I saw the question:
and though I know that there are rare CM elliptic curves, I wonder what kind of curves with higher genus have the CM Jacobians?
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1$\begingroup$ The Fermat curves $x^n+y^n+z^n =0$ are candidates. Their genus grows and their Jacobians have many extra automorphisms. For the Jacobian of a curve to be CM its dimension should match up with the rank of its endomorphism algebra, and that doesn't happen here for large $n$. On the other hand, the Jacobian of the Fermat curve does have CM quotients. This fact is exploited very often; see for instance Section III in jstor.org/stable/2946559?seq=1#page_scan_tab_contents or Lang's book on Complex Multiplication. $\endgroup$– Ariyan JavanpeykarCommented Jan 25, 2016 at 12:32
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2 Answers
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I am not sure whether this answers your question: it is a conjecture of Coleman that for a fixed genus $g$ sufficiently high, there should be only finitely many CM Jacobians of genus $g$. In fact Coleman stated the conjecture for $g\geq 4$, but by now there are counter-examples for $g\leq 7$ (at least). See for instance this paper of B. Moonen.
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$\begingroup$ At any rate, even if Kazuma Morita's proof for the BSD in the CM case (I could not find errors?) is true, his method seems to be ineffective to higher genus theory. $\endgroup$– J.S.R.Commented Jan 25, 2016 at 14:42
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It should be noted that Murabayashi determined that there should be finitely many (whose moduli lie in the rational numbers) for $g=2$ over the complex numbers and (mostly) explicitly determined them.
http://link.springer.com/article/10.1007%2Fs00208-008-0251-2
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$\begingroup$ Thank you. Because even in the case g=2, there are very few CM Jacobian, the property ``CM" seems to be a privilege of elliptic curves (though there are not many). $\endgroup$– J.S.R.Commented Jan 26, 2016 at 3:22