Can someone explain what is the motivation behind the definition of a flat module? I saw the definition but I don't really know why it is important to work with these structures.

  • $\begingroup$ Tensor products conserve almost every property you could ask for. Tensors with flat modules really preserve almost anything you could want. And it turns out that lots of things are flat - for example, over PIDs, flat and torsion free are the same thing. $\endgroup$
    – davidlowryduda
    Mar 22, 2012 at 3:08
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    $\begingroup$ Are you convinced that tensors products are useful ? If you are, then you might want them to behave nicely. $\endgroup$
    – Joel Cohen
    Mar 22, 2012 at 9:07
  • $\begingroup$ @JoelCohen What do you mean by "useful" and behave nicely? $\endgroup$
    – Jr.
    Mar 23, 2012 at 0:27
  • $\begingroup$ In general, a tensor product of submodules is not a submodule of the tensor product (e.g. $0 \neq \Bbb Z/2\Bbb Z \otimes_{\Bbb Z} \Bbb Z$ while $\Bbb Z/2\Bbb Z \otimes_{\Bbb Z} \Bbb Q = 0$). But with flat modules, this doesn't happen! (Here $\Bbb Z/2\Bbb Z$ is not a flat $\Bbb Z$-module). $\endgroup$
    – Watson
    Jan 28, 2017 at 15:39

2 Answers 2


Flatness in commutative algebra satisfies a geometric condition: the fibers of a morphism between two varieties (schemes) don't vary too wildly. A flat morphism $f \colon X \to Y$ of varieties (schemes) can be thought as a continuous family of varieties (schemes) $\{ f^{-1} (y) \}_{y \in Y}$. An important theorem says that if $f \colon X \to Y$ is a flat morphism between two irreducible varieties then its fibers have dimensions equal to $\dim X - \dim Y$.

For example, fix an algebraically closed field $k$, and consider the morphism $f \colon \mathbb{A}^1 \to \mathbb{A}^1$ defined by $x \mapsto x^2$. It corresponds to the ring homomorphism $k[x^2] \hookrightarrow k[x]$. You should be able to prove that $k[x]$ is a flat $k[x^2]$-algebra (it is also free), hence the morphism $f$ is flat. The fibers are almost always made up of two points.

Consider the morphism $g \colon \mathbb{A}^2 \to \mathbb{A}^2$ defined by $(x,y) \mapsto (x, xy)$ (it is an affine chart of the blowing up of the plane). It corresponds to the ring homomorphism $k[x,xy] \hookrightarrow k[x,y]$. The fiber of the point $(0,0)$ is the line $\{ x= 0 \}$ which has dimension $1$, while the fiber of the other points is empty or made up of a single point. You should be able to check that $k[x,y]$ is not flat as $k[x,xy]$-algebra.

  • $\begingroup$ You could make this more precise: for example, you might mention that "a finitely-generated module over a noetherian integral domain is flat if and only if it is locally free of constant rank". But personally I think flatness is a concept from homological algebra and should be distinguished from local-freeness. $\endgroup$
    – Zhen Lin
    Apr 22, 2012 at 0:10

According to the answers to this MO question, flatness should really be thought of as an algebraic condition rather than a geometric one. Mumford [The red book of varieties and schemes, Ch. III, § 10] writes,

The concept of flatness is a riddle that comes out of algebra, but which technically is the answer to many prayers.

For the purposes of homological algebra, it is incredibly useful for a functor to preserve exact sequences, and flatness of a (right) $R$-module $M$ is precisely what is needed to make the functor $M \otimes_R -$ send exact sequences of (left) $R$-modules to exact sequences of abelian groups. For example, if $I$ is a (left) ideal of $R$, then we have an exact sequence of (left) $R$-modules: $$0 \longrightarrow I \longrightarrow R \longrightarrow R / I \longrightarrow 0$$ In general, the tensored sequence is only right exact: $$M \otimes_R I \longrightarrow M \longrightarrow M \otimes_R R / I \longrightarrow 0$$ It's not hard to see that the image of $M \otimes_R I \to M$ is the abelian group $M I$, so when $M$ is flat, the natural map $M \otimes_R I \to M I$ is an isomorphism. Conversely, if this holds for every (left) ideal $I$, then $M$ is flat. Morally, what this is saying is that a flat $R$-module has no "generalised" $R$-torsion; if $R$ is a principal ideal domain, then an $R$-module is flat if and only if it has no $R$-torsion in the ordinary sense. (Think of $M \otimes_R I$ as being a group of formal linear combinations of formal products; one possible way $M \otimes_R I \to M I$ could fail to be an isomorphism is if $m \cdot a = 0$ for some non-zero $m$ in $M$ and non-zero $a$ in $I$.)

Under certain circumstances, flatness coincides with other conditions which are more geometric in nature. For example, if $R$ is a commutative noetherian ring, then the following are equivalent for a finitely-generated $R$-module $M$:

  • $M$ is projective.
  • $M$ is flat.
  • $M$ is locally free, in the sense that there are $f_1, \ldots, f_n$ in $R$ such that $f_1 + \cdots + f_n = 1$ and each $M \otimes_R R[1 / f_i]$ is free.

Actually, we always have the implication projective $\Rightarrow$ flat, even when $M$ is not finitely generated and $R$ is non-commutative/non-noetherian (just so long as the axiom of choice holds!), but the commutative noetherian case is what is usual in algebraic geometry. Being locally free means that $M$ is the module of global sections of a finite-dimensional vector bundle $\tilde{M}$ over $\operatorname{Spec} R$. It is reasonably clear that the rank of $\tilde{M}$ is locally constant on $\operatorname{Spec} R$. We have a partial converse: if $R$ is a noetherian integral domain and the rank of finitely-generated module $M$ is constant on $\operatorname{Spec} R$, then $M$ is locally free and hence flat.

  • $\begingroup$ Mumford's words are really nice! $\endgroup$
    – Andrea
    Apr 22, 2012 at 9:55
  • $\begingroup$ Great answer, very helpful. $\endgroup$
    – Babai
    May 10, 2020 at 23:45

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