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I've been to many talks in Number Theory and for some reason I've yet to fully grasp, we all seem to like Jacobian Varieties a lot. I know that they are Abelian varieties, which give information about their respective curve, but I'm not sure what information exactly. I know of the analytic description of the Jacobian, but I'm still not exactly sure why the Jacobian is so studied.

In his AMS article, What is a motive Barry Mazur seems to suggest that Jacobians encapsulate all cohomology theories. Is this true? How can I see this?

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    $\begingroup$ Kleiman's chapter in FGA explained (available separately on arXiv) contains a great and very detailed historical account (albeit possibly a bit tangential to your main question) of the development of the Picard scheme (as well as details for the construction in a very general setting). $\endgroup$ – R. van Dobben de Bruyn Jul 6 at 20:12
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If you are a number theorist, you presumably like class groups? Let $X$ be a curve defined over $\mathbb{F}_p$, let $J$ be its Jacobian and let $x$ be an $\mathbb{F}_p$ point of $X$. Let $A$ be the coordinate ring of the affine curve $X \setminus \{ x \}$. Then the class group of $A$ is $J(\mathbb{F}_p)$. (And similar statements can be made for deleting more than one point, or deleting points defined over extensions of $\mathbb{F}_p$.)

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  • $\begingroup$ Why do you need to delete a point? $\endgroup$ – Rdrr Jul 8 at 1:18
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    $\begingroup$ Answer 1 to get an affine variety. If someone is truly coming from classical number theory, they may only know class groups of rings, not Pic. Answer 2 if you take $Pic(X)$, you get $\mathbb{Z} \times J(\mathbb{F}_p)$, not $J(\mathbb{F}_p)$. $\endgroup$ – David E Speyer Jul 8 at 10:51
  • $\begingroup$ Could you give a reference for the fact that the class group of $A$ is $J(\mathbb{F}_p)$? Thanks. $\endgroup$ – Ang Jul 15 at 17:18
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    $\begingroup$ @Ang I don't have a reference off the top of my head. We have $\mathrm{Pic}(X \setminus \{ x \}) \cong \mathrm{Pic}^0(X)$, since every divisor on $X \setminus \{ x \}$ can be extended to a degree $0$ divisor on $X$ in a unique way. (Here it matters that $X$ is an $\mathbb{F}_p$ point.) The fact that $J(\mathbb{F}_p) \cong \mathrm{Pic}^0(X)$ is the defining property of the Picard functor. $\endgroup$ – David E Speyer Jul 15 at 17:30
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Suppose $X/\mathbb{Q}$ is a (smooth, projective, geometrically integral) curve of genus $g\geq 2$ and $J/\mathbb{Q}$ its Jacobian variety. If one is interested in determining the (finite, by Faltings) set of rational points $X(\mathbb{Q})$, then it can be useful to compute $J(\mathbb{Q})$ first. The latter is easier because $J(\mathbb{Q})$ is a finitely generated abelian group, and descent theory analogous to elliptic curves allows us to often do this in practice. If we pick a point $P\in X(\mathbb{Q})$ then we have an associated embedding $i_P: X \hookrightarrow J$. In favorable situations studying this embedding allows us to determine $X(\mathbb{Q})$ from $J(\mathbb{Q})$. For example, the method of Chabauty-Coleman gives a very concrete instance of this when the rank of $J(\mathbb{Q})$ is less than $g$ (for a friendly introduction to this method see the nice survey of McCallum-Poonen).

The moral is: by replacing $X$ by $J$, we somehow have made the geometry harder but the arithmetic easier.

The relation with motives can be explained in relatively concrete terms. The $\ell$-adic cohomology groups $H^i(X_{\bar{\mathbb{Q}}},\mathbb{Q}_l)$ are zero if $i\neq 0,1$, $2$ and isomorphic to $\mathbb{Q}_l, \mathbb{Q}_l(-1)$ if $i=0, 2$ respectively. (The minus $-1$ denotes the Tate twist.) So the only interesting degree is $i=1$, and pulling back via $i_P$ will induce an isomorphism $H^1(X_{\bar{\mathbb{Q}}},\mathbb{Q}_l) \simeq H^1(J_{\bar{\mathbb{Q}}},\mathbb{Q}_l)$. This last group (with its Galois action) is isomorphic to the dual of the $\ell$-adic Tate module of $J$. So $J$ and its torsion points encapsulate all the cohomological information of $X$. Similar statements will hold for other Weil cohomology theories: the only interesting degree is $1$ and $i_P$ will induce an isomorphism on $H^1$.

Edit: as pointed out in the comments, the geometry of $J$ is arguably easier than that of $X$. A better moral is thus maybe that we have made the space we're considering larger but richer in structure.

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    $\begingroup$ I must respectfully disagree that replacing $C$ with $J$ "makes the geometry harder." It does increase the dimension, which one could argues makes the geometry harder, but it introduces a group structure, and I'd sugestt that the geometry of a high dimensional group variety (especially one that's compact) is much less difficult than the geometry of lower dimensional varieties having less structure. Or even ignoring the group structure, $J$ has Kodaira dimension 0, while $C$ has Kodaira dimension 1, again suggesting that $J$'s geometry is simpler than $C$'s. $\endgroup$ – Joe Silverman Jul 6 at 20:53
  • $\begingroup$ Thanks for the comment, I'll edit my vague moral to make it more accurate. $\endgroup$ – Jef Jul 6 at 20:57
  • $\begingroup$ What are the favourable situations that allow us to determine $J(\mathbb{Q})$ from $X(\mathbb{Q})$? Also, why does $i_P$ become an isomorphism on $H^1$? $\endgroup$ – Rdrr Jul 8 at 1:18
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As outlined by the other answers, the Jacobian $J_X$ of a curve $X$ defined over $\mathbb{F}_q$ indeed encapsulates all cohomology information of $X$. In particular one can read the zeta function $\zeta_X$ directly on $J_X$: the numerator of $\zeta_X$ is simply the (reciprocal) polynomial of the Frobenius $\pi_q$ acting on $J_X$.

In particular André Weil's original proof of the Hasse-Weil bound for curves used Jacobians (implicitely). That was a big motivation in his Foundations of algebraic geometry: the algebraic construction of Jacobians over any field.

By the way over $\mathbb{C}$ the Abel-Jacobi map shows that the Jacobian of $X$ is intimately related to the study of abelian integrals. I think historically that was the prime motivation to study Jacobians. A fun fact is that modular functions coming from hyperelliptic integrals can be used to solve algebraic equations. Cf the appendix of Mumford's TATA2.

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