Of course "flatness" is a word that evokes a very particular geometric picture, and it seems to me like there should be a reason why this word is used, but nothing I can find gives me a reason!

Is there some geometric property corresponding to "flatness" (of morphisms, modules, whatever) that makes the choice of terminology obvious or at least justifiable?

  • $\begingroup$ What is the flat ness you are having in mind? Flat morphisms of schemes? $\endgroup$
    – user709
    Nov 25, 2009 at 11:53
  • 2
    $\begingroup$ Flat morphisms of schemes would be good, although I think most of the different things called "flat" in algebraic geometry are pretty closely related, so really anything would work. $\endgroup$ Nov 25, 2009 at 12:10
  • 56
    $\begingroup$ The term was introduced by Jean-Pierre Serre. You could ask him or, better, suggest he signs up at MO :P $\endgroup$ Nov 25, 2009 at 12:45
  • 73
    $\begingroup$ @Mariano: A couple of weeks ago I asked Serre about this. He didn't remember why the word flat was used, or if the word was due to him or possibly Cartan/Eilenberg. One point he emphasized is that it was Grothendieck who deserves all credit for the discovery of the importance of flatness in geometry (fibral criteria, families, etc.). For Serre it was a matter of isolating the "right" algebraic notion with which to discuss the various changes of rings (analytic vs. algebraic local rings, completions thereof, general localization) which came up in GAGA and FAC. $\endgroup$
    – BCnrd
    May 23, 2010 at 6:11

11 Answers 11


A lot of people will tell you that flatness means "continuously varying fibres" in some sense, and that flatness was invented to have correspondingly nice consequences, which is true. But there is a way to expect this (vague) interpretation a priori from an alternative, equivalent definition:

An $A$-module $M$ is flat $\iff$ $I \otimes_A M \to IM$ is an isomorphism for every ideal $I$.

I would prefer to present this as the definition of flatness, and present the fact that tensoring with $M$ preserves exact sequences as a theorem. Why?

Thinking "geometrically", $I$ just corresponds (uniquely) to a closed subscheme $Z=Z(I)=$
$=Spec(A/I)\subseteq Spec(A)$. If we think of $M$ in the usual geometric way as a module of generalized functions on $X$ (like sections of a bundle), and $M/IM \simeq M\otimes_A A/I$ as its restriction to $Z$, then the above definition of flatness can be interepreted directly to mean that $M$ restricts nicely to closed subschemes $Z$.

More precisely, it says that what we lose in this restriction, the submodule $IM$ of elements which "vanish on $Z$", is easy to understand: it's just formal linear combinations of elements $i\otimes m$, with no surprise relations among them, i.e. the tensor product $I \otimes_A M$.

In topology, continuous functions "restrict nicely" to points and closed sets (by taking limits), so you can see, without much experience at all, how this definition corresponds in an intuitive way to continuity.

Having this motivation in place, the best thing to do is to check out examples along the lines of Dan Erman's answer to see the analogy with continuity and limits at work.

  • 22
    $\begingroup$ I guess I'm almost a couple years late, but this is really a very nice answer. $\endgroup$ Aug 14, 2011 at 14:29
  • 6
    $\begingroup$ This should be taught in every elementary school $\endgroup$
    – rmdmc89
    Jan 15, 2021 at 19:19
  • $\begingroup$ Is this answer implicitly implying, among other things, that in the case of manifold there is no notion of flat continuous function because every function is "trivially" continuous? $\endgroup$ Mar 16, 2021 at 15:16

The key geometric meaning is that flat families are those families where the fibers vary "continuously". This notion allows one to talk about limits of families of algebraic varieties, which is particularly important in the study of deformation theory/moduli problems. Since the colloquial meaning of flatness also suggests a certain uniformity or lack of variation, one might imagine that this justifies its use in algebraic geometry.

For instance, if you have a flat family of projective varieties, then as Timo points out, the dimension of each fiber is the same. But more is true: the Hilbert polynomial of each fiber is also the same. This allows degeneration techniques. For instance, you can take a flat degeneration of your variety, compute a property about the degeneration, and then lift this information to your original variety.

