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I am reading Schoof's 1995 paper, Counting points on elliptic curves over finite fields, page 236, Proposition 6.1(i). I am trying to understand page 238 (second paragraph) of the proof: if the conductor is $1$ then $j$ invariant is $0$, or $1728$.

In Silverman's book, The Arithmetic of Elliptic Curves, Section III.10 on page 103, a relation between the Automorphism Group and $j$-invariant is given. Is there any relation between the endomorphism ring of an elliptic curve to the $j$-invariant? When $j=0,1728$, how does the endomorphism ring look?

If the endomorphism ring is equivalent to the maximal order of the field, why is the $j$ invariant equal to $0$ or $1728$?

I know that given $E: y^2=x^3+Ax+B$, if $A=0$ then $j=0$ and if $B=0$ then $j=1728$. The endomorphism ring will be isomorphic to an imaginary quadratic field if $E$ is ordinary and to a quaternion algebra if $E$ is supersingular.

Please help me to understand the relation between the endomorphism ring and $j$ invariant.

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2 Answers2

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If the endomorphism ring is equivalent to the maximal order of the field, why is the $j$ invariant equal to $0$ or $1728$?

The main source of all these theorems about endomorphism rings is in Deuring's Die Typen der Multiplikatorenringe elliptischer Funktionenkörper. Unfortunately, I can't find a free copy online, and I don't have a copy either. Another source besides Silverman's books is Washington's (not free) Elliptic Curves: Number Theory and Cryptography.

In any case, in those references, you can find a proof the following: The endomorphism ring of supersingular elliptic curves is a maximal order in a quarternion algebra. Silverman presents it as a starred homework assignment in the book you reference (problem 3.18), but I think going through it all is a bit too off-topic for this site.

As I mentioned in the comments, the $j$-invariant is useful in characterizing isomorphisms between curves, and if two curves are isomorphic (e.g., a curve and its quadratic twist), then they have the same $j$-invariant, but their associated endomorphism rings are isomorphic.

In general though, characterizing the endomorphism group is more complicated than the automorphism group. As Silverman notes in the section you reference:

If an elliptic curve is given by a Weierstrass equation, it is generally a nontrivial matter to determine the exact structure of its endomorphism ring. The situation is much simpler for the automorphism group.

$j$-invariants are relevant for automorphisms because those contain invertible isogenies, but endomorphisms aren't necessarily invertible. $j$-invariants characterize isomorphisms between curves. Thus I wouldn't expect there to be a simple relationship between $j$-invariants and endomorphism rings.

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Over $\mathbb{F}_p$ an elliptic curve and its quadratic twist have isomorphic endomorphism rings. If $p\equiv 1 \pmod{3}$, the six curves with $j=0$ all have their endomorphism ring isomorphic to $\mathbb{Z}[\frac{1+\sqrt{-3}}{2}]$ and if $p\equiv 1\pmod{4}$, the four curves $E$ with $j=1728$ all have $End(E)$ isomorphic to the ring of Gaussian integers $\mathbb{Z}[i]$

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