11

I do not believe you have equality in general. I sketch a counterexample below, which is a geometric version of the fact that if $E/K$ is an elliptic curves such that the quadratic twist $E^{(d)}/K$ has Mordell-Weil rank at least 2 then the Mordell-Weil of $E/K(\sqrt{d})$ is at least two bigger than the Mordell-Weil rank of $E/K$.
Take an elliptic K3 ...

answered Jul 26 '13 at 12:36

Remke Kloosterman

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You can observe that the elliptic curve $E_2: y^2=x^3+(t^3+1)^2$ is a generic fibre of a rational elliptic surface. Over the algebraically closed field $k$ the group of $k(t)$ points on the curve has rank equal to 2 (by the Shioda-Tate formula) and from the classification of possible groups of $k(t)$-rational points by Oguiso-Shioda (case 39 because we have ...

9

The minimal $s$ is $3$.
It is attained by several elliptic K3's,
including $y^2 = x^3 + (t^2-t)^4$ which has IV* fibers at
$t = 0, 1, \infty$ and no other singular fibers.
The comment by Ariyan Javanpeykar gives one argument that
$s$ can be no smaller. (See postscript.
This uses characteristic zero; in small positive characteristic
$s$ can be as small as $...

answered Oct 8 '18 at 14:24

Noam D. Elkies

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If $2$ is a cube mod $p$ then you can take
$(x,y) = (ct^2, t^6-1)$ where $c^3 = -4$.
This works for every odd $p \equiv -1 \bmod 3$,
but since you specified $p \equiv +1 \bmod 3$
the first case is $p=31$, with $c \in \{-3,-15,18\}$.
The next few such $p$ are $43, 127, 223, 283, 307, 439, 499, 643, 691$.
By Cubic Reciprocity, the condition on $p$ is ...

answered Jul 25 '19 at 14:58

Noam D. Elkies

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Where you say "trivial action on $\Gamma_{t0}$" near the end, you mean "carries $\Gamma_{t0}$ into itself" (i.e., doesn't move it), not that the effect on $\Gamma_{t0}$ is the identity.
The global base is a red herring: the core issue here completely concerns the object over the local ring at $t$ (recall that the formation of both the minimal regular proper ...

7

The modular elliptic surfaces are quite rare. E.g., the Mordell-Weil group is finite and the Picard number of the surface equals $h^{1,1}$ (see Shioda's paper). Such elliptic surfaces are called extremal and for a fixed Euler number of the surface and fixed genus of the base curve there are only finitely many extremal elliptic surfaces with non-constant $j$-...

answered Oct 17 '13 at 22:33

Remke Kloosterman

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If a surface is foliated by geodesics (in the induced metric) that are also asymptotic lines, then these curves are also geodesics in the ambient manifold (just look at the definitions). Conversely, if a surface is a $1$-parameter family of geodesics in an ambient manifold, then these curves are also geodesics in the induced metric and are asymptotic lines. ...

answered Apr 3 '17 at 11:53

Robert Bryant

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If you look at the global monodromy action $\pi_1 ( \mathbb P^1 -\{0,1\}) \to SL_2(\mathbb Z)$, you see that $\pi_1 ( \mathbb P^1 -\{0,1\}) = \mathbb Z$ so the image is abelian and therefore has infinite index. On the other hand, if we consider the $j$ map $\mathbb P^1 \to X(1)$, the image of the fundamental group has finite index in $\pi_1 ( X(1) -\{0, 1728,...

6

The structure of $\textrm{Num}(S)$ for a bielliptic surface $S$ is given in the paper by F. Serrano Divisors of bielliptic surfaces and embeddings in $\mathbb{P}^4$, Mathematische Zeitschrift 203 (1990), 527-533.
First, Serrano proves that a basis for $\textrm{Num}(S) \otimes _{\mathbb{Z}} \mathbb{Q} \cong H^2(S, \, \mathbb{Q})$ is given by the fibres $A$...