I think that the geometric meaning of flatness is best understood via simple examples. Consider first $\text{Spec}(k[x,y,t]/(xy-t))\to \text{Spec}(k[t])$ via the natural map. This is a flat family. You can see this geometrically, as the fiber over t is a hyperbola when $t\ne 0$, and as $t$ approaches $0$, the hyperbola gets sharper and sharper and then it "breaks" into two lines when $t=0$.

Constrast this example with $\text{Spec}(k[x,y,t]/(txy-t))\to \text{Spec}(k[t])$. This is not a flat family. Here, when $t\ne 0$, the fiber is always the same hyperbola {xy-1=0}. But, when $t=0$, the fiber is an entire copy of $\text{Spec}(k[x,y])$. This pathological variation of the fibers is encoded by the fact that this is not a flat family.

  • 1
    $\begingroup$ but this does not answer the OP: what does "flat" the way we know it in everyday life have to do with "flat" the way we know it for algebras and schemes. $\endgroup$
    – Jose Capco
    Nov 25, 2009 at 14:19
  • $\begingroup$ @Jose: To address your point, I added a sentence at the end of the first paragraph. $\endgroup$ Nov 25, 2009 at 15:43
  • 17
    $\begingroup$ Dan: I don't like the description of flat families as families that "vary continuously". As you note, an important feature of flat families is that they allow degenerations. For me at least "continuously varying family" should mean that the topology doesn't change, but for instance $xy=0$ and $xy=1$ are topologically different. For intuition purposes, I prefer to think of smooth morphisms as being "continuous families", and flat morphisms as being "not-too-badly-discontinuous families". $\endgroup$ Nov 25, 2009 at 23:57
  • $\begingroup$ Anyway my point is just that one should be careful about specifying exactly what is meant by "continuous" in this situation, as it can easily be misleading. $\endgroup$ Nov 27, 2009 at 13:37
  • $\begingroup$ Rather than 'varying continuously', I would probably say 'nothing sticking out'. In the projective case, this is what the Hilbert polynomial criterion is getting at: all numerical measures of 'size' are the same across all fibres (dimension, degree, ...). (See also this answer I wrote on MSE for a similar question.) $\endgroup$ Aug 2, 2019 at 16:33

I remember the following two quotes about flatness (I forgot who said/wrote this):

  1. For every geometric description of flatness there is a counterexample.
  2. Flatness is one of the few notions in algebraic geometry that were motivated by algebra and not by geometry.

This does not answer your question, I know... :)

  • 23
    $\begingroup$ Your second quote might be Mumford's remark (in his Red Book, Chapter III, $10) : "The concept of flatness is a riddle that comes out of algebra, but which technically is the answer to many prayers". I didn't know the first quote: it is very witty, thanks for posting it. $\endgroup$ Nov 25, 2009 at 14:30
  • 6
    $\begingroup$ I don't like the way people use quote (2), because although it's true that flatness was originally motivated algebraically, it demotivated me for a long time to try thinking of a geometric motivation that is meaningful a priori... now that I have one, I'm much happier, and wish someone had just said that instead! (see my answer) $\endgroup$ Nov 25, 2009 at 22:12
  • 17
    $\begingroup$ Red Book is $10?? where do I get that? :) $\endgroup$
    – Jose Capco
    Nov 26, 2009 at 12:32
  • 18
    $\begingroup$ Jose, you can get the book for about $5 at biblio.mccme.ru/node/1856/shop, but the downside is that you need to read Russian. On the plus side, it is genuinely a red-colored book. $\endgroup$
    – KConrad
    Oct 4, 2010 at 13:39
  • 3
    $\begingroup$ I seem to have stumbled by accident on how to invoke TEX, just try to use dollar signs for actual dollars. $\endgroup$
    – roy smith
    Jun 17, 2013 at 1:43

Here is another possible explanation.

Theorem of Govorov-Lazard: Let $A$ be a ring. Then an $A$-module is flat if and only if $M$ is a filtered colimit of finite free $A$-modules.

