Easiest example where field of definition is not field of moduli There are many examples of varieties over $\overline{\mathbb Q}$ whose field of moduli is $\mathbb Q$ but which can't be defined over $\mathbb Q$. What is the easiest such example? It should be a genus two curve probably, but which one?
Also, what is the easiest example of a variety over $\mathbb C$ whose field of moduli is $\mathbb R$ but which can't be defined over $\mathbb R$?
 A: I believe that the other answer to this question is incorrect (I don't have the reputation to comment on it). The curve
$$y^2=(x^2-1)(x^2-2)(x-i)$$
is in fact defined over $\mathbb{R}$. To see this directly, make the change of variables $x\mapsto ix$ and $y\mapsto \omega y$, where $\omega^2=i$. Also, I'm computing that the conjugate-linear map $g:C\rightarrow C$ in fact squares to the identity, not to the hyperelliptic involution.
Here is an example of a genus 2 curve over $\mathbb{C}$ which has field of moduli (contained in) $\mathbb{R}$ but which is not defined over $\mathbb{R}$. I learned this example from a paper of Shimura ("On the field of rationality for an abelian variety, Nagoya Math. J.").
Let $a$ and $b$ be complex numbers. Consider the curve $C$ defined by the equation $y^2=f(x)$, where
$$f(x)=x^6+ax^5+bx^4+x^3-\overline{b}x^2+\overline{a}x-1$$
One can show that if $a$ and $b$ are chosen in a sufficiently generic way, then the automorphism group of $C$ has order 2, generated by the hyperelliptic involution. It follows that $C$ cannot be defined over any smaller field than $\mathbb{C}$. On the other hand, the birational map
$$(x,y)\mapsto (-x^{-1},ix^{-3}y)$$
induces an isomorphism $\mu$ between $C$ and its conjugate $\overline{C}$ (the curve whose equation is $y^2=\overline{f(x)}$). As in the other answer, this shows that the field of moduli of $C$ is (contained in) $\mathbb{R}$.
The remainder of the analysis in the other answer goes through for this curve $C$. For instance, let $\theta:C\rightarrow C$ be the composite of $\mu$ and the conjugate linear isomorphism $\overline{C}\rightarrow C$. Then $g^2$ is equal to the hyperelliptic involution, so we have an isomorphism $\mathrm{Aut}_{\mathbb{R}}(C)=\mathbb{Z}/4$, etc.
A: EDIT: This answer is incorrect, for the reason indicated in the other answer; it should be consulted for a correct curve.

Here is a recipe for constructing some examples.
Suppose that $f$ is a polynomial of degree 5 over $\Bbb C$ without repeated roots.  Suppose that $f$ has the following two properties:

*

*The only linear fractional transformations in $x$ fixing the union of the zeros of $f(x)$ with $\infty$ is the identity.


*We have $-\overline{f(-\overline x)} = f(x)$.
(For example, you could take $f(x) = (x^2-1)(x^2-2)(x-i)$.)
Consider the resulting hyperelliptic curve $C$ of genus two of the form $y^2 = f(x)$.  The projection $C \to \Bbb P^1$ determined by the canonical bundle on $C$ is projection onto the $x$-coordinate, and the first condition we've imposed on $f$ guarantees that any automorphism of $C$ must fix the $x$-coordinate (as it must fix the six ramification points).  As a result, the only possible such automorphisms are the identity and $y \mapsto -y$.
Thus we have a hyperelliptic curve whose automorphism group is exactly $\Bbb Z/2$.
Now we also find that the second condition on $f$ means that there is an isomorphism of $C$ with its complex conjugate $\overline{C}$, given by $y \mapsto iy$, $x \mapsto -x$.  This means that the field of moduli has to be strictly smaller than $\Bbb C$, and so it must be $\Bbb R$.
However, this curve cannot be defined over $\Bbb R$.  The obstruction, as is typical when the automorphism group is abelian, shows up as an element in $H^2(Gal(\Bbb C/\Bbb R), Aut(C))$.  In terms of group cohomology, this is determined by an extension
$$
1 \to Aut(C) \to G \to Gal(\Bbb C/\Bbb R) \to 1.
$$
Here $G$ is the collection of automorphisms of $C$ which are either $\Bbb C$-linear or conjugate-linear.
In this case, both groups on the outside are $\Bbb Z/2$, and so we want to show that the extension is $\Bbb Z/4$.  To calculate the extension, we take the nontrivial automorphism of $\Bbb C$, lift it to an isomorphism $g: C \to C$ which is conjugate-linear (which we already calculated above), and then form the composite $g^2: C \to C$.  This automorphism is the hyperelliptic involution, and so the extension is nontrivial.  (Here is where the mistake was, I was not careful enough.)
I believe that the same method works over $\Bbb Q(i)$, but you need a sturdier Galois cohomology computation.