answered Jul 31 '14 at 12:40

Francesco Polizzi

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EDIT We show that the answer to the OP's question is yes. Thanks to Will Sawin for his comments.
I use the notation of [Hartshorne, Algebraic Geometry, Chapter V Section 2].
Since $S$ is a ruled surface, there exists a section $C_0$ of minimal self-intersection; set $C_0^2 = -e$. If we write $S=\mathbb{P}(\mathcal{E})$, with $\mathcal{E}$ normalized, then $...

answered May 20 '13 at 15:48

Francesco Polizzi

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[In comments guest2014 amended the question to ask not for a
$D_{14}$ fiber but for $I^*_{14}$, a.k.a. $\tilde D_{18}$]
The elliptic surface
$$
X : y^2 = x^3 + (t^3+2t) x^2 - 2(t^2+1)x + t
$$
over ${\bf C}(t)$ has a $I^*_{14}$ fiber at $t=\infty$.
(Note that the right-hand side is a cubic in $x$ whose discriminant
$4t^4 + 13 t^2 + 32$ has degree only $4$ ...

answered Feb 19 '14 at 3:29

Noam D. Elkies

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If you take the Tate curve $\mathbb G_m/\langle q \rangle$ over $\mathbb Z[[q]]$ and mod out by the subgroup $\mu_n$ and resolve singularities, you should get a copy of the semistable fiber type $I_n$ with lattice $A_{n-1}$. Note that this means the Tate curve, modulo nothing, is of fiber type $I_1$, and hence already has a singular fiber.
However $I_n$ can ...

5

I will sketch a counterexample for the modified question. The idea behind the construction is similar to the counterexample for the original question. Only the geometric details of this construction are trickier and does not integrate so well in my previous answer. For this reason I will post this a new answer.
For this example we take $k=\mathbb{C}$ and $C=...

answered Dec 15 '13 at 10:30

Remke Kloosterman

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One more attempt.
All the fibers of $p\colon S\to C$ dominate $E$, so they are all isogenous to $E$ and by Kodaira's classification of elliptic fibers they all have smooth support. Hence $p$ is a so-called "quasi-bundle'' and by a result of Serrano [F.Serrano,
Isotrivial fibred surfaces
Ann. Mat. Pura Appl. (4) 171 (1996), 63–81] is of the form $(F\...

5

It's always true with $\mathbb{Q}$ coefficients. It follows from a general result of Zucker, who proved Leray degenerates whenever you have a projective map to curve. But in this case, it's simpler to check it by hand. There is only one (two !) differentials to worry about
$$d_2:H^0(\mathbb{P}^1, R^1\phi_*\mathbb{Q})\to H^2(\mathbb{P}^1,\mathbb{Q})$$
But ...

5

Please confer Corollary 2.2 of the following with $d$ equal to $3$ and with $n$ equal to $2$.
Jason Starr
A pencil of Enriques surfaces of index 1 with no section
https://arxiv.org/pdf/math/0602639.pdf
This proves that for every integer $e\geq 2$, for a very general hypersurface $X$ in $\mathbb{P}^2\times \mathbb{P}^1$ in the complete linear system of $\...

4

Any $\mathbb F_2(t)$-point of $K_t$ would give, upon pullback to $E \times E'$, a $\mathbb F_2(E)$-point of $E'$. Because $E$ is ordinary, $a_2(E)\neq 0$, hence $E$ is not isogenous to its quadratic twist $E'$, so any such point arises from an $\mathbb F_2$-point of $E'$. Because $E'$ has two $\mathbb F_2$-points, these are the only ones.

4

For the case with section and $q=0$ see
http://www.math.colostate.edu/~miranda/preprints/weierstrassfibrations.pdf
A similar construction should work in the case (with section, $q$ fixed and $p_g$ sufficiently large) you should get a moduli space together with a morphism to $M_g$.
In the case (with section; $q=p_g=1$) then $p_g$ is ``sufficiently large" ...

answered Feb 2 '15 at 21:07

Remke Kloosterman

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3

Consider first an elliptic fibration with a section over $\mathbb{P}^1$. (In this case none of the singular fibers are multiples of smooth curves.)
Assume that the minimal discriminant has degree $12n$.
Then the number of components of singular fibers which do not intersect the zero section equals $12n-2a-m$, where $a$ is the number of additive fibers and $...

answered Dec 8 '15 at 17:42

Remke Kloosterman

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3

Let $E_0$ be the elliptic curve $y^2 = x^3 + 1$,
and choose $\beta$ in $k = {\bf F}_q$ so that $b = \beta^2$.
Then $W = W_b$ has $\rho=18$ unless the elliptic curve
$$
E_\beta : Y^2 = X^3 + \beta \, (3X + (\beta+1)^2)^2
$$
is isogenous to $E_0$, in which case $\rho(W)$ is
$20$ ("singular") or $22$ ("supersingular")
depending on whether $...