See MO/127769 for applications of this Theorem. A Riemannian manifold is called flat if its curvature is $0$ i.e. locally it looks like affine $n$-space $\mathbb{R}^n$. If $A$ is commutative, the functor $M \mapsto \mathbb{V}(M) := \mathrm{Spec}(\mathrm{Sym}(M))$ from $A$-modules to $A$-schemes maps finite free $A$-modules onto affine spaces over $A$. Hence, it maps flat $A$-modules onto filtered limits of affine spaces (where the transition maps should be "linear"). So we definitely get a (vague) connection between flat modules and flat manifolds. It has been discussed at MO/19308 if there is a notion of curvature in algebraic geometry.


As others have stated above, flatness of a family should mean that the fibres of the family vary somehow continuously. Let state this in terms of a module M over a ring R. Here a fibre of M over a prime P of R is M(P), the k(P)-vector space MP/PAP, where k(P) denotes the quotient field of R/P. If the fibres vary continuously, it should be possible to extend a basis of M(P) to nearby fibres, i.e. that the lift of a k(P)-basis wrt. the canonical map MP -> M(P) should yield a basis of MP over AP, i.e. that the stalk MP is a free module.

And in fact: If M is a finitely presented R-module it is flat if and only if M is locally free, i.e. that stalks are free.

(And that a notion may become less geometric when we turn to non finitely presented modules is something which one may expect anyway.)

  • $\begingroup$ This seems indeed to provide a nice geometric approximation. One point seems still unclear to me: what do you concretely mean by "continuous" variation of the nearby fibers? $\endgroup$
    – user267839
    Feb 12, 2019 at 1:19
  • $\begingroup$ We fix firstly a prime $P$ and consider the fiber $M(P) = M_P/PA_P =M_P \otimes_A A_P/P$. Which fibers do you interpret as "neighbours" of $M(P)$? The $M(Q)$ with $Q \subset P$? And why it is reasonable to say that if the fibers vary "continuously" then the lifting provides the freeness of $A_P$-module $M_P$? How "translates" this kind of "continuity" intuitively to this algebraic statement about freeness of the stalk? $\endgroup$
    – user267839
    Feb 12, 2019 at 1:20
  • $\begingroup$ "Neighbors" should indeed be those primes $Q$ contained in $P$. Then we can localize the module $M_P$ at (the image of) $Q$ and recover $M_Q$. If $M_P$ is a free $A_P$ module, then $M_Q$ will in fact be a free $A_Q$ module (of the same rank!) I'm sure this is an exercise in Hartshorne, but I can't quite find it. At the very least exercise II.5.8 is of a very similar nature. $\endgroup$ Mar 21, 2021 at 16:52

One of the consequences of flatness of morphisms between projective schemes is that the dimension of the fibers stays constant. Maybe this is the reason for the term. I'm not sure whether this makes so much sense though. After all, the alps stay three dimensional all the time, but they don't really count as being "flat". But it probably would even much harder to climb them if they had one more dimension...

At least if you think about something that has 0-dimensional fibers all the time and suddenly aquires one two-dimensional fiber flatness really makes sense in the usual sense. I tried drawing a picture here, but mathoverflow always eats my ascii art.


As I understand, at least a part of the original question is about the WORD flat; Why this word is used. Then, if I am not wrong (but maybe I am), this word was introduced for modules first, and then for schemes just by extension of terminology. Hence, maybe, the choice of word, "flat", should not really contain some geometric intuition about the schemes (rather, it was chosen for some linear algebra intuition for modules, maybe as one can see from the discussion of module flatness in one of the books by S. Gelfand, Y. Manin).

  • $\begingroup$ Maybe it has geometric meaning, but the word was chosen because of algebraic meaning. Maybe not; I do not know for sure. $\endgroup$
    – Sasha
    Oct 15, 2010 at 20:48

See the illustration on the 4th page of Miles Reid's Undergraduate Commutative Algebra (go to Amazon, click on look inside and click the right arrow 4 times).

This illustration shows a module $M$ that's flat over $A/{\rm ann}(M)$.


Very late to this party, but I've recently started studying this, and I've found an answer that is quite satisfying to me (at least satisfying enough for me to continue on). Specifically, we will use the fact that $M$ is a flat $A$-module if and only if $I\otimes_AM\cong IM$ for all ideals $I$ of $A$ (as Andrew does above).

Let $\pi:X\rightarrow Y$ be a morphism of schemes, let $V\hookrightarrow Y$ be any closed subscheme, and let $\mathcal{I}_{V/Y}$ be the quasicoherent sheaf of ideals carving out $V$. Considering $V\hookrightarrow Y$ to be a closed subfunctor, we can considering the corresponding closed subfunctor $V_X\hookrightarrow X$ to be the "preimage" of $V$ in $X$. The quasicoherent sheaf of ideals carving out $V_X\hookrightarrow X$ is exactly the image sheaf (call it $\mathcal{I}_{V_X/X}$) given by the morphism $\pi^*\mathcal{I}_{V/Y}:=\pi^{-1}\mathcal{I}_{V/Y}\otimes_{\pi^{-1}\mathcal{O}_Y}\mathcal{O}_X\rightarrow \mathcal{O}_X$, $i\otimes m\mapsto im$.

Using the fact that $\pi$ is flat is a stalkwise condition and the definition of flatness in terms of ideals, it is clear that $\pi^*\mathcal{I}_{V/Y}\cong \mathcal{I}_{V_X/X}$ for all $V\hookrightarrow Y$ if $\pi$ is flat. That is, a necessary condition for $\pi$ to be flat is that $\pi^*\mathcal{I}_{V/Y}\cong \mathcal{I}_{V_X/X}$ for all $V\hookrightarrow Y$.

It turns out, this condition is also sufficient! Indeed, choose any $[\mathfrak{q}]\in X$, and define $[\mathfrak{p}]:=\pi([\mathfrak{q}])$. Next, consider the canonical morphism $i:\text{Spec}\mathcal{O}_{Y,[\mathfrak{p}]}\rightarrow Y$. It's a general fact that the image of $i$ is contained in any open subset of $Y$ which also contains $[\mathfrak{p}]$ (Vakil 7.3.M), which in particular implies that $i$ is affine (choose any affine open $\text{Spec}A\hookrightarrow Y$. If $[\mathfrak{p}]\in \text{Spec}A$, then $\text{Spec}\mathcal{O}_{Y,[\mathfrak{p}]}\times_Y\text{Spec}A\cong \text{Spec}A_{\mathfrak{p}}\times_{\text{Spec}A}\text{Spec}A\cong \text{Spec}A_{\mathfrak{p}}$, and if $[\mathfrak{p}]\not\in \text{Spec}A$, then this is empty). Next, let $I$ be any ideal in $\mathcal{O}_{Y,[\mathfrak{p}]}$, and let $j:V\hookrightarrow \text{Spec}\mathcal{O}_{Y,[\mathfrak{p}]}$ be the corresponding closed subscheme. Then $j$ is affine, so $i\circ j$ is affine. In particular, $i\circ j$ is qcqs, so $(i\circ j)_*\mathscr{F}$ is quasicoherent on $Y$ for all quasicoherent sheaves on $V$, which implies that $(i\circ j)_*\mathcal{O}_V$ is quasicoherent on $Y$. Now consider the map $\mathcal{O}_Y\rightarrow (i\circ j)_*\mathcal{O}_V$, coming from the composition $V\hookrightarrow \text{Spec}\mathcal{O}_{Y,[\mathfrak{p}]}\rightarrow Y$. Since $(i\circ j)_*\mathcal{O}_V$ is quasicoherent and the quasicoherent sheaves on $Y$ form an abelian category, so $\mathcal{I}:=\text{ker}(\mathcal{O}_Y\rightarrow (i\circ j)_*\mathcal{O}_V)$ is a quasicoherent sheaf of ideals on $Y$. Now, consider the stalk $\mathcal{I}_{[\mathfrak{p}]}$. This is the kernel of $\mathcal{O}_{Y,[\mathfrak{p}]}\rightarrow ((i\circ j)_*\mathcal{O}_V)_{[\mathfrak{p}]},$ where $((i\circ j)_*\mathcal{O}_V)_{[\mathfrak{p}]}=(i_*(j_*\mathcal{O}_V))_{[\mathfrak{p}]}=\Gamma(\text{Spec}\mathcal{O}_{Y,[\mathfrak{p}]},j_*\mathcal{O}_V)$, which is exactly $I$. We have now shown that for any point $[\mathfrak{p}]$ in the image of $\pi$ and any ideal $I$ of $\mathcal{O}_{Y,[\mathfrak{p}]}$, we can choose a quasicoherent sheaf of ideals $\mathcal{I}$ on $Y$ such that $\mathcal{I}_{[\mathfrak{p}]}\cong I,$ and our result follows.

