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Let $f\colon X\to Y$ be a surjective morphism of irreducible complex algebraic varieties with nonsingular fibers. Moreover, assume that each geometric fiber $f^{-1}(y)$ is isomorphic to an affine space ${\Bbb A}^{n(y)}$.

Question 1. What are algebraic-geometrical conditions on $f$ that imply that $f$ is a submersion, that is, the induced maps on the tangent spaces are surjective?

Question 2. The same question in the case when $y\to n(y)$ is a constant function.

Mikhail Borovoi
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1 Answers1

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The correct notion of a submersion in algebraic geometry is a smooth morphism. The key statement for your situation is that a flat morphism locally of finite presentation with smooth fibers is smooth (ref Stacks 01V8). Assuming "variety" means at least a scheme of finite type over a field, your map is locally of finite presentation (01T8 + 01TX) and it has smooth fibers by assumption. Thus all we need to do to guarantee smoothness is to show that your morphism is flat.

In your second case, if you know that $X$ is Cohen-Macaulay and $Y$ is regular, this follows from Miracle Flatness (see this answer of Emerton). Without this assumption in the second case, or in the first case, I am struggling to think of big general results that will be useful to you. Perhaps perusing some criteria for flatness will be helpful to you.

KReiser
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  • Many thanks for the detailed answer! Yes, by a complex algebraic variety I mean a separated integral scheme of finite type over the field $\Bbb C$. – Mikhail Borovoi Apr 21 '21 at 20:07
  • Do I understand correctly that a smooth morphism of algebraic varieties over an algebraically closed field of characteristic $0$ is automatically surjective on tangent spaces of closed points? – Mikhail Borovoi Apr 21 '21 at 20:19
  • I have edited my question: the word "geometric" is added. Do I understand correctly that your answer is still valid for the edited question? – Mikhail Borovoi Apr 21 '21 at 20:21
  • Yes to both. See here for smooth implies submersion (with no assumptions on the characteristic). For the addition of the word geometric, just note that if the fiber over $y$ is $\Bbb A^{n(y)}{k(y)}$, then the geometric fiber is just $\Bbb A^{n(y)}{\overline{k(y)}}$. – KReiser Apr 21 '21 at 20:34
  • Thank you, this is very helpful! – Mikhail Borovoi Apr 25 '21 at 17:55
  • Now I would like to have a reference for the following statement: 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. I can refer to the proof here. However, I would prefer to refer to a book, or a published paper, or the Stacks Project, or an arXiv preprint. – Mikhail Borovoi Apr 25 '21 at 18:02