Size of automorphism group of element of Y_0(5).

Konciljev56

Konciljev56

Answered question

2022-09-07

Size of automorphism group of element of Y 0 ( 5 )
The modular curve Y 0 ( 5 ) parametrizes elliptic curves E with an isogeny of degree five. So an element of Y 0 ( 5 ) can be interpreted as
E ϕ E
Suppose we are working over a finite field. An automorphism of these curves is an automorphism of the curve E, that sends the kernel G of the map ϕ to itself, over some algebraic closure. The elements of Y 0 ( 5 ) do admit automorphisms, because any curve has the hyperelliptic involution.
Because G is of order five, it can have at most four automorphisms. If the curve E has six automorphisms, it is clear, by comparing the sizes of the groups, that only the identity and the hyperelliptic involution respect the subgroup G. If E has four automorphisms, this kind of reasoning doesn't work. Is there a similar result if E has four automorphisms? To me it is not clear if there are always only two automorphisms that map G to itself, or that there could be four.

Answer & Explanation

kappastud98u

kappastud98u

Beginner2022-09-08Added 10 answers

Step 1
Suppose that E has CM by Z[i]. The prime 5 splits in Z[i], as 5 = π π ¯ ,, where π = 2 + i.
Step 2
Let G be the subgroup of π-torsion in E, or the subgroup of π ¯ -torsion. These are each of order 5, and are preserved under mult. by all of Z[i], and in particular by the automorphism group { ± 1 , ± i }. Conversely, if G is preserved under { ± 1 , ± i }, then in fact is preserved under all of Z[i], and so is easily checked to be either the π or π ¯ torsion subgroup.
So of the 6 subgroups of order 5, two of them are preserved the full automorphism group, and the other four aren't.

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