The Divergence of 'Rank' in Mordell-Weil Theory: Rational vs. Finite Fields — Epoche C1
Two groups of order six The curve defined by $y^2 = x^3 + 1$ has exactly six points with rational coordinates: the three pairs $(0, \pm 1)$, $(2, \pm 3)$ and $(-1, 0)$, together with a point at infinity that serves as the identity of the group law. Reduce the same equation modulo $5$ and count solutions in the five-element field $\mathbb{F}_5$, and one finds six points again. The two groups have the same order and are in fact isomorphic. They are nevertheless instances of two quite different phenomena, and unpicking why is the subject of this essay. The reason the sameness is accidental is that over $\mathbb{Q}$ the six points exhaust the curve because a deep theorem says they must, while over $\mathbb{F}_5$ the count is bounded by the field. Change the curve slightly and the rational group can become infinite; no change of curve can make the finite-field group infinite. That contrast is what the word "rank" is supposed to capture, and the original version of this essay concluded from it that rank simply does not apply over finite fields. That conclusion is too strong in one direction and not strong enough in another, and both corrections are made below. What the Mordell–Weil theorem says, and what rank measures Take a curve given by $y^2 = x^3 + Ax + B$ with $A$ and $B$ rational and no repeated roots on the right. The rational points on such a curve form an abelian group: given two of them, the line through them meets the curve in a third rational point, and reflecting that point in the $x$-axis gives the sum. Mordell proved in 1922 that this group is finitely generated — that some finite list of rational points suffices to produce every other by repeated addition and subtraction. Weil extended the result in his 1929 thesis to abelian varieties over number fields, which is why the theorem carries both names. Finite generation is exactly the hypothesis of the structure theorem for finitely generated abelian groups, which says that any such group splits as a free part and a finite part. So $$ E(\mathbb{Q}) \;\cong\; \mathbb{Z}^{r} \oplus E(\mathbb{Q})_{\mathrm{tors}}, $$ where the torsion subgroup $E(\mathbb{Q})_{\mathrm{tors}}$ collects the points $P$ for which some multiple $nP$ is the identity, and $r$ — the rank — counts the independent generators of infinite order. Equivalently $r = \dim_{\mathbb{Q}} \left( E(\mathbb{Q}) \otimes_{\mathbb{Z}} \mathbb{Q} \right)$: tensoring with $\mathbb{Q}$ annihilates every torsion element, because a point of order $n$ becomes $P \otimes 1 = P \otimes (n \cdot \tfrac{1}{n}) = nP \otimes \tfrac{1}{n} = 0$, and leaves the free part as a vector space of dimension $r$. Rank is notoriously hard to compute. There is no algorithm known to be guaranteed to terminate and return it, because the standard descent procedure bounds the rank from above by the size of a Selmer group and the gap between that bound and the true rank is measured by an object, the Tate–Shafarevich group, which is not known in general to be finite. Elkies exhibited in 2006 a curve of rank at least $28$; whether ranks are bounded at all remains open. Why the finite-field group is finite, and how large it is Over a field with $q$ elements the finiteness needs no theorem. There are $q$ possible values of $x$, each giving a quadratic in $y$ with at most two roots, so with the point at infinity the count is at most $2q+1$. That crude bound is off by a factor of two, and the exact statement is Hasse's theorem, proved in 1936, which says $$ \left| \#E(\mathbb{F}_q) - (q+1) \right| \;\le\; 2\sqrt{q}. $$ The heuristic behind $q+1$ is that for each $x$ the quantity $x^3+Ax+B$ is a non-zero square about half the time, giving two points, and a non-square about half the time, giving none, so the average is one point per value of $x$; the theorem says the deviation from that average is at most $2\sqrt{q}$. Its proof is worth a sentence, because it explains the square root. Over $\mathbb{F}_q$ the map $\phi$ raising each coordinate to the $q$-th power fixes exactly the points with coordinates in $\mathbb{F}_q$, so $\#E(\mathbb{F}_q)$ is the number of solutions of $\phi(P) = P$, which is the degree of the map $\phi - 1$. The degrees of the maps $m + n\phi$ form a positive definite quadratic form in the integers $m$ and $n$, namely $m^2 + mna + n^2 q$ where $a$ is the trace of $\phi$. A positive definite binary quadratic form has non-positive discriminant, so $a^2 - 4q \le 0$, which is Hasse's bound with $\#E(\mathbb{F}_q) = q + 1 - a$. Counting the running example over $\mathbb{F}_5$ makes this concrete. The non-zero squares modulo $5$ are $1$ and $4$. Evaluating $x^3+1$ at $x = 0,1,2,3,4$ gives $1, 2, 4, 3, 0$: the values $1$ and $4$ are non-zero squares and contribute two points each, the values $2$ and $3$ are non-squares and contribute none, and the value $0$ contributes one. That is five affine points, and six with the point at infinity, so $a = 0$. Over $\mathbb{F}_7$, where the non-zero squares are $1, 2$ and $4$, the same evaluation gives $1, 2, 2, 0, 2, 0, 0$, producing $2+2+2+1+2+1+1 = 11$ affine points and $12$ in all, so $a = -4$ and $|{-4}| \le 2\sqrt{7} \approx 5.29$. Since the group is finite, every element has finite order, there is no free part, and the free rank is zero for every curve over every finite field. That much of the original essay is correct and is also, on its own, close to trivial. The right analogue of rank is two, not zero The original essay concluded that the notion of rank does not apply over finite fields. There is a better answer, and it is the one that carries the arithmetic content. Rank in the Mordell–Weil setting counts free generators; the corresponding invariant for a finite abelian group is the minimum number of generators, equivalently the number of invariant factors. For elliptic curves that number is always one or two: $$ E(\mathbb{F}_q) \;\cong\; \mathbb{Z}/m\mathbb{Z} \times \mathbb{Z}/n\mathbb{Z}, \qquad m \mid n, \qquad m \mid q-1 . $$ Two generators su