I have a general question about the motivation behind to definition the smooth morphisms as we know it from algebraic geometry. The most common definition of a smooth morphism $: X \to Y$ between two smooth Noetherian schemes $X,Y$ is:

$f$ is smooth if and only if

(i) $f$ is flat and locally of finite presentation

(ii) for every $y \in Y$ the fiber $X \times_Y k(y)$ is smooth variety over $k(y)$

I read recently that the motivation of smooth morphisms in algebraic geometry arise as a kind of imitation attempt of a maps called "submersions" in differential geometry. Namely if $X,Y$ are smooth manifolds then a submerison $s: X \to Y$ is a surjective, proper $C^{\infty}$ map for which for every $x \in X$ the induced differention $D_xf: T_x X \to T_y Y$ is surjective. The Ehresmann's lemma says that such submersion is moreover a locally trivial fibration. That seems to coincide with the intuition that flaness is something like a continuous family of neighbored fibers.

My Question is if there exist a definition of a smooth morphism $: X \to Y$ in algebraic geometry world, which emphesize more immediately that the motivation for smoothness in algeom arise from submersions in differential geometry?

Let look again at definition above. (ii) seems reasonable, since this tells that every fiber of $f$ is smooth, ie morally a manifold. But that reason that the point (i) arise immediately from the differential geometry isn't immediately clear if one not believes that flatness makes families "continuous". Morally "continuity of fibers" (= flatness) should be a consequence (like by of Ehresman lemma in differential geo), not an immediate "part" of definition.

Can the flatness requirement be replaced in algeobraic definition by requirement that the induced differention $D_xf: T_x X \to T_y Y$ is surjective? And is this equivalent of flatness in algebraic setting?

The reason is that I conjecture that this could be true is that we can surely define the tangent space of $X$ at every $x$ pure algebraically as the dual of the stalk $\Omega_{X,x}$ or equivalently as $\{ \phi \in \operatorname{Hom} (\operatorname{Spec}k[\epsilon], X) \ \vert \text{ Im } \phi = \{x\} \}$.

The Question is if in algebraic setting the surjectivity of algebraic $D_xf: T_x X \to T_y Y$ at every $x$ is equivalent to flatness of $f$?

  • $\begingroup$ There are definitions of smoothness that do not require flatness, see e.g. Tag 00T2. It's not to so hard to show that your condition implies flatness at least when $X$ and $Y$ are smooth varieties (this also follows from miracle flatness, see Tag 00R4). I suspect your condition is not quite right in general, but I don't know an example off hand. $\endgroup$ – R. van Dobben de Bruyn Nov 1 '20 at 4:00

One of the many equivalent definitions of smoothness of a morphism $f\colon X\to Y$ of varieties over a field $k$ is that $f$ is smooth if and only if it is formally smooth. The latter means the following: given any square-zero extension of $k$-algebras $S\to R$ and a commuting square $\require{AMScd}$ \begin{CD} \mathrm{Spec}(R) @>x>> X\\ @V V V @VV f V\\ \mathrm{Spec}(S) @>y>> Y \end{CD} of $k$-schemes, there exists at least one diagonal map $\mathrm{Spec}(S)\to X$ which splits the square into two commuting triangles.

Let's unpack what this means in the particular case $R=k$ and $S=k[\varepsilon]$ the ring of dual numbers over $k$ (i.e. with $\varepsilon^2=0$). As the OP already noted, a morphism $\mathrm{Spec}(k[\varepsilon])\to Y$ is the same as a $k$-rational tangent vector in $Y$. The base point of such a tangent vector is the composite $\mathrm{Spec}(k)\to\mathrm{Spec}(k[\varepsilon])\to Y$.

So what formal smoothness tells us in this case is that for any $k$-rational point $x$ of $X$ and any $k$-rational tangent vector $\overrightarrow y$ of $Y$ based at $f(x)$, there is at least one $k$-rational tangent vector $\overrightarrow x$ of $X$, based at $x$, such that $f(\overrightarrow x)=\overrightarrow y$. In other words, the map $D_xf\colon T_xX\to T_{f(x)}Y$ is surjective.

