety. Let $X\subset\mathbb{P}^n\times T$ be a closed subscheme, and for each fiber, we define the Hilbert Polynomial in the usual way and denote it by $P_t$. $X$ is flat over $T$ if and only if the Hilbert polynomial is independent of $T$.

That last one is HUGE. It’s really where this stuff starts to come together. It tells us immediately that there are no flat families over an irreducible base such that one fiber is a smooth plane conic and another is a smooth plane cubic: degree has to be constant. As does arithmetic genus.

As an example, the space of plane conics. Now, a degree two homogeneous polynomial in three variables looks like $a_0x^2+a_1xy+a_2y^2+a_3xz+a_4yz+a_5z^2=0$. The coefficients naturally belong to a projective space $\mathbb{P}^5$. So then we have a family $X\subset \mathbb{P}^2\times\mathbb{P}^5$, where $X$ is defined to be the vanishing of the polynomial as written, with $x,y,z$ coordinates on the $\mathbb{P}^2$ and the $a_i$ the coordinates on $\mathbb{P}^5$. So we get a family $X\to \mathbb{P}^5$. So is this family flat? Well, all we need to check is that the Hilbert polynomial is constant. As every fiber is a curve, we can write it as $an+b$ for some $a,b$. Now, they’re all conics, though they may be degenerate, so we get that the degree is 2. Way back, we saw that the Hilbert polynomial of a plane curve (no hypothesis on smoothness!) depends only on the degree. As it happens, this also holds for nonreduced subschemes of the plane, for instance, $x^2=0$. So the Hilbert Polynomial doesn’t depend on $t\in\mathbb{P}^5$, so the family is flat. We can repeat this exercise for plane curves of ANY degree, and get the same answer.

We’re going to continue in this direction for awhile, and we’ll forget the distinction between a morphism and a family, so we will speak of flat families and flat morphisms interchangeably.