Thread: Special Linear Group

Hy. Im looking for a proof that SL_n(IZ)->SL_n(IZ/mIZ) is onto. Can you help me?

Originally Posted by housi

Hy. Im looking for a proof that is onto. Can you help me?

Have I interpreted the question correctly? If so, then I think that the result is false.

For , denote by its image in the quotient ring I assume that denotes the set of all matrices with entries in with determinant and that the map from to is the natural quotient map given by

Now take m = 7 and let Then but any lifting of A to a matrix in will have determinant of the form 8+28k (mod 49), which can never be equal to 1.

Edit. Stupid mistake: 8 + 28k can be equal to 1 (mod 49). In fact, 8 + 12×28 = 7^3 + 1; and in fact is a lifting of that matrix A to an element of So it looks to me as though the result may be true after all. But it's not at all obvious!

Originally Posted by Opalg

Have I interpreted the question correctly? If so, then I think that the result is false.

Yes you have interpretet the question correctly! More precisely i think i need this result to prove (pro-finite completion) in the case (this is actually what i want to prove). With the solution of the congruence subgroup problem one can get (projective limes) where denotes the kernel of . Because it shouldn't be difficult to prove it remains to prove . Thus i need to show that this natural projection is onto. Maybe one can solve this using matrices of the type where has 1 at the place and 0 else. But have no idea how to show...