To see this, we demonstrate a natural correspondence between endomorphisms of and and show that invertible endomorphisms correspond to invertible matrices. Let be any ring homomorphism. It is clear that is determined by its action on and , since
Suppose then that and . Then
Now, is surjective if both and are in its image. But if and only if there is some such that
Solving this pair of equations for we see that we must have and thus . Similarly, if and only if . Thus is surjective precisely when , i.e. precisely when the matrix representation of has determininant . This then gives a map from to that is obviously a ring isomorphism. This concludes the proof.
has a simple and well-known set of generators as a group:
Note that for any integer . We now prove this fact.
Define the subgroup by , the subgroup of generated by and . If , define if and are in the same -coset.
Our objective is to show that , which we can do by showing that each , where is the identity transformation of . This demonstration is essentially an application of the Euclidean algorithm. For suppose
Assume, by applying if necessary, that , and choose such that . Then , so that
Continuing this process, we eventually see that
But , so we have . Applying either or as appropriate, we get
Thus, we are done if we show that all such forms with are in the same -coset as . The case where is obvious. For the other cases, note that
and is obviously the composition of these two.
This result is often phrased by saying that the matrices
|Date of creation||2013-03-22 16:31:38|
|Last modified on||2013-03-22 16:31:38|
|Last modified by||rm50 (10146)|