3

This is too long for a comment, so I'm making it an answer. Magma now has machinery for computing ranks and generators over function fields (see here) which only work for rational elliptic surfaces (using the connection with the Neron-Severi group). I used the L-function machinery (see here) to compute the analytic ranks for $p = 7$, $p = 19$ and $p = 31$. ...

2

I am expanding naf's comments to make a self-contained community wiki answer. By an elliptic fibration we mean a smooth projective relatively minimal surface $f: X \to C$ with general fiber given by an elliptic curve.
Main Proposition. If $C$ is a smooth projective curve and $S \subset C$ a finite subset, then the number of nonisotrivial elliptic surfaces ...

2

Let us start from any elliptic surface $X_0 \longrightarrow \Sigma$ with $\chi(X_0) >0$ and whose
fibre at the point $x_i$ is of type $I_0$ (i.e., smooth) for all $i \in \{1, \ldots, p\}$. For instance, a general elliptic fibration over $\Sigma$ without multiple fibres satisfies these requirements.
Then with a sequence of $p$ logarithmic transformations ...

answered Jun 7 '17 at 8:41

Francesco Polizzi

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2

You could follow Suwa's construction of the Kodaira surfaces from his paper Compact quotients of $C^2$ by affine transformation groups, and define various surfaces which are quotients of $k^2$ by groups of affine transformations of the special form that Suwa arrives at in his paper. I don't know any applications, but it is a very nice paper.
If I remember ...

2

Here is an alternative approach.
Since $S$ is an elliptic fibration, the rational curves of the ruling must cover $C$ and hence $C$ is also rational, i.e., $C\simeq \mathbb P^1$.
Next do a base change by the induced map from a (general) fiber of $p$, say $F$, to $E$. We get a new ellipticly ruled surface: $\rho: T=S\times_EF\to F$.
Now consider the ...

answered May 20 '13 at 18:00

Sándor Kovács

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2

Your question is local, so we may take $S$ to be the ring of integers in a local field $K$. If the point $P\in E(K)$ is torsion or not does not matter in my answer here. I think here of an ellpitic curve over a number field.
I think the best theoretical tool to study fibres of type $I_n$ is by using Tate's uniformisation. This works over $K$ if the ...

2

This is the journal that is now called the Münster Journal of Mathematics. Only recent volumes, from 2008, are online. The complete journal, including the volume 33 you are looking for, has actually been digitized by Hathitrust, but for copyright reasons the access to it is restricted (search only). The only way to go seems to consult a library, here is a ...

answered Oct 7 '15 at 17:51

Carlo Beenakker

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2

Suppose that $\omega$ is the invariant differential of an elliptic quotient $C/\sigma$ of $C$. In particular, $\omega$ is non-zero. If $\omega$ lies in the image of $\lambda$, then there exists a point $P = (x(t), y(t))$ on $E_D(\mathbb{Q}(t))$ such that $\lambda(P) = \omega$. Let $\rho_1(P)$ denote the morphism in $\operatorname{Mor}_{\mathbb{Q}}(C/\sigma, ...

answered Jul 11 '15 at 1:48

Stanley Yao Xiao

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2

In general, not, as the monodromy group does not need to be abelian. Say, take a typical elliptic surface: the monodromy must kill all $H_1$ of the fiber, which is $\Bbb Z^2$; a cyclic group can only kill $\Bbb Z$.
I could not understand your particular example.

1

The first claim can be wrong if the elliptic surface has singular fibers with multiple components. This increases $R^2$ at those points, giving it a punctual subsheaf, giving it more global sections. We can associate these with the classes of fibral divisors that have zero intersection with the class of a fiber, say.
Your second claim follows from the ...

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