Thus, we see that a smooth map of varieties induces surjective maps on tangent spaces. However, in full generality (permitting singular varieties), the two conditions are not equivalent. For example, consider the case when $X$ is the singular affine curve with equation $xy=0$, and $Y=\mathrm{Spec}(k)$ is the point. Since the tangent space of $Y$ is trivial, the structure morphism $X\to Y$ must induce a surjection on tangent spaces. On the other hand, $X\to Y$ is certainly not a smooth map since $X$ is not smooth! The failure of formal smoothness can be seen directly: the map $k[\delta]/(\delta^3)\to k[\varepsilon]$ is a square-zero extension, but the $k[\varepsilon]$-valued point $(x,y)=(\varepsilon,\varepsilon)$ of $X$ does not lift to a point valued over $k[\delta]/(\delta^3)$.

So for general $k$-varieties, smoothness implies surjectivity on tangent spaces, but not conversely. On the other hand, I think that formal smoothness does what you asked for in the first question: it's a definition which is very similar to that of a submersion which captures the ``correct'' definition of smoothness.

On the other hand, if you assume that all your varieties are smooth, then the story becomes much simpler: a morphism $f\colon X\to Y$ of smooth $k$-varieties is smooth if and only if it induces a surjection on tangent spaces at each point of $X$. We've seen the left-to-right implication already. For the converse implication, suppose that $f\colon X\to Y$ induces a surjection on tangent spaces. To show that $f$ is smooth, it suffices to prove that $f$ is flat and that the relative cotangent sheaf $\Omega^1_{X/Y}$ is locally free.

The first of these -- flatness of $f$ -- is addressed by R. van Dobben de Bruyn's comment (using miracle flatness).

For the second of these, we have an exact sequence $$ f^*\Omega^1_{Y/k} \to \Omega^1_{X/k} \to \Omega^1_{X/Y} \to 0 \,. $$ Surjectivity of $f$ on tangent spaces is equivalent to saying that $f^*\Omega^1_{Y/k} \to \Omega^1_{X/k}$ is an injection on fibres. This makes $\Omega^1_{X/Y}$ into a coherent sheaf, all of whose fibres have the same dimension, which then implies that it is locally free (this is e.g. Lemma 10.77.3 in the Stacks project).

  • $\begingroup$ Do you know a reference for the following statement (proved in your answer)? Let $\varphi\colon X\to Y$ be a surjective morphism of smooth $\Bbb C$-varieties. Then $\varphi$ is smooth if and only if the map on $\Bbb C$-points is a submersion, that is, $\varphi$ induces surjections on the tangent spaces. $\endgroup$ – Mikhail Borovoi Apr 25 at 19:46
  • $\begingroup$ I don't know a precise reference off the top of my head, but the place I'd look first would be Griffiths and Harris' Principles of Algebraic Geometry, which develops a lot of the theory of algebraic geometry from a complex-analytic perspective. I don't have a copy with me right now, so can't say for certain whether they treat smoothness and submersions. Let me know if this doesn't work, and I can ask a few colleagues for other suggestions. $\endgroup$ – Alexander Betts Apr 26 at 7:43

This is not directly an answer to your question, but a detail on the definition of submersion that may be relevant, or maybe will help to see the question differently:

Submersions are not usually required to be proper. And then Ehresmann's theorem says that proper submersions are locally trivial.

Separating these two conditions, a submersion is "locally trivial around points of the domain" (this is the statement of the submersion normal form), while a proper submersion is "locally trivial around each fibre" (the statement of Ehresmann).

So, $f$ being a submersion, guarantees that every fibre of $f$ is smooth. And additionally $f$ being proper guarantees that its fibres fit into a locally trivial fibration.

From what I understood from the question, it looks like if (ii) is what guarantees that every fibre is smooth, then (i) should play a similar role to properness in differential geometry - either ensuring some sort of local triviality, or ensuring that enough properties that follow from local triviality are satisfied.


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