Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 19 / 2

We use the basic dual isogeny identities in the endomorphism ring of an elliptic curve, extending the dual of the zero map to zero:Here $[n]$ denotes the multiplication-by-$n$ endomorphism, and the degree of the zero map is set to zero. To see the source of additivity, identify $E$ with its Picard group of degree-zero line bundles using the origin. Dualization is pullback on this group. Pullback of a degree-zero line bundle along the sum of two homomorphisms is the tensor product of its pullbacks along those homomorphisms, giving additivity; functoriality of pullback reverses composition. The last identity is the defining degree identity for a dual isogeny. Every nonzero elliptic-curve endomorphism is an isogeny.

The natural inclusion $\mathbb{Z}\hookrightarrow \mathrm{End}(E)$ remains injective even in positive characteristic: for nonzero $n$, the map $[n]$ has degree $n^2$ and is not zero. Therefore an equality of multiplication endomorphisms uniquely specifies its integer coefficient.

Apply the dual isogeny degree identity to $1-\varphi$. Its dual is $1-\hat{\varphi }$, soRearrangement givesThe right side is multiplication by an integer, proving the assertion. In particular the trace of an elliptic-curve endomorphism isThe argument includes $\varphi =0$ and $\varphi =1$ with the zero-map convention. Multiplying $\hat{\varphi }=[t]-\varphi$ by $\varphi$ also yields the useful characteristic relation

The $q$-power Frobenius isogeny sends affine coordinates to $(x^q,y^q)$, has degree $q$, and fixes exactly the points defined over ${\mathbb{F}}_q$, including the origin. ConsequentlyThe degree is $q$ because the $q$-power Frobenius on a smooth curve has function-field inseparable degree $q$. By the permitted separability of $1-\varphi$, this separable isogeny has exactly $\mathrm{deg}(1-\varphi )$ geometric kernel points; there is no inseparable multiplicity to remove. Part(i) therefore givesThis relates the elliptic-curve point count over a finite field to the integer trace, rather than treating the point count as an independently assumed definition of the trace.

The nonzero quadratic residues modulo five are $1,4$. The values of $x^3+x+1$ and the numbers of possible $y$ areThere are eight affine points and one point at infinity. Thus $\#E({\mathbb{F}}_5)=9$ and part(ii) gives $\boxed{\mathrm{tr}(\varphi )=5+1-9=-3}$. The curve is nonsingular, since its short-Weierstrass discriminant $-16(4+27)$ is nonzero modulo five.

The dual operation reverses products, so $\widehat{{\varphi }^2}=(\hat{\varphi }{)}^2$. Also $\varphi \hat{\varphi }=[5]$ and $\varphi +\hat{\varphi }=[-3]$. Squaring the latter identity givesTherefore $\boxed{\mathrm{tr}({\varphi }^2)=-1}$. Over ${\mathbb{F}}_{25}$ the Frobenius is ${\varphi }^2$, of degree $25$, soMore generally the Frobenius trace recurrence has $t_0=2$, $t_1=t$ and $t_j=t{t}_{j-1}-q{t}_{j-2}$; the square calculation is its first nontrivial case.