Thus, $\pi:X\rightarrow Y$ is a flat morphism of schemes if and only if for all closed subschemes $V\hookrightarrow Y$ (carved out by a quasicoherent sheaf of ideals $\mathcal{I}_{V/Y}$), the functions that vanish on $V_X\hookrightarrow X$ are exactly those of the form $\pi^{-1}\mathcal{I}_{V/Y}\otimes_{\pi^{-1}\mathcal{O}_Y}\mathcal{O}_X.$ That is, the map $\pi^{-1}\mathcal{I}_{V/Y}\otimes_{\pi^{-1}\mathcal{O}_Y}\mathcal{O}_X\rightarrow \mathcal{I}_{V_X/X},$ $i\otimes m\mapsto im$, is an isomorphism. This is always surjective, so this condition is that there are no extra "surprise" relations between the $i$'s and $m$'s (some language borrowed either again Andrew, or from Vakil (24.4.2), I do not know who wrote it first). That is, saying that $\pi:X\rightarrow Y$ is flat is equivalent to saying that $X$ sits nicely (or maybe, "flatly"?) over $Y$.

  • $\begingroup$ @BenMcKay Very true, thank you! Unfortunate typo, fixed. $\endgroup$ Nov 10, 2023 at 16:18

This is not really an answer, more of a suggestion of where to look for one. In a locally trivial fibre-bundle, the cohomology groups of the fibres with real coeffs naturally form a flat vector bundle over the base. This is because the integer-valued cohomology gives a lattice which one can use to define the parallel sections. (I guess this is called the Gauss-Manin connection?)

From my outsider's perspective on algebraic geometry, I always imagined that the cohomology groups (of the relevant theory) associated to members of a flat family behaved in a similar way. Perhaps to make this true, one should talk of the alternating sum of cohomology groups as an element of K-theory in the base, or something similar. I've no idea if this can be made rigourous, but various consequences of flatness seem to fit into this mould. E.g., the constancy of the fibre dimension mentioned in Timo Schürg's answer, the fact that the holomorphic Euler characteristic is constant etc.

I would love to hear from algebraic geometers whether this has any rigourous sense to it!

  • $\begingroup$ This is not a complete response to your comments, but -- one of the main points of flatness is that it allows degenerations, and in particular it allows the topology of the fibers to change. For example you can have flat families with some fibers being smooth elliptic curves (topologically a torus) and other fibers being singular elliptic curves (topologically a "pinched" torus), and the (singular) cohomology will not be constant. $\endgroup$ Nov 26, 2009 at 0:08
  • $\begingroup$ On the other hand, if we have an algebraic morphism which is smooth and proper in the sense of algebraic geometry, and if we consider that morphism as an analytic morphism, then it will be a proper submersion in the sense of differential geometry. Then Ehresmann's theorem says that proper submersion = locally trivial fiber bundle, and thus it follows that the topology of the fibers is constant (for the pedants: assuming the base is connected), and then we get the Gauss-Manin connection and all that other good stuff. But I don't know of any analogous thing for flat morphisms. $\endgroup$ Nov 26, 2009 at 0:38
  • $\begingroup$ Thanks for your comments Kevin. I certainly didn't mean to suggest that flat families would be locally trivial. (Differential geometers like degenerations too!) I was more wondering if there was any sort of analogy between the two situations. E.g., for what families can one find a flat bundle on the base (more correctly one should probably look for a complex of sheaves, what ever flat means there) in analogy with Gauss-Manin? $\endgroup$
    – Joel Fine
    Nov 26, 2009 at 8:55

Best elementary explanation of this that I have seen is in the book of Eisenbud and Harris (Geometry of Schemes) Section II.3. They describe what limits of families mean and how it is related to flatness.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.