1
$\begingroup$

It is well known that Kodaira gave an essentially topological classification of the possible singular fibers of elliptic fibrations according to their type:

https://en.wikipedia.org/wiki/Elliptic_surface#Kodaira.27s_table_of_singular_fibers

Les us assume that we have two elliptic fibrations $\pi\colon X\to S$, $\pi'\colon X'\to S'$ and consider two singular fibers $X_s\hookrightarrow X$, $X'_{s'}\hookrightarrow X'$ over closed points such that the Kodaira type of $X_s$ is the same as the Kodaira type of $X'_{s'}$.

What I'm asking is whether it is always true or not that $X_s$ and $X'_{s'}$ are isomorphic as schemes with their naturally induced closed subscheme structure.

EDIT: By an elliptic fibration $\pi\colon X\to S$, I understand:

$X$ is a smooth algebraic surface. $S$ is a complete smooth algebraic curve over an algebraically closed field, $\eta$ is its generic point. $\pi$ is a projective morphism, and the generic fiber $X_\eta$ is a geometrically integral smooth algebraic curve of arithmetic genus $1$.

Improve this question
$\endgroup$
5
  • $\begingroup$ Not in general: Just take two non-isomorphic smooth fibres. These have the same Kodaira type $I_0$. $\endgroup$
    – Daniel Loughran
    Mar 16, 2016 at 12:05
  • $\begingroup$ @Daniel Loughran: My question is just about special fibers, and these are singular curves. $\endgroup$
    – user6319
    Mar 16, 2016 at 12:13
  • 1
    $\begingroup$ What is your definition of "special fibre"? For me, for some scheme over a spectrum of a DVR, the special fibre is the fibre over the closed point; this can certainly be smooth. What are your $S$ and $S'$? Smooth projective curves over an algebraically closed field? $\endgroup$
    – Daniel Loughran
    Mar 16, 2016 at 12:17
  • 1
    $\begingroup$ Anyway, for singular fibres, assuming that $S$ and $S'$ are smooth projective curves over an algebraically closed field $k$ of characteristic $0$, the answer should be yes. Most cases consist of a collection of smooth genus $0$ curves meeting in some configuration of points. As there is a unique smooth genus $0$ curve over $k$ up to isomorphism (namely $\mathbb{P}^1$), the result is clear. The remaining cases are that of a plane cubic nodal or cuspidal curve. Again, such a curve is unique over $k$ up to isomorphism, hence the result follows. $\endgroup$
    – Daniel Loughran
    Mar 16, 2016 at 12:22
  • $\begingroup$ @Daniel Loughran: Thanks for your comments. I have modified the question accordingly to ask for "singular fibers". $\endgroup$
    – user6319
    Mar 16, 2016 at 13:29

1 Answer 1

Reset to default
11
$\begingroup$

No, this is not the case. Indeed, there is exactly one type of singular fibre on the list for which the scheme structure is not unique up to isomorphism, namely type $I_0^\ast$ in Kodaira's notation. The point, of course, is that here we have a $\mathbf P^1$ component which intersects 4 other componenents, and automorphisms of $\mathbf P^1$ don't act transitively on sets of 4 points.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Good point, I had over-looked this case. $\endgroup$
    – Daniel Loughran
    Mar 16, 2016 at 13:16
  • $\begingroup$ @The Salmon of Ignorance: Thanks for your answer. Please could you be so kind as to elaborate a bit further your answer and/or provide a reference, especially with respect to the role played by the nilpotent elements of the double component $2\Gamma_2$ in $I_0^*=\Gamma_1+2\Gamma_2+\Gamma_3+\Gamma_4+\Gamma_5$. $\endgroup$
    – user6319
    Mar 17, 2016 at 14:02

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.