I think it should be a standard procedure to construct such things, can anyone give a reference or give a hint? Can this be done over any base scheme?
[Math] (nontrivial) isotrivial family of elliptic curves
ag.algebraic-geometryelliptic-curvesmoduli-spacesnt.number-theory
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This "answer" will basically restate the comments of Marty and Jason Starr.
Any variety covered by schemes of the form $\mathrm{Spec}(\mathbf Z[M_i])$, or any torified variety, is rational. And indeed Severi conjectured at one point that $M_g$ is rational for any $g$! But we know a lot about the Kodaira dimension of $M_g$ by work of Harris--Mumford, Farkas, Eisenbud, Verra, ... in particular we know that Severi's conjecture is maximally false. We have $\kappa(M_g) = -\infty$ for $g \leq 16$, we don't know anything for $17 \leq g \leq 21$, and $\kappa(M_g) \geq 0$ for $g \geq 22$. In fact we know that $M_{g}$ is of general type for $g = 22$ and $g \geq 24$. But if one only wants an example of a $g$ for which $M_g$ is not rational, then I think one can find examples much earlier (maybe $g \approx 6,7$?).
See http://arxiv.org/abs/0810.0702 for a survey by Farkas.
Moreover, the cohomology of a torified scheme can only contain mixed Tate motives. In particular, this implies properties like: the number of $\mathbf F_q$-points is a polynomial function of $q$. As Marty says there are no problems in genus zero; these might be honest-to-God $\mathbf F_1$ schemes. The cohomology of $M_{1,n}$ contains only mixed Tate motives when $n \leq 10$, but for $n \geq 11$ one finds motives associated to cusp forms for $\mathrm{SL}(2,\mathbf Z)$. (The number 11 arises as one less than the smallest weight of a nonzero cusp form; the discriminant form $\Delta$.) This implies that the polynomiality behaviour changes drastically -- the number of $\mathbf F_q$-points is now given by Fourier coefficients of modular forms, which are far more complicated and contain lots of arithmetic information.
The connection with birational geometry is also very visible here. The cohomology classes on $M_{1,n}$ associated to the cusp forms of weight $n+1$ are of type $(n,0)$ and $(0,n)$ in the Hodge realization, since given such a cusp form one can explicitly write down a corresponding differential form in coordinates. So by the very definition of Kodaira dimension we can not have $\kappa = -\infty$ anymore.
Bergström computed the Euler characteristic of $M_{2,n}$ in the Grothendieck group of $\ell$-adic Galois representations by point counting techniques. That is, he found formulas for the number of $\mathbf F_q$-points of $M_{2,n}$ by working really explicitly with normal forms for hyperelliptic curves. These formulas turned out to be polynomial in $q$ for $n \leq 7$ (and conjecturally for $n \leq 9$), just as they would be if we knew that the cohomology of $M_{2,n}$ contained only mixed Tate motives. By thinking a bit about the stratification of topological type, one concludes that this holds also for $\overline M_{2,n}$ when $n \leq 7$, which is smooth and proper over the integers. Then a theorem of van den Bogaart and Edixhoven implies that the cohomology of $\overline M_{2,n}$ is all of Tate type and with Betti numbers given by the coefficients of the polynomials.
There are similar results by Bergström for $M_{3,n}$ when $n \leq 5$ and Bergström--Tommasi for $M_4$. But the general phenomenon is that increasing either $g$ or $n$ will rapidly take you out of the world of $\mathbf F_1$-schemes, at least if this is taken to mean "commutative monoids".
However, I don't know enough $\mathbf F_1$--geometry to say what the answer is if one takes $\mathbf F_1$-schemes in the sense of Borger. The first nontrivial case to answer would be: are the motives associated to cusp forms for $\mathrm{SL}(2,\mathbf Z)$ defined over $\mathbf F_1$ in his set-up? To be clear, I don't believe that this is true, but I know almost nothing about $\lambda$-schemes.
Let me also make two small remarks: (i) Everything I have written above is necessary conditions for being defined over $\mathbf F_1$. (ii) The fact that $M_{g,n}$ is smooth over the integers says that the cohomology of $M_{g,n}$ is at least quite "special", even though it is not defined over $\mathbf F_1$. Smoothness is a strong restriction on the Galois representations that can occur in the cohomology.
The argument that allows you to show that an elliptic curves defined as you say is a group scheme, and even a commutative one is the construction of a functorial and natural isomorphism $E(T) \rightarrow Pic_{E/S}^0(T)$ for every $S$-scheme $T$. This allows to see the functor $T \mapsto E(T)$ as a funtor in group, and since this functor is representable by $E$, this gives a structure of group scheme on $E$, which is the one you are looking for.
Essentially this map is defined as follows: one attached to a point in $E(T)$, that is a $T$-section of $E_T$, the invertible sheaf of divisor this section minus the trivial section $e_T$ (obtained by base change from the section $e$ which is part of the definition). To prove that this map is an isomorphism, one essentially reduces, using base change theorems for direct image in coherent cohomology, to the case of a field, where it becomes a consequence of Riemann-Roch. Note that by definition, the trivial section $e_T$ is send to to the trivial sheaf, so that $e$ is the neutral section of the group scheme structure on $E$, as desired.
Of course there are many details to deal with to make this argument a complete one, but this is done with great care in the beginning of the (unique) book by Katz and Mazur.
Best Answer
Hint: use quadratic twists.
Edit: So as not to drag things out, I hope it's okay if I just give you a standard example. Let
$E_0: y^2 = x^3 + Ax + B$
be your favorite elliptic curve over $\mathbb{Q}$ (i.e., any will do). Consider the elliptic curve $E: t y^2 = x^3 + Ax + B$
over the rational function field $\mathbb{Q}(t) = \mathbb{Q}(\mathbb{P}^1)$. Spreading this out as a scheme over $\mathbb{P}^1_{/\mathbb{Q}}$, we see that there are two singular fibers, at $t = 0$ and $t = \infty$. Discarding these we get an elliptic curve over $\mathbb{A}^1 \setminus \{0\}$ which is isotrivial -- the $j$-invariant over every fiber is $j(E_0)$ -- but nontrivial: the isomorphism classes of the fibers are in bijection with $H^1(\mathbb{Q},\mathbb{Z}/2\mathbb{Z}) \cong \mathbb{Q}^{\times}/\mathbb{Q}^{\times 2}$.
It's a good bet that you'll find this example somewhere in the chapter on elliptic surfaces in Silverman's Advanced Topics in the Arithmetic of Elliptic Curves.
Can it be done over any base scheme? Unless I misunderstand, of course not, e.g. not over the spectrum of a field.