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 $1$.

$\endgroup$
5
  • $\begingroup$ Not in general: Just take two non-isomorphic smooth fibres. These have the same Kodaira type $I_0$. $\endgroup$ Commented Mar 16, 2016 at 12:05
  • $\begingroup$ @Daniel Loughran: My question is just about special fibers, and these are singular curves. $\endgroup$
    – user6319
    Commented 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? $\endgroup$ Commented 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. $\endgroup$ Commented 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
    Commented Mar 16, 2016 at 13:29

1 Answer 1

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.

$\endgroup$
2
  • $\begingroup$ Good point, I had over-looked this case. $\endgroup$ Commented 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
    Commented Mar 17, 2016 at 14:02

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .