# Intuition for when modules are flat

I am having trouble building my intuition for when a module $M$ over a commutative ring $R$ is flat. Are there any standard examples of modules that are flat over $R$? Are there any standard examples of modules that aren't? Is there a general method that I can use to quickly determine whether a module should morally be flat or not?

The standard answer I hear when I ask this question is that there is a somewhat geometric criterion: for a flat module $M$, the dimension of the fibers of $M$ vary continuously over primes $p$ of $R$. I.e. the dimension of $M_p/pM_p$ over $A_p/pA_p$ should vary continuously with our choice of $p$.

Maybe I do not have enough intuition for fibers yet, but I do not find this to be very helpful. Even in basic cases, I am unsure of how to use this. For example, if I have an integral extension of a ring $R$, say $A=k[x,y]/(y^2-x^3) \subset B = k[x,y,x/y]/(y^2-x^3)$, is $B$ flat over $A$?

Another case of interest I have is when $A \subset B$ is an integral extension of rings. Is there a good way to tell if $B$ is flat over $A$?

• Flat modules are immoral thugs :P – Zelos Malum Oct 29 '15 at 4:56
• @ZelosMalum But I never know how to tell whether a module is an immoral thug or not =/ – BigMathTimes Oct 29 '15 at 4:58
• What is the definition of flatness you are using? The definition I'm accustomed to has a rather nice and intuitive interpretation in my mind, though I must confess I have rarely ever found it all that practical to use. – zibadawa timmy Oct 29 '15 at 19:41

There are a lot of nice theorems about flat modules over general rings. For example, over a domain, any flat module is torsion-free, and the converse holds over a principal ideal domain. (Note that the notion of being torsion-free does not make sense over rings that are not domains). Qiaochu mentioned some other nice theorems. Often, there is a bit of interplay between the algebra and the geometry, and it depends on your situation which approach you want to use.

But the fibre dimension criterion is not as useless as you think! For example, consider the example that you give of $k[t^2,t^3] \subseteq k[t]$ (I assume this is the example you mean; for this you have adjoin $\frac{y}{x}$, not $\frac{x}{y}$, because $x$ corresponds to $t^2$ and $y$ to $t^3$). Then you probably know that this gives an isomorphism away from the origin, i.e. when we invert $t$ it becomes an isomorphism. Thus, over all primes not equal to $(t^2,t^3)$, the fibre dimension is $1$.

However, at the prime $(t^2,t^3)$, we can just compute: $$\dim_k k[t]/(t^2,t^3) = 2,$$ with basis given by $1, t$ (note that for a maximal ideal $\mathfrak m$, we do need to localise, because $A/\mathfrak m$ is already a field). Thus, this extension of rings is not flat.

But I wouldn't say that there is a general method for testing whether an extension of rings is flat. In geometric situations (i.e. where both rings are finite type algebras over a field), often the best way to decide such questions is by looking at the geometry. One of the nicest theorems I know is the following (see Hartshorne, Thm III.9.9):

Theorem. Let $T$ be an integral Noetherian scheme. Let $X \subseteq \mathbb P^n_T$ be a closed subscheme. For each point $t \in T$, we consider the Hilbert polynomial $P_t \in \mathbb Q[z]$ of the fibre $X_t$, considered as a closed subscheme of $\mathbb P^n_{\kappa(t)}$. Then $X$ is flat over $T$ if and only if the Hilbert polynomial $P_t$ is independent of $t$.

What this says is that not only the dimension of a finitely generated module is constant in the fibres, but more generally if we have a projective morphism (a finite ring morphism is a special case of this), then the entire Hilbert polynomial is constant if and only if the morphism is flat.

You don't necessarily have to know precisely what this means to appreciate it: the Hilbert polynomial captures numerical invariants like the (Krull) dimension and the degree (for a finite morphism, this is the (linear) dimension of the module fibre), so for a flat family the fibre dimensions cannot jump. Often when you want to prove something is not flat, you have some geometric reason for it.

As an example of this, consider the ring homomorphism \begin{align*} \phi \colon k[x,z] &\to k[x,y]\\ x &\mapsto x,\\ z &\mapsto xy. \end{align*} Maximal ideals of $k[x,z]$ are of the form $\mathfrak m = (x-a, z-b)$, and for $a \neq 0$, $x$ becomes invertible in $B/\mathfrak mB$, so $$B/\mathfrak mB = k[x,y]/(x-a,xy-b) = k[x,y]/(x-a,y-\tfrac{b}{a}) = k,$$ so the fibre is just a point (Spec of a field). On the other hand, when $a = 0$ and $b = 0$, we get $$B/\mathfrak mB = k[x,y]/(x,xy) = k[x,y]/(x) = k[y],$$ so the fibre is one-dimensional. Geometrically, this is saying that the map \begin{align*} k^2 &\to k^2\\ (x,y) &\mapsto (x,xy) \end{align*} collapses the $y$-axis onto the origin, whereas everywhere else it is an isomorphism.

Thus, we conclude that $\phi$ is not flat, since some of its fibres have (Krull) dimension $1$, whereas most of them have dimension $0$.

• I just realised that the example I give at the end is not quite an example of the theorem I quote, for this map is not given by a closed immersion into a projective space. So it does not constitute a proof that the morphism is not flat (however, it really isn't). But the motivation still stands: flat morphisms have a lot of constancy in families. – Remy Oct 29 '15 at 6:50
• Sorry for reviving this post after so long. Does flatness require the vector space dimensions you consider to be equal? You say that the first morphism is not flat because all of its fibres have (vector space) dimension 1 except at one point, where it has dimension 2. I have also asked this here: math.stackexchange.com/questions/2564785/… – Nao Dec 13 '17 at 21:41

A standard class of examples is that every localization $S^{-1} R$ is flat. More generally, Lazard's criterion says that the flat modules are precisely the filtered colimits of free modules.

Flatness is a local condition, so in some sense it suffices to answer the question of when a module over a local ring is flat. A finitely generated flat module over a local ring is free, so for finitely generated modules (which includes the case of finite ring extensions) flatness is equivalent to being locally free.

• Yeah, I'm familiar with the fact that localizations are flat, but not faithfully flat. Also that flatness can be checked locally. I didn't know (didn't piece together from what I know) that being locally free is equivalent to be locally flat / flat. That's a nice characterization. – BigMathTimes Oct 29 '15 at 6:12
• Although there's something to be careful about here, I think. Discussion at MO: mathoverflow.net/questions/33522/flatness-and-local-freeness – Qiaochu Yuan Oct 29 '15 at 6:14