In many places the existence of automorphism is acknowledged as one of the reasons why fine moduli spaces cannot exist. A typical example is the following.


Consider a curve $C$ with a nontrivial automorphism, for instance a hyperelliptic curve with its involution $\phi$. Now let $B$ be any scheme with a free action of (in this case) $\mathbb{Z}/(2)$.


Let $D$ be the quotient of $B \times C$ by the diagonal action of $\mathbb{Z}/(2)$ and let $B'$ be the quotient of $B$ by the action of $\mathbb{Z}/(2)$.


Consider the map $f \colon D \rightarrow B'$. Then the fibers of $f$ are all isomorphic to $C$, but for a suitable choice of $B$ the family $f$ is not isomorphic to the product $C \times B'$.


How can I get an explicit example of the last assertion?



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> How can I produce $B$ as above such that $f$ is not isomorphic to the trivial family?
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How to use automorphisms to produce isotrivial non trivial families
Here is an explicit example. Denote by $E$ the elliptic curve $E=\mathbb C/(\mathbb Z+i\mathbb Z)$. Now let $B=C=E$. The involution that we will consider is 
$$(z,w)\in (E\times E)\to (z+\frac{1}{2},-w).$$


Notice now that this quotient is not a direct product anymore. You can see this by calculating $H\_1(E\times E)/\mathbb Z\_2$ and verifying that it is not $\mathbb Z^4$. 
Alternatively, you can notice that the quotient does not have a non-vanishing holomorphic volume form. Indeed if such form existed it could be lifted to $E\times E$ to a form $cdz\wedge dw$ ($c\ne 0$), which would give a contradiction since $dz\wedge dw$ is anti-invariant. 


