(December 8, 2005)

SL2( )

SL2( /p)

Paul Garrett

[email protected] http://www.math.umn.edu/˜garrett/

In the discussion of the action of SL2 ( p ) or GL2 ( one begins with the surjectivity of the natural map

p)

on the p-power projective limit of modular curves,

SL2 ( ) −→ SL2 ( /p) It is important to understand the simplicity of this and related results.

Claim: Let R be a principal ideal domain. Let M be a maximal ideal. Then the natural map SL2 (R) −→ SL2 (R/M )

is surjective

Proof: Let q be the quotient map R −→ R/M . First, given u, v not both 0 in R/M , we will find relatively prime c, d in SL2 (R) such that q(c) = u and q(d) = v. Consider the case that v 6= 0 in R/M . Since q : R −→ R/M is surjective, there is 0 6= d ∈ R such that q(d) = v. Consider the conditions on c ∈ R c = u mod M

c = 1 mod Rd

As d 6∈ M , by the maximality of M there are x ∈ R and m ∈ M such that xd + m = 1. Let c = xdu + m. From xd + m = 1 we have xd = 1 mod m and m = 1 mod d, so this expression for c does satisfy the system of congruences. In particular, q(c) = u, and since c = 1 mod d it must be that gcd(c, d) = 1. For v = 0 in R/M , necessarily u 6= 0, and we reverse the roles of c, d in the previous paragraph. Thus, we have relatively prime c, d in R whose images mod M are u, v. In a PID, given s, t there are a, b such that gcd(s, t) =as − bt. Here, the coprimality of c, d implies that there are a, b in R such that ad − bc = 1. a b That is, ∈ SL2 (R), and c d ∗ ∗ a b mod M = u v c d Thus, given

r s u v

r s u v

in SL2 (R/M ), we have

a b c d

−1

=

r s u v

1 w d −b = 0 1 −c a

mod M

(where w = sa − br mod M )

since the right-hand side is in SL2 (R/M ). Let t ∈ R be such that q(t) = w. Then −1 1 0 1 −t a b r s mod M = 0 1 0 1 c d u v So

1 t 0 1

a c

r s b = u v d

This gives the surjectivity.

mod M ///

1