endomorphisms of the jacobian of a general hyperelliptic curve Let $C$ be a curve of genus $g$. If $C$ is very general, we know that the Jacobian $JC$ of $C$ is simple and thus $End(JC)=\mathbb{Z}$.
Do we know something about $End(JC)$ if $C$ is a very general hyperelliptic curve?
 A: I will assume you are working over the complex numbers (note that for instance, the statement of your first paragraph is false over the algebraic closure of a finite field).
It is sufficient to exhibit a single example in each genus.  A theorem of Yuri Zarhin shows that if $P(x) \in \mathbb{Q}[x]$ is an irreducible polynomial of degree $d \geq 5$ with Galois group $S_d$ or $A_d$ (note that by Hilbert Irreducbility most polynomials over $\mathbb{Q}$ have this property!) then the endomorphism ring of the Jacobian of $y^2 = P(x)$ is $\mathbb{Z}$.  The paper is:

MR1748293 (2001a:11097)
  Zarhin, Yuri G.(1-PAS)
  Hyperelliptic Jacobians without complex multiplication.
  Math. Res. Lett. 7 (2000), no. 1, 123–132. 

Added: let me give a little more detail as to why the existence of one hyperelliptic curve (of any given genus $g \geq 2$) with this property implies that a very general hyperelliptic curve has this property.  First, it is known that a very general principally polarized abelian variety has endomorphism ring $\mathbb{Z}$: for each strictly larger endomorphism ring $R$ there is a closed ("Shimura") subvariety of the Siegel space $\mathcal{A}_{g,1}$ which is the locus of all ppavs with endomorphism ring containing $R$.  Every $R$ is a $\mathbb{Z}$-order in a finite-dimensional semisimple $\mathbb{Q}$-algebra, so there are at most (and in fact are) countably infinitely many possibilities.  Therefore the locus of ppavs of dimension $g$ with endomorphism ring $\mathbb{Z}$ is what is called (by me, at least) an algebraic stratum: it is the complement in an irreducible variety $V$ of a countably union of proper, Zariski-closed subvarieties.  In general, when one says "a very general point of $V$ has property $P$" one means that the set of points of $V$ which has property $P$ is an algebraic stratum.  Notice that over $\mathbb{C}$ every algebraic stratum is nonempty, e.g. because its complement has "measure zero".
We are interested in hyperelliptic Jacobians, i.e., we are intersecting the algebraic stratum $A$ of $\mathcal{A}_{g,1}$ consisting of abelian varieties with $\operatorname{End} A \cong \mathbb{Z}$ with the closed subset $H$ of hyperelliptic Jacobians.  In order to be sure this gives an algebraic stratum on $H$, it is necessary and sufficient to check that for each proper closed subvariety $W_R$ of $\mathcal{A}_{g,1}$ consisting of abelian varieties with endomorphism ring containing a strictly larger ring $R$ than $\mathbb{Z}$, then $H_R = H \cap W_R$ is still proper in $H$.  But this is accomplished by finding even a single element of $H$ which does not lie on any $W_R$, and Zarhin's theorem does this (and more).  
A: The following is the main result of Zarhin's paper Hyperelliptic Jacobian's without complex multiplication:

Let $K$ be a field of characteristic
  $0$, let $n \geq 5$ and let $f(x)$ be
  a polynomial of degree $n$ in $K[x]$
  whose Galois group is $S_n$ or $A_n$.
  Then the endomorphism ring of the
  Jacobian of  $$y^2 = f(x),$$
  considered over the algebraic closure
  of $K$, is $\mathbb{Z}$.

Since a degree $n$ polynomial with coefficients in $\mathbb{Q}$ has Galois group $S_n$, except on a thin set, this shows that there are lots of hyperelliptic curves, defined over $\mathbb{Q}$, whose automorphism ring over $\overline{\mathbb{Q}}$ is $\mathbb{Z}$. I say "lots" rather than "very general", because I don't want to misuse a technical term, but I expect this is good enough for your purposes.
The first page of Zarhin's paper gives many references to earlier results, which may also be of use to you.
