Is this the right forum for this? I need some help getting started with a couple of proofs. Can anyone get me started in the right direction?
I am confused by the map,
Because the sets,
are all finite and cannot be expressed as the ideals of .
What you want to show is the isomorphism,
Using the map defined above. Then it would make sense.
Do you understand me?
Okay here is the proof on what I said.
First we are given a function,
We need to show that this map is well-defined.
Assume there are two ways to express the same element of
Since is in the same coset we can express it as,
for some integer .
Now we show that they map this element into the same ordered coset pair.
Because is an element of and .
Thus the map is well-defined.
We now show that this map is a homomorphism between these two groups.
First if and
Thus, we need to see whether or not,
---> True for all.
Thus it is indeed a homomorphism.
Now, we see if this function is injective.
So, and for some integers then, for some integer ; note the important fact that otherwise we cannot conclude that has this form. That means that is in the same coset as . Thus, this show the function is indeed injective.
The last thing is whether any element of can be obtained from a map.
Does there exists a such that can be mapped into,
For any ?
Meaning, a such that and . Note the last two equations are saying,
And we can find such an !.
That is the Chinese Remainder Theorem.
If we chose,
Then, is a simultaneous solution to the congruence.
Thus, we proved isomorphism!
Note, the last two congruences has solutions to them because, .
There is a more elegant way to prove this, without solving that congruence via Chinese Remainder Theorem, though I did not find a proper map.
We know that,
Is a one-to-one map.
Then if we can find a map,
That is also one-to-one.
Then, you can conclude that the sets has the same cardinality.
Now there is a theorem, that if a function is injective and maps a finitie set into another finite set of equal cardinality then that function is also surjective.
So if you can find a one-to-one mapping function backswards then you automatically know that your injective is a bijection which assures us of a solution to those congruences.